How many different nine-digit numbers can be | vedclass

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Question

(C) The given number is $223355888$. The digits are $2, 2, 3, 3, 5, 5, 8, 8, 8$. There are $9$ digits in total. The positions are $1, 2, 3, 4, 5, 6, 7

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The given number is $223355888$. The digits are $2, 2, 3, 3, 5, 5, 8, 8, 8$. There are $9$ digits in total. The positions are $1, 2, 3, 4, 5, 6, 7, 8, 9$. The even positions are $2, 4, 6, 8$,which are $4$ in number. The odd digits in the given number are $3, 3, 5, 5$. These $4$ odd digits must occupy the $4$ even positions. The number of ways to arrange these is $\frac{4!}{2! 	imes 2!} = \frac{24}{4} = 6$. The remaining $5$ digits are $2, 2, 8, 8, 8$,which must occupy the $5$ odd positions $(1, 3, 5, 7, 9)$. The number of ways to arrange these is $\frac{5!}{2! 	imes 3!} = \frac{120}{2 	imes 6} = 10$. Therefore,the total number of ways is $6 	imes 10 = 60$.

How many different nine-digit numbers can be formed from the digits of the number $223355888$ by rearranging the digits such that the odd digits occupy even places?

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