SOLUTION: 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?Thanks!

Algebra ->  Probability-and-statistics -> SOLUTION: 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?Thanks!      Log On


   



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?Thanks!
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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?
----------------
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.