How many different nine-digit numbers can be | vedclass

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Question

(C) The given number is $223355888$. The digits are $2, 2, 3, 3, 5, 5, 8, 8, 8$.Total digits = $9$. The positions are $1, 2, 3, 4, 5, 6, 7, 8, 9$.Even

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 = $9$. The positions are $1, 2, 3, 4, 5, 6, 7, 8, 9$.Even places are $2, 4, 6, 8$ (total $4$ places).Odd digits are $3, 3, 5, 5$ (total $4$ digits).Even digits are $2, 2, 8, 8, 8$ (total $5$ digits).Step $1$: Arrange the $4$ odd digits in the $4$ even places. The number of ways is $\frac{4!}{2! 	imes 2!} = \frac{24}{4} = 6$.Step $2$: Arrange the $5$ even digits in the remaining $5$ odd places. The number of ways is $\frac{5!}{2! 	imes 3!} = \frac{120}{2 	imes 6} = 10$.Total number of ways = $6 	imes 10 = 60$.

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

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