CSAT Solved Papers/ 2022/Q76

2022 CSAT — Q76

Quant Counting & combinatorics 2.5 marks Easy

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?

  1. A 105
  2. B 525
  3. C 945 Answer
  4. D 1050

Worked rationale

The password is three characters in a fixed pattern: vowel, then the digit, then consonant (it must start with a vowel and end with a consonant, leaving the digit in the middle).

Counts available:

  • vowels: 55 (A,E,I,O,UA, E, I, O, U)
  • non-zero digits: 99 (11–99)
  • consonants: 26−5=2126 - 5 = 21

By the multiplication principle:

5×9×21=945.5 \times 9 \times 21 = 945.

Answer: (c) 945.

Why the other options miss

  • A
    dropped the digit slot: uses 5×215 \times 21 only, leaving the digit factor out.
  • B
    wrong digit count: uses 5×5×215 \times 5 \times 21, taking 55 digits instead of 99.
  • D
    counted zero as a digit: takes 1010 digits (00–99, i.e. 5×10×215 \times 10 \times 21), ignoring the “non-zero” clause.

Specialist insight

The position pattern is fully forced (vowel–digit–consonant), so there is no arrangement factor — just one choice per slot: 5×9×215 \times 9 \times 21. The two traps are using 1010 digits instead of 99 (the “non-zero” clause, trap d) and forgetting the digit slot entirely (trap a). Read the constraints into the slot counts before multiplying.

The trap, in one line

Pattern vowel–digit–consonant: 5×9×21=9455 \times 9 \times 21 = 945 (digit is non-zero ⇒9\Rightarrow 9, not 1010) == (c).

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