# 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, 1, 3, 3) and 4 odd places(first, third, fifth and seventh).

At these places the odd digits can be arranged in $$4!\over 2!2!$$ = 6

Then at the remaining 3 places, the remaining three digits(2, 2, 4) can be arranged in $$3!\over 2!$$ = 3 ways

Therefore,  The required number of numbers = 6 $$\times$$ 3 = 18

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