Question 48935
how many numbers of five different digits, each number to contain 3 odd and 2 even digits, can be formed from the digits 1,2,3,4,5,6,7,8 and 9?
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number of ways to select 3 odd = 5C3=5*4/1*2=10
number of ways to select 2 even = 4C2=4*3/1*2=6
number of ways to permute the 5 digits = 5!

Answer: 10*6*5!=7200

Cheers,
Stan H.