Question 1162948
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Use the combination function nCr. Order does not matter.


We have three scenarios
A) 3 boys, 2 girls
B) 4 boys, 1 girl
C) 5 boys, 0 girls
in which there are more boys than girls


For scenario A
n = 7 boys, r = 3 selections, nCr = 7C3 = 35 ways to pick the boys 
n = 6 girls, r = 2 selections, nCr = 6C2 = 15 ways to pick the girls
35*15 = 525 ways to form a committee of 3 boys, 2 girls. Let P = 525 since we'll use it later.


Scenario B
n = 7 boys, r = 4 selections, nCr = 7C4 = 35 ways to pick the boys 
n = 6 girls, r = 1 selection, nCr = 6C1 = 6 ways to pick the girls
35*6 = 210 ways to form a committee of 4 boys, 1 girl. Let Q = 210 since we'll use it later.


Scenario C
n = 7 boys, r = 5 selections, nCr = 7C5 = 21 ways to pick the boys 
n = 6 girls, r = 0 selection, nCr = 6C0 = 1 way to do this
21*1 = 21 ways to form a committee of 5 boys, 0 girls. Let R = 21 since we'll use it later.


Add the results P, Q and R
P+Q+R = 525+210+21 = 756



Answer: 756


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