Permutations and Combinations: Tricky Questions Solved (5)

Question

How many four-digit numbers can be formed using the digits $0, 1, 2, 3, 4, 5$ without repetition such that the number is divisible by $5$?

Solution — Step by Step

Why This Works

The trick is to handle the two cases for divisibility by 5 separately, because the constraint “first digit ≠ 0” interacts differently with each case.

When the last digit is $0$, the $0$ is “used up” so the first digit can be any of the remaining non-zero digits. When the last digit is $5$, the $0$ is still available, but it can’t go in the first position — so we deduct.

Always: identify constraints, split into cases that simplify each constraint, count separately, then add (mutually exclusive cases) or use inclusion-exclusion (overlapping cases).

Alternative Method

Total 4-digit numbers from $\{0,1,2,3,4,5\}$ without repetition: first digit has $5$ choices (not $0$), then $5 \times 4 \times 3 = 60$ for the remaining. Total $= 5 \times 5 \times 4 \times 3 = 300$.

Of these, divisible by $5$: those ending in $0$ or $5$. Compute as in cases. Same result.

Common Mistake

The most common trap: forgetting that “first digit ≠ 0” when the last digit is $5$. Students compute $5 \times 4 \times 3 = 60$ for both cases (treating them symmetrically) and get $120$. That overcounts numbers like $0235$, which is really a 3-digit number.

Another trap: counting the last-digit-$0$ case as $5 \times 5 \times 4 = 100$ — they forget that when $0$ is fixed at the end, only $5$ digits ($1$ to $5$) are available for the first three positions, and the order matters with no repetition.

When in doubt, write a small example: how many 2-digit numbers from $\{0, 1, 2, 5\}$ are divisible by $5$? Last digit $0$ or $5$. Last $= 0$: first from $\{1, 2, 5\}$ — 3 numbers. Last $= 5$: first from $\{1, 2\}$ — 2 numbers. Total $5$. List them: $10, 20, 50, 15, 25$. ✓