The number of ways to form a 7-digit number u | 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 $(2, 2, 4)$.In a $7$-digit number,the odd

Vedclass · Accepted Answer

$18$

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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 $(2, 2, 4)$.In a $7$-digit number,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 odd places,we arrange the $4$ odd digits $(1, 1, 3, 3)$ in $4$ odd places. The number of ways is $\frac{4!}{2!2!} = \frac{24}{4} = 6$.Next,we arrange the $3$ even digits $(2, 2, 4)$ in $3$ even places. The number of ways is $\frac{3!}{2!} = 3$.Therefore,the total number of ways $= 6 	imes 3 = 18$.

The number of ways to form a $7$-digit number using the digits $1, 2, 3, 4, 3, 2, 1$ such that odd digits always occupy odd places is:

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