Find the number of five digit numbers that can be formed using the digits 1,2,3,4,5,6,7,8,9 in which one digit appears once and two digits appear twice (e.g.41174 is one such number but 75355 is not)
The three digits can be chosen from 9 digits in $\binom{9}{1}\binom{8}{1}\binom{7}{1}$ ways.These three can be arranged in $\frac{5!}{2!2!}$ways.So total ways are
$\binom{9}{1}\binom{8}{1}\binom{7}{1}\frac{5!}{2!2!}=15120$ but the correct answer is 7560.
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