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UPSC CSE CSAT 2022 · Paper II (CSAT) · Question 76 · Counting, permutations & probability

एक शून्येतर अंक, अंग्रेजी वर्णमाला से एक स्वर और एक व्यंजन (कैपिटल में) पासवर्ड बनाने में इस तरह…

UPSC Civil Services (Preliminary) Examination 2022, General Studies Paper II (CSAT), Question 76 — Counting, permutations & probability.

CSAT 2022 · Q76

Counting, permutations & probability 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). ·

Cite this question: UPSC CSE CSAT 2022, Q76 — minimalistias.com/pyqs/csat/2022/q76

In the UPSC syllabus: Counting, permutations & probability › Counting figures & cases See it and tick it covered →

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