How many distinct 9-digit numbers can be form | vedclass

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Question

(C) The given number is $223355888$. The digits are: $2, 2, 3, 3, 5, 5, 8, 8, 8$.Total digits = $9$.Odd digits are: $3, 3, 5, 5$ (Total $4$ digits).Ev

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The given number is $223355888$. The digits are: $2, 2, 3, 3, 5, 5, 8, 8, 8$.Total digits = $9$.Odd digits are: $3, 3, 5, 5$ (Total $4$ digits).Even digits are: $2, 2, 8, 8, 8$ (Total $5$ digits).Positions are: $1, 2, 3, 4, 5, 6, 7, 8, 9$.Even positions are: $2, 4, 6, 8$ (Total $4$ positions).Odd positions are: $1, 3, 5, 7, 9$ (Total $5$ positions).We need to place $4$ odd digits in $4$ even positions and $5$ even digits in $5$ odd positions.Number of ways to arrange $4$ odd digits $(3, 3, 5, 5)$ in $4$ positions = $\frac{4!}{2!2!} = \frac{24}{4} = 6$.Number of ways to arrange $5$ even digits $(2, 2, 8, 8, 8)$ in $5$ positions = $\frac{5!}{2!3!} = \frac{120}{2 	imes 6} = 10$.Total number of ways = $6 	imes 10 = 60$.

How many distinct $9$-digit numbers can be formed by rearranging the digits of the number $223355888$ such that all odd digits occupy even positions?

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