The number of 9-digit numbers that can be for | vedclass

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Question

(C) The given number is $123412341$. The digits are ${1, 1, 1, 2, 2, 3, 3, 4, 4}$.There are $5$ odd digits ${1, 1, 1, 3, 3}$ and $4$ even digits ${2,

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The given number is $123412341$. The digits are ${1, 1, 1, 2, 2, 3, 3, 4, 4}$.There are $5$ odd digits ${1, 1, 1, 3, 3}$ and $4$ even digits ${2, 2, 4, 4}$.$A$ $9$-digit number has $9$ places: $1, 2, 3, 4, 5, 6, 7, 8, 9$.The even places are $2, 4, 6, 8$ (total $4$ places).The odd places are $1, 3, 5, 7, 9$ (total $5$ places).According to the condition,the $4$ even digits must occupy the $4$ even places.The number of ways to arrange the even digits ${2, 2, 4, 4}$ in $4$ even places is $\frac{4!}{2!2!} = \frac{24}{4} = 6$.The number of ways to arrange the $5$ odd digits ${1, 1, 1, 3, 3}$ in $5$ odd places is $\frac{5!}{3!2!} = \frac{120}{6 	imes 2} = 10$.Total number of such $9$-digit numbers $= 6 	imes 10 = 60$.

The number of $9$-digit numbers that can be formed using all the digits of the number $123412341$ such that the even digits occupy only even places is $..........$

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