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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 :

Option 1)
$180$

Option 2)
$160$

Option 3)
$175$

Option 4)
$162$

Answers (1)

S solutionqc

$1,1,2,2,2,2,3,4,4=9 \; digit$

We have total 4 even place and 3 odd digit = $1,1,3$

Number of way placing odd digit at even places = $^{4}C_{3}\frac{3!}{2!}=4\times3=12$

Number of ways placing even digits $=\frac{6!}{4!2!}=\frac{6\times5}{2}=15$

So total number of ways = $12\times15=180$

Option 1)

$180$

Option 2)

$160$

Option 3)

$175$

Option 4)

$162$

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