All possible numbers are formed using the dig | vedclass

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Question

(A) The total number of digits is $9$. The digits are $1, 1, 2, 2, 2, 2, 3, 4, 4$.Odd digits are $1, 1, 3$ (total $3$ odd digits).Even digits are $2,

Vedclass · Accepted Answer

$180$

Vedclass · Answer

(A) The total number of digits is $9$. The digits are $1, 1, 2, 2, 2, 2, 3, 4, 4$.Odd digits are $1, 1, 3$ (total $3$ odd digits).Even digits are $2, 2, 2, 2, 4, 4$ (total $6$ even digits).There are $4$ even places ($2^{nd}, 4^{th}, 6^{th}, 8^{th}$ positions) in a $9$-digit number.We need to place $3$ odd digits in these $4$ even places.The number of ways to choose $3$ places out of $4$ is $^4C_3 = 4$.The number of ways to arrange the $3$ odd digits $(1, 1, 3)$ in these chosen places is $\frac{3!}{2!} = 3$.The remaining $6$ places must be filled by the $6$ even digits $(2, 2, 2, 2, 4, 4)$.The number of ways to arrange these $6$ even digits is $\frac{6!}{4!2!} = \frac{720}{24 	imes 2} = 15$.Total number of such numbers $= ^4C_3 	imes \frac{3!}{2!} 	imes \frac{6!}{4!2!} = 4 	imes 3 	imes 15 = 180$.

All possible numbers are formed using the digits $1, 1, 2, 2, 2, 2, 3, 4, 4$ taken all at a time. The number of such numbers in which the odd digits occupy even places is

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