Question
(JEE Main 2024 style) How many 4-digit numbers can be formed from the digits such that no digit is repeated and the number is even?
Solution — Step by Step
For an even number, the last digit must be even — so the units place must be one of . That’s 3 choices.
Place an even digit in the units position: 3 ways.
After placing the units digit, 5 digits remain for the other three positions.
Thousands, hundreds, tens places are filled from the remaining 5 digits — and order matters, so:
ways.
Total four-digit even numbers.
Answer: .
Why This Works
The key principle: tackle the most constrained position FIRST. Units is constrained (must be even), so we pick it before the unconstrained positions. This avoids overcounting and double-counting.
If we had filled thousands first (no constraint), then hundreds, then tens, we would have committed to specific digits before knowing whether an even one was still available — possible but messier.
Strategy rule: In permutation problems with one position constrained, fix that position FIRST, then permute the rest. Saves seconds and avoids casework.
Alternative Method — Total Minus Odd
Total 4-digit numbers with no repetition from 6 digits .
Odd numbers: units digit , 3 choices. Remaining: . So odd count .
Even count . Same answer.
This works because half the digits are even and half are odd — the symmetry guarantees a 50-50 split.
Common Mistake
Students often start by filling the thousands place first (with 6 choices) and don’t separate out the even-digit constraint. They end up with , the total count — and forget to halve it.
Another classic: forgetting that “no digit repeated” is given. Without that condition, the answer would be . Always re-read for the “no repetition” clause.
JEE Main has at least one P&C problem per shift, often built around digit constraints (even, divisible by 5, between 1000 and 5000, etc.). The “fix the constrained slot first” template handles 80%+ of these. CBSE Class 11 boards include similar problems for 4 marks.