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 $(2^{nd}, 4^{th}, 6^{th}, 8^{th})$ and $5$ odd places $(1^{st}, 3^{rd}, 5^{th}, 7^{th}, 9^{th})$.We need to place $3$ odd digits in $4$ even places. This is not possible as we have $3$ odd digits and $4$ even places. However,the question implies we must place the $3$ odd digits into the $4$ even places. Since we have $3$ odd digits $(1, 1, 3)$,we choose $3$ places out of $4$ even places in $^4C_3$ ways.The number of ways to arrange $1, 1, 3$ in these $3$ places is $\frac{3!}{2!} = 3$.The remaining $6$ digits $(2, 2, 2, 2, 4, 4)$ must be placed in the remaining $6$ places. The number of ways to arrange them is $\frac{6!}{4!2!} = 15$.Total 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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