Question:
How many five-digit numbers can be formed using the integers 3, 4, 5 and 6 with exactly one digit appearing twice?
How many five-digit numbers can be formed using the integers 3, 4, 5 and 6 with exactly one digit appearing twice?
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Whenever a digit repeats $r$ times, divide by $r!$ in the permutations: $\dfrac{n!}{r!}$ for a single repeated digit.
Updated On: Sep 1, 2025
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Correct Answer: 240
Solution and Explanation
Step 1: Choose the repeated digit. There are 4 choices (one of 3,4,5,6).
Step 2: Arrange the multiset. Each number uses the multiset $\{a,a,b,c,d\}$ with all $a,b,c,d$ distinct.
Number of arrangements $=\dfrac{5!}{2!}=60$.
Step 3: Multiply choices. Total numbers $=4\times 60=240$.
Step 2: Arrange the multiset. Each number uses the multiset $\{a,a,b,c,d\}$ with all $a,b,c,d$ distinct.
Number of arrangements $=\dfrac{5!}{2!}=60$.
Step 3: Multiply choices. Total numbers $=4\times 60=240$.
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