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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.
Please help

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Choose two numbers to appear twice and then one number to appear once. Now we have five numbers. Choose two places for the smaller number that appears twice and then the two places for the larger number that appears twice. $$\binom{9}{2}\binom{7}{1}\binom{5}{2}\binom{3}{2}=7560$$

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You need to divide the answer you've found by $2$ again because the numbers that appear twice are also interchangeable, i.e., while doing $\binom{9}{1}\binom{8}{1}\binom{7}{1}$, chosing one of the digits that appear twice first, and then another is as same as chosing the second one first.

For example we choose $1$,$2$ and $3$ where $1$ and $2$ appear twice. Then chosing $1$ first and then $2$ and then $3$ gives the same result as chosing $2$ first and then $1$ and then $3$ (by considering the order of chosing the numbers that appear twice first and then the number that appears once).

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Your answer ($7560$) is off by a factor of two, because when counting the number of ways in which three digits can be chosen, you forgot to divide by $2$ to account for the fact that the digits that appear twice are interchangeable.

In other words, for a number consisting of the digits $a$, $b$, and $c$, where $a$ appears once, you've also counted it as a number consisting of the digits $a$, $c$, and $b$.

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