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Question

Find the total number of nine digit numbers that can be formed using the digits 2,2,3,3,5,5,8,8,8 so that the odd digit occupy the even places.

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Solution

Given digit are 2,3,3,5,5,8,8,8

There are 4 even places and 5 odd places odd digits are 3,3,5,5

odd number are placed in these 4 places in 4!2!2! ways

even digits are 2,2,8,8,8

These even digits are placed in 5 odd places in 5!3!2! ways

No. of required a digit numbers are 4!2!2!5!3!2!

=4×3!×5×4×3×2×12×2×3!×2

=60

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