NIMCET 2025 — Mathematics PYQ

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Question

NIMCET 2025 — Mathematics PYQ

NIMCET | Mathematics | 2025

The number of 3-digit integers that are multiples of $6$ which can be formed by using the digits $\{1, 2, 3, 4, 5, 6\}$ without repetition is:

Choose the correct answer:

Explanation

Step 1: Divisibility Rules for 6

A number is a multiple of $u n d e f in e d$ if it is:

Even: The last digit must be $u n d e f in e d$ or $u n d e f in e d$ .
Divisible by 3: The sum of the digits must be a multiple of $u n d e f in e d$ .

Step 2: Find Sets of 3 Digits whose Sum is a Multiple of 3

We need to pick 3 distinct digits from $u n d e f in e d$ such that their sum $u n d e f in e d$ .

Sum = 6: $u n d e f in e d$
Sum = 9: $u n d e f in e d$
Sum = 12: $u n d e f in e d$
Sum = 15: $u n d e f in e d$

Total sets = $u n d e f in e d$ .

Step 3: Count Even Arrangements for each Set

For each set, we must arrange the digits such that the last digit is even ( $u n d e f in e d$ ).

$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (2 numbers)
$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (4 numbers)
$u n d e f in e d$ : Even digits: None. (0 numbers)
$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (4 numbers)
$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (2 numbers)
$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (6 numbers)
$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (2 numbers)
$u n d e f in e d$ : Even digits: $u n d e f in e d$ . Arrangements: $u n d e f in e d$ (4 numbers)

Step 4: Total Count

Total $u n d e f in e d$

Final Answer:

The total number of such $u n d e f in e d$ -digit integers is 24. (Option D)