Question 462398: I am still having trouble trying to solve this problem. Everytime I do it I come up with a different answer. Can someone please show me how to go beyond this step?
8.In how many different ways can a committee of 9 members be chosen from
6 freshmen, 5 sophomores, and 4 juniors if the committee must consist of at least one freshman?
Ans:
#'s in the order of (frosh;soph;juniors)
The limits are 6 frosh;5 soph;4 juniors
1:5;3 (this means 1 frosh; 5 soph; 3 juniors): # of ways = 6*5*4C3 = 6*5*4 = 120
1;4;4::: # of ways = 6*5C4*4C4 = 30
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! If it must have one freshman so now we have a committee of 8 to choose from 14 students.
15C9=5005 ways with everyone.
9C9=1 way with no frosh.
5005-1=5004 different ways can a committee of 9 members be chosen from 6 freshmen, 5 sophomores, and 4 juniors if the committee must consist of at least one freshman.
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Ed
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