Question
How many 3-digit numbers have all distinct (different) digits?
Solution — Step by Step
A 3-digit number has three positions: hundreds (H), tens (T), and units (U).
- The number must be between 100 and 999.
- All three digits must be different from each other.
We’ll count using the Fundamental Principle of Counting: if there are ways to do the first thing and ways to do the second thing (independent), there are ways to do both.
The hundreds digit (H) can be 1 through 9 (it cannot be 0, because that would make it a 2-digit number or smaller).
Choices for H = 9 (digits 1, 2, 3, 4, 5, 6, 7, 8, 9)
The tens digit (T) can be any digit from 0–9 except the digit already used for H.
Since one digit is used by H, there are remaining digits.
Choices for T = 9 (digits 0–9, minus whatever H is)
The units digit (U) can be any digit from 0–9 except the two digits already used for H and T.
Since two digits are already used, there are remaining digits.
Choices for U = 8
Total 3-digit numbers with all distinct digits:
Why This Works
The key insight is that the hundreds digit is more restricted than the other two: it cannot be 0 (or the number would not be a 3-digit number), AND it must be distinct from T and U.
The tens digit has 9 choices (any of 10 digits except the hundreds digit — including 0).
The units digit has 8 choices (any of 10 digits except the two already used).
The multiplication principle applies because each choice is made independently (in sequence, with only the constraint of distinctness from previously chosen digits).
Alternative Method — Total Minus Repetitions
Total 3-digit numbers: 999 − 99 = 900 (from 100 to 999 inclusive)
Now, count 3-digit numbers where at least two digits are the same:
Let:
- A = numbers where H = T (but U can be anything)
- B = numbers where H = U (but T can be anything)
- C = numbers where T = U (but H can be anything)
By inclusion-exclusion, this gets complicated quickly. The first method (direct counting) is much simpler. But as a check:
Numbers with all same digits (like 111, 222, …, 999): 9 numbers. Numbers with at least two same digits = 900 − 648 = 252.
We can verify: numbers with exactly two same digits = 252 − 9 = 243. (Not needed for the answer, but useful as a cross-check.)
The direct method gives 648 — this is the correct and simplest approach.
Common Mistake
The most common error is treating the hundreds digit the same as the other two and writing . This is wrong because 10 choices for H would include 0, giving numbers like 012, 023 — these are not 3-digit numbers; they are 2-digit numbers with a leading zero.
The correct count for H is 9 (digits 1–9), then 9 for T (any digit except H), then 8 for U (any digit except H and T). This gives .
In JEE and CBSE Class 11 permutation problems, the first digit of a multi-digit number is almost always a special case because it cannot be 0. Always handle the first digit separately before counting other digits. Write: “H cannot be 0 → 9 choices; then count remaining positions.”