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

Options

A. 180
B. 175
C. 160
D. 162

Answer

A

Explanation

Given digits are 1, 1, 2, 2, 2, 2, 3, 4, 4.
According to the question, odd numbers 1, 1, 3 should occur at even places only.

`therefore` The number of ways to arrange odd numbers ot even places are `""^(4)C_(3)xx(3!)/(2!)`
and the number of ways to arrange remaining even numbers are `(6!)/(4!2!)`.
So, total number of 9-digit numbers, that can be formed using the given digits are
`""^(4)C_(3)xx(3!)/(2!)xx(6!)/(4!2!)=4xx3xx15=180`