Permutation and CombinationHard
Question
How many nine digit numbers can be formed using the digits 2, 2, 3, 3, 5, 5, 8, 8, 8 so that the odd digits occupy even positions?
Options
A.7560
B.180
C.16
D.60
Solution
Even place
There are four even places and four odd digit number so total number of filling is
rest are also occupy is 
ways
Hence total number of ways =
= 60
There are four even places and four odd digit number so total number of filling is
rest are also occupy is waysHence total number of ways =
= 60Create a free account to view solution
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