How many numbers can be formed with the digit | vedclass

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(C) The given digits are $1, 2, 3, 4, 3, 2, 1$. There are $7$ digits in total.Out of these,the odd digits are $1, 3, 3, 1$ (total $4$ digits) and the

Vedclass · Accepted Answer

$18$

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(C) The given digits are $1, 2, 3, 4, 3, 2, 1$. There are $7$ digits in total.Out of these,the odd digits are $1, 3, 3, 1$ (total $4$ digits) and the even digits are $2, 4, 2$ (total $3$ digits).The odd places are $1^{st}, 3^{rd}, 5^{th},$ and $7^{th}$ (total $4$ places).The even places are $2^{nd}, 4^{th},$ and $6^{th}$ (total $3$ places).Since odd digits must occupy the odd places,the $4$ odd digits $(1, 1, 3, 3)$ can be arranged in $4$ odd places in $\frac{4!}{2!2!} = \frac{24}{4} = 6$ ways.The $3$ even digits $(2, 2, 4)$ can be arranged in $3$ even places in $\frac{3!}{2!1!} = \frac{6}{2} = 3$ ways.Therefore,the total number of ways to form the numbers is $6 	imes 3 = 18$.

How many numbers can be formed with the digits $1, 2, 3, 4, 3, 2, 1$ so that odd digits always occupy the odd places?

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