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

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(B) The total number of digits is $8$. The odd places are $1, 3, 5, 7$ ($4$ places) and the even places are $2, 4, 6, 8$ ($4$ places).The given digits

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) The total number of digits is $8$. The odd places are $1, 3, 5, 7$ ($4$ places) and the even places are $2, 4, 6, 8$ ($4$ places).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 do not occupy odd places. This means the $3$ odd digits must be placed in the $4$ available 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 $1$ remaining even place and the $4$ odd places) to fill with the remaining $5$ even digits $(2, 2, 2, 4, 4)$.The number of ways to arrange these $5$ digits 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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