The number of different nine-digit numbers th | vedclass

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Question

(C) The given number is $223355888$,which has $9$ digits: $2, 2, 3, 3, 5, 5, 8, 8, 8$. There are $4$ odd digits $(3, 3, 5, 5)$ and $5$ even digits $(2

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The given number is $223355888$,which has $9$ digits: $2, 2, 3, 3, 5, 5, 8, 8, 8$. There are $4$ odd digits $(3, 3, 5, 5)$ and $5$ even digits $(2, 2, 8, 8, 8)$. In a $9$-digit number,the positions are $1, 2, 3, 4, 5, 6, 7, 8, 9$. The even positions are $2, 4, 6, 8$ (total $4$ positions). The odd positions are $1, 3, 5, 7, 9$ (total $5$ positions). We must place the $4$ odd digits in the $4$ even positions. The number of ways to arrange $3, 3, 5, 5$ is $\frac{4!}{2!2!} = \frac{24}{4} = 6$. We must place the $5$ even digits in the $5$ odd positions. The number of ways to arrange $2, 2, 8, 8, 8$ is $\frac{5!}{2!3!} = \frac{120}{2 	imes 6} = 10$. Therefore,the total number of such $9$-digit numbers is $6 	imes 10 = 60$.

The number of different nine-digit numbers that can be formed by rearranging the digits of the number $223355888$ such that the odd digits occupy even positions is:

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