ITPEC FE Morning April 2019 Question 2
ITPEC FE Morning April 2019 — Question 2 of 80
Permutations with restrictions (leading zero) — counting four-digit numbers from distinct digits where the leading digit cannot be zero.
We need four-digit decimal numbers using 4 different digits from 0–9, with a non-zero leading digit.
Step-by-step:
- 1st digit: 9 choices (1–9, since 0 is not allowed as a leading digit)
- 2nd digit: 9 choices (0–9 minus the digit already used)
- 3rd digit: 8 choices
- 4th digit: 7 choices
Total = 9 × 9 × 8 × 7 = 4536
Why not others:
- (a) 4320 — incorrect calculation, likely a distractor
- (b) 4436 — close but wrong arithmetic
- (d) 5040 — this is P(10, 4) = 10 × 9 × 8 × 7, which counts all 4-digit permutations including those starting with 0; a classic trap
Key rule: When forming numbers with distinct digits, always handle the leading-digit restriction first, then count the remaining positions.
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