# 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?

Given Numbers = 1, 2, 3, 4, 3, 2, 1

So we  can form 7 digits  numbers.

Here 1th , 3th , 5th , 7th are , Odd  places which we have  to fill odd  numbers.

In Given question odd numbers  are = 1, 3, 3, 1

Number  of  places  to arrange  them  $=\frac{4 !}{2 ! \times 2 !}=\frac{4 \times 3}{2}=6 \text { ways }$

We  have  2th , 4th, 6th are even place.

Even numbers are  = 2, 4 , 2

Number  of ways  to arrange  them $=\frac{3 !}{2 ! }=\frac{3 \times 2}{2}=3 \text { ways }$

Total number  of  ways = 3 × 6 = 18 ways.

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