A probability problem by shreyas shastry
Probability
Level
3
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digit so that the odd digits occupy even positions ,that is two,fourth,sixth and eight positions.
The answer is 60.
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1 solution
Solution: Let us consider only odd numbers in 4 positions. So, 2 odd number in 4 positions. Odd numbers in Even position can be occupied in 4! / 2! x 2! = 6 ways. Now assume remaining even digits in 5 position. So, 2 even numbers in 5 positions. Hence, Even numbers in Odd positions can be occupied in 5! / 2! x 3! = 10 ways. In total 9 digits can be arranged in 6 x10 ways i.e., 60 ways. :)