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CSAT · 43 questions

Counting, permutations & probability

Every UPSC CSAT question on this topic, 2016–2026, newest first. Tap an option to check yourself; the answer and explanation open below it.

Counting, permutations & probability questions per year: 2016: 4, 2017: 3, 2018: 4, 2019: 4, 2020: 2, 2021: 3, 2022: 8, 2023: 11, 2024: 1, 2025: 0, 2026: 3 Asked in 10 of 11 years · most in 2023 (11)
Timed · 30 minutes a set Practise Counting, permutations & probability as timed tests 2 sets of about 20 questions with UPSC's CSAT marking. Start practising

UPSC syllabus: “Basic numeracy (numbers and their relations, orders of magnitude, etc.) (Class X level), Data interpretation (charts, graphs, tables, data sufficiency etc. — Class X level);” Sub-topics in the detailed syllabus →

Showing 1–30 of 43, newest first.

CSAT 2026 · Q9

Easy Provisional key

QUEUE शब्द के वर्णों का स्थान अदल-बदल कर कितने शब्द बनाए जा सकते हैं, यदि Q के बाद सदैव U आता है? इस प्रकार बने शब्द का कोई अर्थ होना आवश्यक नहीं है।

Read this question in English

How many words can one form by shuffling the letters of the word QUEUE, if Q is always followed by U? The words thus formed need not necessarily have any meaning.

(a) 6 · (b) 8 · (c) 10 · (d) 12

उत्तर और व्याख्या · explanation in English

Answer: (d) 12

Glue Q and U together as one block 'QU'. Then arrange four units — QU, U, E, E — where the two E's are identical: 4!/2! = 12.

  1. QUEUE has Q, U, U, E, E.
  2. Tie Q to a U as a block: the units are QU, U, E, E.
  3. Arrangements = 4! ÷ 2! (two identical E's) = 24 ÷ 2 = 12.
  4. Check: Q occurs once, so each word has exactly one block position — no word is counted twice.

Remember · 'X always followed by Y' → treat XY as one block, then divide by factorials of repeated letters.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2026 · Q15

Medium Provisional key

एक मेज़ का ऊपरी पृष्ठ (top of a table) आयताकार है और उसके आयाम 6′ × 10′ हैं। मेज़ के ऊपरी पृष्ठ के दो आयताकार भागों को नीले रंग से रंगा गया है; इन दोनों भागों के आयाम 2·5′ × 8′ हैं और उनमें से प्रत्येक की सटीक दो भुजाएँ, मेज़ के ऊपरी पृष्ठ के दो किनारों के उभयनिष्ठ हैं। यदि मेज़ को ज़मीन पर स्थिर रखा जाता है और मेज़ के ऊपरी पृष्ठ के शेष भाग को सफ़ेद रंग से रंगा जाता है, तो ऊपर से देखने पर कितने भिन्न प्रतिरूप (पैटर्न) संभव हैं?

Read this question in English

The top of a table is rectangular and its dimensions are 6′ × 10′. Two rectangular portions of the table top are painted in blue colour; both these portions have dimensions 2.5′ × 8′ and each of them has exactly two sides common with two edges of the table top. If the table is fixed to the ground and the remaining portion of the table top is painted in white, how many different patterns are possible when observed from above?

(a) 2 · (b) 4 · (c) 6 · (d) 8

उत्तर और व्याख्या · explanation in English

Answer: (b) 4

Each 2·5′ × 8′ strip must lie with its 8′ side along a 10′ edge and sit in a corner. Two strips cannot share the same 10′ edge without overlapping, so one strip goes to each long edge, at its left or right end: 2 × 2 = 4 patterns.

  1. The 8′ side cannot lie along the 6′ edge (8 > 6), so it lies along a 10′ edge and the 2·5′ side along a 6′ edge.
  2. Two sides common with two table edges means the strip sits in a corner (it cannot touch two opposite edges, since 2·5 < 6 and 8 < 10).
  3. Two strips along the same 10′ edge would overlap (8 + 8 > 10), so one strip lies along each 10′ edge; they do not meet since 2·5 + 2·5 = 5 < 6.
  4. Each strip can be at the left or the right end of its edge: 2 × 2 = 4.
  5. The table is fixed, so these four layouts (both left, both right, and the two diagonal ones) all look different from above.

Remember · Fix the orientation first (which side fits which edge), then count positions; a fixed table means mirror images count separately.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2026 · Q23

Medium Provisional key

चार प्रकार के बाट (weights) हैं, नामतः 1 kg, 2 kg, 5 kg और 10 kg. यदि मापते समय, कम-से-कम आठ, किन्तु ग्यारह से अनधिक 1 kg के बाटों का उपयोग किया जाना है, तो 20 kg के वज़न को अधिकतम कितने भिन्न प्रकार से मापा जा सकता है?

Read this question in English

There are four types of weights, namely 1 kg, 2 kg, 5 kg and 10 kg. What is the maximum number of different ways one can measure 20 kg, if at least eight but not more than eleven weights of 1 kg are to be used while measuring?

(a) 7 · (b) 8 · (c) 9 · (d) 10

उत्तर और व्याख्या · explanation in English

Answer: (b) 8

Fix the number of 1 kg weights at 8, 9, 10 or 11 and make up the rest with 2, 5 and 10 kg weights. The counts are 3, 1, 3 and 1, giving 8 ways.

  1. Eight 1 kg weights: 12 kg left → 10 + 2; 5 + 5 + 2; six 2s → 3 ways.
  2. Nine 1 kg weights: 11 kg left from 2, 5 and 10 kg weights. 11 is odd, so exactly one 5 is needed, plus three 2s → 1 way.
  3. Ten 1 kg weights: 10 kg left → 10; 5 + 5; five 2s → 3 ways.
  4. Eleven 1 kg weights: 9 kg left, odd again → 5 + 2 + 2 → 1 way.
  5. Total = 3 + 1 + 3 + 1 = 8.

Remember · Counting combinations of coins or weights: fix the most restricted item, then list the rest systematically from the largest.

Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2024 · Q49

Medium

प्रथम सात धनपूर्ण संख्याओं के समुच्चय में से तीन संख्याएँ x, y, z इस प्रकार चुनी जाती हैं कि x > 2y > 3z है। इस प्रकार के कितने भिन्न त्रिक (x, y, z) संभव हैं?

Read this question in English

Three numbers x, y, z are selected from the set of the first seven natural numbers such that x > 2y > 3z. How many such distinct triplets (x, y, z) are possible?

(a) One triplet · (b) Two triplets · (c) Three triplets · (d) Four triplets

उत्तर और व्याख्या · explanation in English

Answer: (d) चार त्रिक

Since x ≤ 7, 2y ≤ 6 and so 3z < 6, which forces z = 1 and y = 2 or 3. That gives (5, 2, 1), (6, 2, 1), (7, 2, 1) and (7, 3, 1): four triplets.

  1. x ≤ 7 and x > 2y, so 2y ≤ 6, i.e. y ≤ 3.
  2. 2y > 3z with 2y ≤ 6 means 3z < 6, so z = 1; then 2y > 3 gives y = 2 or 3.
  3. y = 2: x > 4, so x = 5, 6 or 7, giving three triplets.
  4. y = 3: x > 6, so x = 7, giving one triplet.
  5. Total = 4 triplets: (5, 2, 1), (6, 2, 1), (7, 2, 1), (7, 3, 1).

Remember · In chained inequalities, start from the tightest bound (here x ≤ 7) and work down to the smallest variable.

Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q4

Medium

राज के पास एक डिब्बे में दस जोड़े लाल जूते, नौ जोड़े सफ़ेद जूते और आठ जोड़े काले जूते हैं । यदि वह पहनने हेतु एक जोड़ा लाल जूता लेने के लिए डिब्बे में से यादृच्छिक रूप से एक-एक कर (बिना उसे वापस रखे) जूते निकालता है, तो उसे अधिकतम कितने प्रयास करने होंगे ?

Read this question in English

Raj has ten pairs of red, nine pairs of white and eight pairs of black shoes in a box. If he randomly picks shoes one by one (without replacement) from the box to get a red pair of shoes to wear, what is the maximum number of attempts he has to make?

(a) 27 · (b) 36 · (c) 44 · (d) 45

उत्तर और व्याख्या · explanation in English

Answer: (d) 45

In the worst case Raj first takes out every white and black shoe, and then every red shoe for one foot. A pair he can wear needs one left and one right shoe, so the pair is certain only on the next draw: 18 + 16 + 10 + 1 = 45.

  1. The box has 20 red shoes (10 left, 10 right), 18 white shoes and 16 black shoes.
  2. Worst case: all 18 white and all 16 black shoes come out first: 34 draws and still no red pair.
  3. Next, all 10 red shoes of one foot (say all left shoes) come out: 44 draws and still no pair to wear.
  4. The 45th draw has to be a red right shoe, which completes a red pair. Maximum attempts = 34 + 10 + 1 = 45.
  5. Check: ignoring left and right would give 34 + 2 = 36, the trap option; a wearable pair needs one shoe of each foot.

Remember · Worst-case questions: use up every unhelpful draw first, including same-foot or same-type items, then add one.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q5

Medium

कोई बल्लेबाज केवल एक रन लेते हुए व चौके और छक्के मारते हुए कितने तरीकों से ठीक-ठीक 25 रन बना सकता है, जबकि रन बनाने का कोई भी अनुक्रम हो सकता है ?

Read this question in English

In how many ways can a batsman score exactly 25 runs by scoring single runs, fours and sixes only, irrespective of the sequence of scoring shots?

(a) 18 · (b) 19 · (c) 20 · (d) 21

उत्तर और व्याख्या · explanation in English

Answer: (b) 19

Count the combinations of sixes, fours and singles that add up to 25; the order of shots does not matter. Fix the number of sixes, count how many fours can fit, and let singles make up the rest: 7 + 5 + 4 + 2 + 1 = 19.

  1. Let the batsman hit x sixes, y fours and z singles: 6x + 4y + z = 25. Once x and y are chosen, z is fixed.
  2. x = 0: 4y ≤ 25, so y = 0 to 6 → 7 ways.
  3. x = 1: 4y ≤ 19, so y = 0 to 4 → 5 ways.
  4. x = 2: 4y ≤ 13, so y = 0 to 3 → 4 ways.
  5. x = 3: 4y ≤ 7, so y = 0 or 1 → 2 ways.
  6. x = 4: 4y ≤ 1, so y = 0 → 1 way. Total = 7 + 5 + 4 + 2 + 1 = 19.

Remember · For 'irrespective of order' counts, fix the biggest unit, count options for the next; the smallest unit fills the rest automatically.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q6

Medium

चार पत्र और चार लिफ़ाफ़े हैं और ठीक-ठीक एक पत्र को सही पते वाले ठीक-ठीक एक लिफ़ाफ़े में डालना है । यदि पत्रों को लिफ़ाफ़ों में यादृच्छिक रूप से डाला जाता है, तो निम्नलिखित कथनों पर विचार कीजिए :

  1. 1.यह संभव है कि ठीक-ठीक एक पत्र ग़लत लिफ़ाफ़े में जाए ।
  2. 2.ऐसे केवल छह तरीके हैं जिनमें केवल दो पत्र ही सही लिफ़ाफ़ों में जा सकते हैं ।

उपर्युक्त कथनों में से कौन-सा/से सही है/हैं ?

Read this question in English

There are four letters and four envelopes and exactly one letter is to be put in exactly one envelope with the correct address. If the letters are randomly inserted into the envelopes, then consider the following statements:

1. It is possible that exactly one letter goes into an incorrect envelope.

2. There are only six ways in which only two letters can go into the correct envelopes.

Which of the statements given above is/are correct?

(a) 1 only · (b) 2 only · (c) Both 1 and 2 · (d) Neither 1 nor 2

उत्तर और व्याख्या · explanation in English

Answer: (b) केवल 2

If three letters are in their correct envelopes, the only envelope left belongs to the fourth letter, so exactly one wrong letter is impossible. For exactly two correct, choose the two correct letters in 6 ways and swap the other two in 1 way: 6 ways.

  1. Statement 1: if 3 letters are correct, the 4th letter is left with only its own envelope, so it is correct too. Exactly one wrong letter cannot happen.
  2. Statement 2: choose the 2 letters that go correctly: C(4, 2) = 6 ways.
  3. The other 2 letters must both be wrong, which is possible in only 1 way (they swap envelopes).
  4. Ways = 6 × 1 = 6, so statement 2 is correct.
  • ✗ 1. With three letters correctly placed, the last envelope is the fourth letter's own; one misplaced letter alone is impossible.
  • ✓ 2. C(4, 2) = 6 choices of the two correct letters, and the remaining two can be wrong in just one way (a swap): 6 ways.

Remember · Exactly one item misplaced is never possible. 'Exactly k correct' = C(n, k) × ways to put all the rest wrong.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q19

Medium

संख्या 11223344 के अंकों को पुनर्व्यवस्थित कर भिन्न 8-अंकों की कितनी संख्याएँ बनाई जा सकती हैं, इस प्रकार कि विषम अंक विषम स्थानों पर हों और सम अंक सम स्थानों पर हों ?

Read this question in English

How many distinct 8-digit numbers can be formed by rearranging the digits of the number 11223344 such that odd digits occupy odd positions and even digits occupy even positions?

(a) 12 · (b) 18 · (c) 36 · (d) 72

उत्तर और व्याख्या · explanation in English

Answer: (c) 36

The four odd positions must hold 1, 1, 3, 3 and the four even positions must hold 2, 2, 4, 4. Each set can be arranged in 4!/(2! × 2!) = 6 ways, so there are 6 × 6 = 36 numbers.

  1. Odd digits 1, 1, 3, 3 fill the 4 odd positions; even digits 2, 2, 4, 4 fill the 4 even positions.
  2. Arrangements of 1, 1, 3, 3 = 4!/(2! × 2!) = 24/4 = 6.
  3. Arrangements of 2, 2, 4, 4 = 6 in the same way.
  4. Total = 6 × 6 = 36.

Remember · Arrangements with repeated items = n!/(p! × q! …); choices for independent slots multiply.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q30

Easy

ABCD कोई वर्ग है । AB और CD प्रत्येक पर एक बिंदु; और BC और DA प्रत्येक पर दो भिन्न बिंदु चुने जाते हैं । इन छह बिन्दुओं में से किन्हीं तीन बिन्दुओं को शीर्ष ले कर कितने भिन्न त्रिभुज खींचे जा सकते हैं ?

Read this question in English

ABCD is a square. One point on each of AB and CD; and two distinct points on each of BC and DA are chosen. How many distinct triangles can be drawn using any three points as vertices out of these six points?

(a) 16 · (b) 18 · (c) 20 · (d) 24

उत्तर और व्याख्या · explanation in English

Answer: (c) 20

Three points fail to make a triangle only if they lie on one straight line. No side of the square carries more than two of the six points, so no three are in a line and every choice of three points gives a triangle: C(6, 3) = 20.

  1. Points: 1 on AB, 1 on CD, 2 on BC and 2 on DA — six in all.
  2. A set of three points forms no triangle only when all three lie on one side; no side has more than two points.
  3. Number of triangles = C(6, 3) = (6 × 5 × 4)/(3 × 2 × 1) = 20.

Remember · Triangles from points = C(n, 3) minus collinear triples; check each line for three or more points.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q36

Medium

किसी डिब्बे में 14 काली गेंदें, 20 नीली गेंदें, 26 हरी गेंदें, 28 पीली गेंदें, 38 लाल गेंदें और 54 सफ़ेद गेंदें हैं । निम्नलिखित कथनों पर विचार कीजिए :

  1. 1.यदि डिब्बे में से यादृच्छिक रूप से कोई सी n गेंदें इस प्रकार निकाली जाएँ कि उनमें कम-से-कम एक रंग का एक पूरा समूह अवश्य हो, तो n की लघुतम संख्या 175 है ।
  2. 2.यदि डिब्बे में से यादृच्छिक रूप से कोई सी m गेंदें इस प्रकार निकाली जाएँ कि उनमें हर रंग की कम-से-कम एक गेंद अवश्य हो, तो m की लघुतम संख्या 167 है ।

उपर्युक्त कथनों में से कौन-सा/से सही है/हैं ?

Read this question in English

A box contains 14 black balls, 20 blue balls, 26 green balls, 28 yellow balls, 38 red balls and 54 white balls. Consider the following statements:

1. The smallest number n such that any n balls drawn from the box randomly must contain one full group of at least one colour is 175.

2. The smallest number m such that any m balls drawn from the box randomly must contain at least one ball of each colour is 167.

Which of the above statements is/are correct?

(a) 1 only · (b) 2 only · (c) Both 1 and 2 · (d) Neither 1 nor 2

उत्तर और व्याख्या · explanation in English

Answer: (c) 1 और 2 दोनों

For a full group, the unluckiest draw takes every colour one ball short of complete: 13 + 19 + 25 + 27 + 37 + 53 = 174, so 175 balls guarantee a full group. For one ball of each colour, the unluckiest draw takes every ball except the smallest colour: 180 − 14 = 166, so 167 balls guarantee all six colours.

  1. Total balls = 14 + 20 + 26 + 28 + 38 + 54 = 180.
  2. Statement 1: the worst case leaves every colour one short: 13 + 19 + 25 + 27 + 37 + 53 = 174 balls with no full group. The next ball completes one, so n = 175. Correct.
  3. Statement 2: the worst case is all balls except the smallest colour (14 black): 180 − 14 = 166 balls without black. The next ball must be black, so m = 167. Correct.
  • ✓ 1. 174 balls can still leave every colour one ball short of full; the 175th ball must complete a group.
  • ✓ 2. 166 balls can all be non-black; the 167th must be black, giving all six colours.

Remember · Guarantee questions: build the unluckiest draw that still fails, then add one.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q47

Medium

अंकों 1, 2, 3 और 4 से, इन अंकों में से किसी अंक को बिना दोहराए, बनी उन सभी 4-अंकों की संख्याओं का, जो 2000 से कम हैं, योगफल क्या है ?

Read this question in English

What is the sum of all 4-digit numbers less than 2000 formed by the digits 1, 2, 3 and 4, where none of the digits is repeated?

(a) 7998 · (b) 8028 · (c) 8878 · (d) 9238

उत्तर और व्याख्या · explanation in English

Answer: (a) 7998

Numbers below 2000 must start with 1, and the other three places take 2, 3 and 4 in 3! = 6 orders. The thousands digits give 6 × 1000 = 6000; in each other place each of 2, 3 and 4 appears twice, adding 18 × 111 = 1998. The total is 7998.

  1. Below 2000 with the digits 1, 2, 3, 4 used once each, the first digit must be 1; 2, 3, 4 fill the rest in 3! = 6 ways.
  2. Thousands place: 6 × 1000 = 6000.
  3. In each of the hundreds, tens and units places, each of 2, 3, 4 appears 2 times: (2 + 3 + 4) × 2 = 18.
  4. Their contribution = 18 × (100 + 10 + 1) = 18 × 111 = 1998.
  5. Total = 6000 + 1998 = 7998.

Remember · Sum of numbers formed by permutations: count how often each digit sits in each place, then multiply by the place values (111…).

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q48

Easy

एक वृत्त के रूप में व्यवस्थित 12 वस्तुओं में से दक्षिणावर्त दिशा में 10 क्रमागत वस्तुओं के चयनों की संख्या क्या है ?

Read this question in English

What is the number of selections of 10 consecutive things out of 12 things in a circle taken in the clockwise direction?

(a) 3 · (b) 11 · (c) 12 · (d) 66

उत्तर और व्याख्या · explanation in English

Answer: (c) 12

Going clockwise, a block of 10 consecutive things is fixed by its starting point, and each of the 12 things can be the start. So there are 12 selections.

  1. Taken clockwise, a run of 10 consecutive things is decided by where it starts.
  2. Each of the 12 things can be the starting point, and different starts give different sets.
  3. Number of selections = 12.
  4. Check: each selection leaves out 2 neighbouring things, and a circle of 12 has exactly 12 neighbouring pairs.

Remember · In a circle of n items, the number of blocks of k consecutive items (k < n) is n.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q65

Medium

किसी परीक्षा में, चार प्रश्न-पत्रों, नामतः P, Q, R और S में से प्रत्येक प्रश्न-पत्र के लिए अधिकतम अंक 100 हैं । विद्यार्थियों द्वारा प्राप्त किए गए अंक पूर्णांकों में हैं । कोई भी विद्यार्थी n विभिन्न तरीकों से 99% प्राप्तांक ला सकता है । n का मान क्या है ?

Read this question in English

In an examination, the maximum marks for each of the four papers namely P, Q, R and S are 100. Marks scored by the students are in integers. A student can score 99% in n different ways. What is the value of n?

(a) 16 · (b) 17 · (c) 23 · (d) 35

उत्तर और व्याख्या · explanation in English

Answer: (d) 35

99% of 400 is 396, so the student loses exactly 4 marks in all across the four papers. Sharing 4 lost marks among 4 papers (0 or more each) can be done in C(7, 3) = 35 ways.

  1. Total maximum = 4 × 100 = 400; 99% of 400 = 396.
  2. Let the marks lost in P, Q, R, S be w, x, y, z (each 0 or more) with w + x + y + z = 4; no paper can exceed 100.
  3. Number of solutions = C(4 + 3, 3) = C(7, 3) = 35.
  4. Check by cases: (4,0,0,0) type 4 ways, (3,1,0,0) 12, (2,2,0,0) 6, (2,1,1,0) 12, (1,1,1,1) 1 → 35.

Remember · Count marks lost instead of marks scored; n identical units in r boxes can be shared in C(n + r − 1, r − 1) ways.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q66

Easy

किसी ध्वज को लाल, हरे या पीले रंगों में से कुछ या सभी रंगों का उपयोग कर चार क्षैतिज पट्टियों द्वारा परिरूपित करना है । कितने ऐसे विभिन्न तरीकों से इसे किया जा सकता है, इस प्रकार की कोई भी दो आसन्न पट्टियाँ एक ही रंग की न हों ?

Read this question in English

A flag has to be designed with 4 horizontal stripes using some or all of the colours red, green and yellow. What is the number of different ways in which this can be done so that no two adjacent stripes have the same colour?

(a) 12 · (b) 18 · (c) 24 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (c) 24

The top stripe can take any of 3 colours, and each later stripe any colour except the one just above it (2 choices). So there are 3 × 2 × 2 × 2 = 24 flags.

  1. Stripe 1: 3 choices.
  2. Stripes 2, 3 and 4: each must differ from the stripe above it, so 2 choices each.
  3. Total = 3 × 2 × 2 × 2 = 24.
  4. 'Some or all of the colours' allows two-colour flags such as red-green-red-green; they are already included.

Remember · For 'no two adjacent alike', choose the first freely, then (colours − 1) for each next item.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2023 · Q68

Medium

P, Q, R, S और T पाँच व्यक्ति हैं, जिनमें से प्रत्येक व्यक्ति को एक कार्य सौंपना है । कार्य-1 न तो P को, न ही Q को सौंपा जा सकता है । कार्य-2 या तो R को, या S को ही सौंपा जाना है । कार्य कितने प्रकार से सौंपे जा सकते हैं ?

Read this question in English

There are five persons P, Q, R, S and T each one of whom has to be assigned one task. Neither P nor Q can be assigned Task-1. Task-2 must be assigned to either R or S. In how many ways can the assignment be done?

(a) 6 · (b) 12 · (c) 18 · (d) 24

उत्तर और व्याख्या · explanation in English

Answer: (d) 24

Task-2 goes to R or S (2 ways). Task-1 cannot go to P or Q or to whoever took Task-2, so it goes to the other of R and S or to T (2 ways). The remaining three tasks go to the remaining three persons in 3! = 6 ways: 2 × 2 × 6 = 24.

  1. Five persons, five tasks, one task each.
  2. Task-2: R or S → 2 ways.
  3. Task-1: not P, not Q, and not the person given Task-2 → 2 ways (the other of R and S, or T).
  4. Tasks 3, 4 and 5 go to the 3 persons left: 3! = 6 ways.
  5. Total = 2 × 2 × 6 = 24.

Remember · Fill the most restricted positions first, then arrange the rest freely.

Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q17

Easy

150 प्रतियोगियों वाली किसी शतरंज टूर्नामेंट में जब-जब कोई खिलाड़ी बाजी हारता है, उसे बाहर कर दिया जाता है। यह निश्चित किया गया है कि कोई भी बाजी बराबरी (टाई/ड्रॉ) पर निर्णीत नहीं होगी। इस पूरे टूर्नामेंट में कितनी बाजियाँ खेली गईं?

Read this question in English

In a tournament of Chess having 150 entrants, a player is eliminated whenever he loses a match. It is given that no match results in a tie/draw. How many matches are played in the entire tournament?

(a) 151 · (b) 150 · (c) 149 · (d) 148

उत्तर और व्याख्या · explanation in English

Answer: (c) 149

Every match has exactly one loser, and every loser is eliminated. The tournament ends when one player is left, so 149 players are eliminated — and that takes exactly 149 matches.

  1. Each match produces exactly one loser, who is eliminated.
  2. To leave a single winner, 150 − 1 = 149 players must be eliminated.
  3. So 149 matches are played.
  4. Check: with 4 players, 2 semi-finals + 1 final = 3 = 4 − 1 matches.

Remember · Knockout tournament with n players: always n − 1 matches, whatever the byes.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q18

Easy

3-अंक की कितनी धनपूर्ण संख्याएँ (अंकों का प्रयोग दुबारा किए बिना) इस प्रकार होंगी कि संख्या का प्रत्येक अंक विषम हो और संख्या 5 से विभाज्य हो?

Read this question in English

How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?

(a) 8 · (b) 12 · (c) 16 · (d) 24

उत्तर और व्याख्या · explanation in English

Answer: (b) 12

A number divisible by 5 with an odd units digit must end in 5. The other two places are filled from 1, 3, 7, 9 without repetition: 4 × 3 = 12 numbers.

  1. Divisible by 5 means the units digit is 0 or 5; it must be odd, so it is 5.
  2. The hundreds digit can be any of the remaining odd digits 1, 3, 7, 9: 4 choices.
  3. The tens digit can be any of the 3 odd digits still unused: 3 choices.
  4. Total = 4 × 3 = 12.

Remember · Fill the restricted place first (here the units digit), then count the choices for the free places.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q24

Easy

वर्ण A, B, C, D और E इस तरह व्यवस्थित किए गए हैं कि A और E के बीच यथातथ्य दो वर्ण हैं। इस तरह कितनी व्यवस्थाएँ संभव हैं?

Read this question in English

The letters A, B, C, D and E are arranged in such a way that there are exactly two letters between A and E. How many such arrangements are possible?

(a) 12 · (b) 18 · (c) 24 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (c) 24

In a row of five, A and E can stand only at places 1 and 4 or 2 and 5, in either order: 4 ways. The other three letters then fill the remaining places in 3! = 6 ways, giving 24.

  1. Number the places 1 to 5. Exactly two letters between A and E means A and E occupy places (1, 4) or (2, 5): 2 choices of places.
  2. A and E can swap within the chosen places: 2 × 2 = 4 ways.
  3. B, C and D fill the other three places in 3! = 6 ways.
  4. Total = 4 × 6 = 24.

Remember · Place the constrained pair first (positions × internal order), then arrange the remaining items freely.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q27

Easy

A, B और C तीन स्थान इस प्रकार हैं कि A से B के लिए तीन भिन्न रास्ते हैं, B से C के लिए चार भिन्न रास्ते हैं और A से C के लिए तीन भिन्न रास्ते हैं। इन रास्तों का प्रयोग कर कोई व्यक्ति कितने भिन्न मार्गों से A से C तक जा सकता है?

Read this question in English

A, B and C are three places such that there are three different roads from A to B, four different roads from B to C and three different roads from A to C. In how many different ways can one travel from A to C using these roads?

(a) 10 · (b) 13 · (c) 15 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (c) 15

One can go directly from A to C (3 ways) or go through B (3 × 4 = 12 ways). These are alternatives, so they add: 3 + 12 = 15.

  1. Direct routes from A to C: 3.
  2. Routes through B: 3 roads from A to B × 4 roads from B to C = 12.
  3. Total = 3 + 12 = 15.

Remember · Legs taken one after another multiply; alternative routes add.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q55

Medium

एक संख्या-आधारित ताला है, जिसके लिए 3-अंक की वैयक्तिक अभिज्ञान संख्या (पिन) है। पिन में 1 से 7 तक अंक हैं। कोई अंक दोबारा नहीं आता। पिन के अंक बाएँ से दाहिने तरफ घटते हुए क्रम में हैं। पिन के किन्हीं दो अंकों के बीच कम-से-कम 2 का अंतर है। अधिकतम कितने प्रयासों में पिन का निश्चित पता लगाया जा सकता है?

Read this question in English

There is a numeric lock which has a 3-digit PIN. The PIN contains digits 1 to 7. There is no repetition of digits. The digits in the PIN from left to right are in decreasing order. Any two digits in the PIN differ by at least 2. How many maximum attempts does one need to find out the PIN with certainty?

(a) 6 · (b) 8 · (c) 10 · (d) 12

उत्तर और व्याख्या · explanation in English

Answer: (c) 10

Each set of three digits from 1–7 that are pairwise at least 2 apart gives exactly one PIN (written in decreasing order). There are 10 such sets, so in the worst case 10 attempts are needed.

  1. Once three digits are chosen, the decreasing order fixes the PIN, so count the digit sets.
  2. The digits must be from 1–7 with every two at least 2 apart. List by the largest digit: 753, 752, 751, 742, 741, 731, 642, 641, 631, 531 — 10 PINs.
  3. Quick count: choosing 3 of 7 numbers with no two consecutive = C(7 − 3 + 1, 3) = C(5, 3) = 10.
  4. In the worst case the right PIN is the last one tried, so up to 10 attempts are needed.

Remember · Choosing k of n numbers with no two consecutive: C(n − k + 1, k).

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q76

Easy

एक शून्येतर अंक, अंग्रेजी वर्णमाला से एक स्वर और एक व्यंजन (कैपिटल में) पासवर्ड बनाने में इस तरह प्रयुक्त किए जाने हैं कि हर पासवर्ड स्वर से शुरू हो और व्यंजन पर समाप्त हो। ऐसे कितने पासवर्ड बनाए जा सकते हैं?

Read this question in English

One non-zero digit, one vowel and one consonant from English alphabet (in capital) are to be used in forming passwords, such that each password has to start with a vowel and end with a consonant. How many such passwords can be generated?

(a) 105 · (b) 525 · (c) 945 · (d) 1050

उत्तर और व्याख्या · explanation in English

Answer: (c) 945

The password has three characters: the vowel first, the consonant last and the digit in the middle. That gives 5 × 9 × 21 = 945 passwords.

  1. Each password has three characters. It starts with the vowel and ends with the consonant, so the digit sits in the middle.
  2. Choices: 5 vowels (A, E, I, O, U) × 9 non-zero digits × 21 consonants.
  3. Total = 5 × 9 × 21 = 945.

Remember · Fix the forced positions first; each remaining position multiplies by its number of choices.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q77

Medium

किसी मेज पर 9 प्याले इस तरह सजाकर रखे हैं कि उनकी पंक्तियों और कॉलमों की संख्या समान है। इनमें से 6 प्यालों में कॉफी और 3 प्यालों में चाय है। इन्हें कितनी प्रकार से इस तरह रखा जा सकता है कि प्रत्येक पंक्ति में कम-से-कम एक कॉफी का प्याला हो?

Read this question in English

There are 9 cups placed on a table arranged in equal number of rows and columns out of which 6 cups contain coffee and 3 cups contain tea. In how many ways can they be arranged so that each row should contain at least one cup of coffee?

(a) 18 · (b) 27 · (c) 54 · (d) 81

उत्तर और व्याख्या · explanation in English

Answer: (d) 81

The 9 cups form a 3 × 3 grid. Choosing the 3 places for tea can be done in C(9, 3) = 84 ways; only the 3 choices that fill a whole row with tea leave a row without coffee. So 84 − 3 = 81.

  1. Equal rows and columns with 9 cups means a 3 × 3 grid.
  2. Cups of the same drink are alike, so an arrangement is fixed by the 3 places holding tea: C(9, 3) = 84.
  3. A row lacks coffee only when all 3 tea cups fill that row: 3 such arrangements.
  4. Arrangements with coffee in every row = 84 − 3 = 81.

Remember · ‘At least one’ counts are easiest as total minus the bad cases.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2022 · Q79

Easy

0·XY रूप की कितनी संख्याएँ हैं, जहाँ X और Y भिन्न शून्येतर अंक हैं?

Read this question in English

What is the number of numbers of the form 0·XY, where X and Y are distinct non-zero digits?

(a) 72 · (b) 81 · (c) 90 · (d) 100

उत्तर और व्याख्या · explanation in English

Answer: (a) 72

X can be any of the 9 non-zero digits and Y any of the other 8, giving 9 × 8 = 72 different decimals.

  1. X can be any of 1–9: 9 choices.
  2. Y must be non-zero and different from X: 8 choices.
  3. Total = 9 × 8 = 72, and each ordered pair gives a different number.

Remember · Distinct choices from the same pool: n × (n − 1) ordered pairs.

Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2021 · Q64

Medium

एक बिसात (शतरंज बोर्ड) पर एक सीधे पथ की लंबाई में विकर्णों पर 6 क्रमागत वर्गों को कितने विभिन्न प्रकार से चयनित किया जा सकता है ?

Read this question in English

On a chess board, in how many different ways can 6 consecutive squares be chosen on the diagonals along a straight path?

(a) 4 · (b) 6 · (c) 8 · (d) 12

उत्तर और व्याख्या · explanation in English

Answer: (b) 6

The diagonals of a chess board are its two main diagonals of 8 squares each. A run of 6 consecutive squares fits in 8 − 6 + 1 = 3 positions on each, giving 6 ways.

  1. 'The diagonals' of a chess board are the two main diagonals, each 8 squares long.
  2. On a line of 8 squares, 6 consecutive squares can start at square 1, 2 or 3 — 8 − 6 + 1 = 3 ways.
  3. Two diagonals: 3 × 2 = 6 ways.
  4. Note: counting every diagonal line of 6 or more squares would give 18, which is not an option — confirming the main-diagonal reading.

Remember · Runs of k consecutive cells in a line of n cells = n − k + 1.

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2021 · Q66

Easy

अंकों के रूप में 2, 2, 3, 3, 3 का प्रयोग करते हुए, 30000 से बड़ी कितनी भिन्न संख्याएँ बन सकती हैं ?

Read this question in English

Using 2, 2, 3, 3, 3 as digits, how many distinct numbers greater than 30000 can be formed?

(a) 3 · (b) 6 · (c) 9 · (d) 12

उत्तर और व्याख्या · explanation in English

Answer: (b) 6

A five-digit number from these digits exceeds 30000 only if it starts with 3. The remaining digits 2, 2, 3, 3 can be arranged in 4!/(2! × 2!) = 6 distinct ways.

  1. A number made from 2, 2, 3, 3, 3 is greater than 30000 only if its first digit is 3.
  2. The remaining digits 2, 2, 3, 3 can be arranged in 4! ÷ (2! × 2!) = 24 ÷ 4 = 6 ways.
  3. So 6 distinct numbers are greater than 30000.
  4. Check: 32233, 32323, 32332, 33223, 33232, 33322.

Remember · Arrangements with repeated items = n! ÷ (product of the factorials of the repeat counts).

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2021 · Q69

Easy

एक पंक्ति में 6 व्यक्ति हैं । एक अन्य व्यक्ति को उनमें से 3 व्यक्तियों से इस प्रकार हाथ मिलाना है कि वह दो क्रमागत व्यक्तियों से हाथ नहीं मिलाएगा । ऐसे कितने भिन्न संभाव्य संयोजनों में हाथ मिलाए जा सकते हैं ?

Read this question in English

There are 6 persons arranged in a row. Another person has to shake hands with 3 of them so that he should not shake hands with two consecutive persons. In how many distinct possible combinations can the handshakes take place?

(a) 3 · (b) 4 · (c) 5 · (d) 6

उत्तर और व्याख्या · explanation in English

Answer: (b) 4

Choosing 3 of 6 people in a row with no two next to each other leaves only {1, 3, 5}, {1, 3, 6}, {1, 4, 6} and {2, 4, 6}. That is 4 combinations, matching C(4, 3).

  1. Number the persons 1 to 6 and choose 3 with no two adjacent.
  2. Possible sets: {1, 3, 5}, {1, 3, 6}, {1, 4, 6}, {2, 4, 6}.
  3. So there are 4 combinations.
  4. Check: C(n − k + 1, k) = C(6 − 3 + 1, 3) = C(4, 3) = 4.

Remember · Ways to pick k non-adjacent items from n in a row = C(n − k + 1, k).

Question and answer: UPSC's official GS Paper II (2021, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q60

Medium

मूल्यवर्गों ₹ 50, ₹ 100, ₹ 200, ₹ 500 और ₹ 2,000 के साथ, एक समय में कम-से-कम तीन मूल्यवर्गों को लेते हुए, कितनी विभिन्न धनराशियाँ बनाई जा सकती हैं?

Read this question in English

How many different sums can be formed with the denominations ₹ 50, ₹ 100, ₹ 200, ₹ 500 and ₹ 2,000 taking at least three denominations at a time?

(a) 16 · (b) 15 · (c) 14 · (d) 10

उत्तर और व्याख्या · explanation in English

Answer: (a) 16

Choosing 3, 4 or all 5 of the five denominations gives 10 + 5 + 1 = 16 selections. Each denomination is more than the sum of all the smaller ones, so no two selections can give the same amount — all 16 sums are different.

  1. Ways to choose 3, 4 or 5 of 5 denominations: ⁵C₃ + ⁵C₄ + ⁵C₅ = 10 + 5 + 1 = 16.
  2. 100 > 50, 200 > 50 + 100, 500 > 50 + 100 + 200 and 2,000 > 850, so different selections always give different totals.
  3. Number of different sums = 16.
  4. Check: the 16 sums run from ₹ 350 (50 + 100 + 200) to ₹ 2,850 (all five) with no repeats.

Remember · Count selections with nCr, then check whether two selections can give the same total before answering.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2020 · Q67

Easy

शब्द ‘DELHI’ के सभी अक्षरों का प्रयोग करते हुए कितने विभिन्न 5-अक्षरी शब्द (जिनका कोई अर्थ हो या न हो) बनाए जा सकते हैं, जबकि प्रत्येक शब्द D के साथ शुरू होता है और I के साथ समाप्त होता है?

Read this question in English

How many different 5-letter words (with or without meaning) can be constructed using all the letters of the word ‘DELHI’ so that each word has to start with D and end with I?

(a) 24 · (b) 18 · (c) 12 · (d) 6

उत्तर और व्याख्या · explanation in English

Answer: (d) 6

With D fixed first and I fixed last, only E, L and H are left to arrange in the three middle places. Three distinct letters can be arranged in 3! = 6 ways.

  1. Positions 1 and 5 are fixed: D _ _ _ I.
  2. E, L and H fill the three middle places in 3! = 3 × 2 × 1 = 6 ways.
  3. Check: DELHI, DEHLI, DLEHI, DLHEI, DHELI, DHLEI — six words.

Remember · Fix the letters whose places are given, then arrange only the remaining ones: n distinct letters give n! orders.

Question and answer: UPSC's official GS Paper II (2020, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q14

Easy

चार समांतर रेखाओं के एक समुच्चय से, जो कि चार समांतर रेखाओं के एक अन्य समुच्चय से प्रतिच्छेदी है, बनाए जा सकने वाले समांतर चतुर्भुजों की संख्या है

Read this question in English

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of four parallel lines, is

(a) 18 · (b) 24 · (c) 32 · (d) 36

उत्तर और व्याख्या · explanation in English

Answer: (d) 36

Every choice of two lines from each family encloses exactly one parallelogram, so the count is 4C2 × 4C2 = 6 × 6 = 36.

  1. A parallelogram needs two lines from the first set and two from the second.
  2. Ways to choose 2 lines out of 4 = 4C2 = 6, for each set.
  3. Parallelograms = 6 × 6 = 36.

Remember · m parallel lines crossing n parallel lines form mC2 × nC2 parallelograms.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

CSAT 2019 · Q56

Medium

मान लीजिए, आपके पास तीन मूल्य-वर्गों, ₹ 1, ₹ 10 और ₹ 50, में रुपया मुद्रा पर्याप्त मात्रा में है । आप ₹ 107 के एक बिल का भुगतान कितने विभिन्न तरीकों से कर सकते हैं ?

Read this question in English

Suppose you have sufficient amount of rupee currency in three denominations: ₹ 1, ₹ 10 and ₹ 50. In how many different ways can you pay a bill of ₹ 107?

(a) 16 · (b) 17 · (c) 18 · (d) 19

उत्तर और व्याख्या · explanation in English

Answer: (c) 18

Count by the number of ₹50 notes. With none, ₹107 can include 0 to 10 ten-rupee notes (11 ways); with one, ₹57 can include 0 to 5 (6 ways); with two, ₹7 uses only ₹1 coins (1 way). Total 18.

  1. No ₹50 note: ₹107 = 10a + b with a = 0, 1, …, 10 → 11 ways.
  2. One ₹50 note: ₹57 = 10a + b with a = 0, 1, …, 5 → 6 ways.
  3. Two ₹50 notes: ₹7 = 10a + b needs a = 0 → 1 way.
  4. In each case ₹1 coins make up the balance, so each choice is one way.
  5. Total = 11 + 6 + 1 = 18.

Remember · For currency combinations, fix the largest denomination, count choices for the next; the smallest always fills the rest.

Question and answer: UPSC's official GS Paper II (2019, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). Permalink ·

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