Counting, permutations & probability

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.

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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.

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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.

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Answer: (d) Four triplets

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.

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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.

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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.

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Answer: (b) 2 only

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.

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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.

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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.

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Answer: (c) Both 1 and 2

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.

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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…).

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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).

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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.

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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.

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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.

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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.

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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).

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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).

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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.

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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.

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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.

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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.

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