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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