8-digit numbers are formed using the digits 1 | vedclass

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(B) In an $8$-digit number,there are $4$ odd places $(1^{st}, 3^{rd}, 5^{th}, 7^{th})$ and $4$ even places $(2^{nd}, 4^{th}, 6^{th}, 8^{th})$.The give

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) In an $8$-digit number,there are $4$ odd places $(1^{st}, 3^{rd}, 5^{th}, 7^{th})$ and $4$ even places $(2^{nd}, 4^{th}, 6^{th}, 8^{th})$.The given digits are $1, 1, 2, 2, 2, 3, 4, 4$. The odd digits are $1, 1, 3$ (total $3$ digits) and the even digits are $2, 2, 2, 4, 4$ (total $5$ digits).We are given that odd digits must not occupy odd places. This means the $3$ odd digits must be placed in the $4$ even places.The number of ways to arrange the $3$ odd digits $(1, 1, 3)$ in $4$ even places is given by $\frac{4!}{2!} = 12$.After placing the $3$ odd digits,we have $5$ remaining places (the $4$ odd places plus the $1$ remaining even place) and $5$ remaining digits $(2, 2, 2, 4, 4)$.The number of ways to arrange these $5$ digits in the remaining $5$ places is $\frac{5!}{3!2!} = \frac{120}{6 	imes 2} = 10$.Therefore,the total number of such $8$-digit numbers is $12 	imes 10 = 120$.

$8$-digit numbers are formed using the digits $1, 1, 2, 2, 2, 3, 4, 4$. The number of such numbers in which the odd digits do not occupy odd places is:

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