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Question
How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?
Solution
There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in\[\frac{4!}{2!2!}\]ways.
The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in\[\frac{3!}{2!}\]ways.
By fundamental principle of counting:
Required number of arrangements =\[\frac{4!}{2!2!}\]\[\times\]\[\frac{3!}{2!}\]= 18
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