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

A

180

B

175

C

160

D

162

Text Solution

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The correct Answer is:
A
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`
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