How many different non-digit numbers can be formed from the digits of the number 223355888 by rearrangement of the digits so that the odd digits occupy even places?

Question

Dr. Abubacker · Accepted Answer

(c)
In a nine digits number, there are four even places for the four odd digits 3, 3, 5, 5
$	herefore$ Required number of ways $= \frac{4!}{2!2!}.\frac{5!}{2!3!} = 60$

Q. How many different non-digit numbers can be formed from the digits of the number 223355888 by rearrangement of the digits so that the odd digits occupy even places?

Explanation

Reviewed by Dr. Abubacker

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