Question 849699: In how many different ways can a committee of 8 members be chosen from 5 freshmen, 4 sophomores, and 3 juniors if the committee must consist of 3 freshmen, 3 sophomores, and 2 juniors
In how many different ways can a committee of 8 members be chosen from 5 freshmen, 4 sophomores, and 3 juniors if the committee must consist of at least one junior
In how many ways can three movies be chosen from 12 movies as the
year’s top 3 movies
? 2.
(1 point) In how many ways can three movies be chosen from 12 movies as the
year’s #1, #2, and #3
movies? 3.
(1 point) A coin is flipped 12 times and the sequence of heads and tails is observed. In how many ways can the sequence consist of exactly 7 heads and 5 tails? 4.
(1 point) Refer to the given graph. Assume one can move only down or to the right. How many different routes are there that start at A, pass through B, and end at C?
6x5 box graph B is 2nd row 4th row
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! I will answer the ones I can.
In how many different ways can a committee of 8 members be chosen from 5 freshmen, 4 sophomores, and 3 juniors if the committee must consist of 3 freshmen, 3 sophomores, and 2 juniors (5 choose 3) * (4 choose 3) * (3 choose 2) = 
In how many different ways can a committee of 8 members be chosen from 5 freshmen, 4 sophomores, and 3 juniors if the committee must consist of at least one junior. There is (11 choose 8) complete ways to choose this committee. Consider the cases where there is no junior. Then we can have these cases
[4 freshman, 4 sophomore] =(5 choose 4)* (4 choose 4)
[5 freshman ,3 sophomore] =(5 choose 5)* (4 choose 3)
Putting this all together we have (11 choose 8) - ((5 choose 4)*(4 choose 4) + (5 choose 5)*(4 choose 3)) =  ways.
In how many ways can three movies be chosen from 12 movies as the
year’s top 3 movies
(12 choose 3) = 
(1 point) In how many ways can three movies be chosen from 12 movies as the
year’s #1, #2, and #3
movies?
12 P 3 = 12!/9! = 12*11*10 = 
(1 point) A coin is flipped 12 times and the sequence of heads and tails is observed. In how many ways can the sequence consist of exactly 7 heads and 5 tails? 4.
We choose 7 of them to be heads out of 12, there is no reason to order the other 5 tails as we are choosing (5 choose 5) ways which is just 1.
(12 choose 7) = 
(1 point) Refer to the given graph. Assume one can move only down or to the right. How many different routes are there that start at A, pass through B, and end at C?
6x5 box graph B is 2nd row 4th row
No graph given. Hard to tell even with your explanation.
Hope this helps!
Devin
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