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?

Asked by Aaryan | 1 year ago |  50

##### Solution :-

Given:

The digits 1, 2, 3, 4, 3, 2, 1

The total number of digits are 7.

There are 4 odd digits 1,1,3,3 and 4 odd places (1,3,5,7)

So, the odd digits can be arranged in odd places in = $$\dfrac{4!}{(2! 2!)}$$ ways.

The remaining even digits 2,2,4 can be arranged in 3 even places in = $$\dfrac{ 3!}{2!}$$ Ways.

Hence, the total number of digits =$$\dfrac{4!}{(2! 2!)}$$ × $$\dfrac{ 3!}{2!}$$

= 3×2×1×3×1

= 18

Hence, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.

Answered by Sakshi | 1 year ago

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