The number of numbers that can be formed with | vedclass

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

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) The given digits are $1, 2, 3, 4, 3, 2, 1$. There are $7$ digits in total.The odd digits are $1, 3, 3, 1$ (total $4$ digits).The even digits are $2, 4, 2$ (total $3$ digits).The odd places are $1^{st}, 3^{rd}, 5^{th}, 7^{th}$ (total $4$ places).The even places are $2^{nd}, 4^{th}, 6^{th}$ (total $3$ places).Since odd digits must occupy odd places,we arrange $1, 3, 3, 1$ in $4$ odd places. The number of ways is $\frac{4!}{2!2!} = \frac{24}{4} = 6$.Next,we arrange the even digits $2, 4, 2$ in $3$ even places. The number of ways is $\frac{3!}{2!} = 3$.Therefore,the total number of ways is $6 	imes 3 = 18$.

The number of numbers that can be formed with the help of the digits $1, 2, 3, 4, 3, 2, 1$ so that odd digits always occupy odd places,is

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