The number of different 9-digit numbers that | vedclass

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Question

(B) The given number is $445577888$. The digits are: $4, 4, 5, 5, 7, 7, 8, 8, 8$. There are $4$ odd digits $(5, 5, 7, 7)$ and $5$ even digits $(4, 4,

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) The given number is $445577888$. The digits are: $4, 4, 5, 5, 7, 7, 8, 8, 8$. There are $4$ odd digits $(5, 5, 7, 7)$ and $5$ even digits $(4, 4, 8, 8, 8)$. There are $9$ positions in total. The even positions are $2, 4, 6, 8$. There are $4$ such positions. The $4$ odd digits must occupy these $4$ even positions. The number of ways to arrange $5, 5, 7, 7$ in $4$ positions is $\frac{4!}{2!2!} = \frac{24}{4} = 6$. The remaining $5$ positions (odd positions $1, 3, 5, 7, 9$) must be occupied by the $5$ even digits $(4, 4, 8, 8, 8)$. The number of ways to arrange $4, 4, 8, 8, 8$ in $5$ positions is $\frac{5!}{2!3!} = \frac{120}{2 	imes 6} = 10$. Therefore,the total number of ways is $6 	imes 10 = 60$.

The number of different $9$-digit numbers that can be formed by rearranging the digits of the number $445577888$ such that the odd digits occupy even positions is:

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