How many numbers can be formed using the digi | vedclass

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(B) The given digits are $1, 1, 2, 2, 3, 3, 4$. There are $4$ odd digits $(1, 1, 3, 3)$ and $3$ even digits $(2, 2, 4)$.There are $7$ positions in tot

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) The given digits are $1, 1, 2, 2, 3, 3, 4$. There are $4$ odd digits $(1, 1, 3, 3)$ and $3$ even digits $(2, 2, 4)$.There are $7$ positions in total. The odd positions are $1^{st}, 3^{rd}, 5^{th},$ and $7^{th}$ (total $4$ positions).The $4$ odd digits can be arranged in these $4$ odd positions in $\frac{4!}{2! 	imes 2!} = 6$ ways.The remaining $3$ even digits $(2, 2, 4)$ can be arranged in the remaining $3$ even positions in $\frac{3!}{2!} = 3$ ways.Therefore,the total number of such numbers is $6 	imes 3 = 18$.

How many numbers can be formed using the digits $1, 2, 3, 4, 3, 2, 1$ such that odd digits always occupy odd positions?

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