6.Permutation and Combination
hard

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

A

$180$

B

$175$

C

$162$

D

$160$

(JEE MAIN-2019)

Solution

Number of such numbers ${ = ^4}{C_3} \times \frac{{3!}}{{2!}} \times \frac{{6!}}{{2!4!}} = 180$

Standard 11
Mathematics

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