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How many distinct 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 (each digit can be used at MOST once)?
A$6!$
B$P(6, 4) = 360$
C$\binom{6}{4} = 15$
D$6^4 = 1296$
Answer & Solution
Correct answer: B. $P(6, 4) = 360$
1. Each 4-digit number is an ORDERED selection of 4 from 6 distinct digits (no repetition).
2. This is a PERMUTATION: $P(6, 4) = \dfrac{6!}{(6-4)!} = \dfrac{6!}{2!} = \dfrac{720}{2} = 360$.
3. Equivalently: 6 choices for first digit, 5 for second, 4 for third, 3 for fourth $= 6 \cdot 5 \cdot 4 \cdot 3 = 360$.
4. Option A is $6!$ which uses ALL 6 digits. Option C is the unordered selection — doesn't account for digit position. Option D allows REPETITION (which we excluded).
_Source: NCERT Class 11 Mathematics, Ch 6, §6.3 + Examples, p. 6–7._
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