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
Best Answer
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. $$\binom{9}{2}\binom{7}{1}\binom{5}{2}\binom{3}{2}=7560$$