SOLUTION: There are 13 males and 18 females in the math class. If chosen completely at random, how many different ways can 3 males and 2 females be chosen to do problems on the board.

Algebra ->  Permutations -> SOLUTION: There are 13 males and 18 females in the math class. If chosen completely at random, how many different ways can 3 males and 2 females be chosen to do problems on the board.       Log On


   

Question 715558: There are 13 males and 18 females in the math class. If chosen completely at random, how many different ways can 3 males and 2 females be chosen to do problems on the board.
Answer by DrBeeee(684) About Me  (Show Source):
You can put this solution on YOUR website!
Use permutations.
To select 3 males from 13 is
(1) 13P3 = 13*12*11 = 1716
To select 2 females from 18 is
(2) 18P2 = 18*17 = 306
The total number of possible of 3 males AND 2 females is the product of (1) and (2) or
(3) total = 525096 groups