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4 girls and 8 boys are standing together. In how many ways can a 7- person committee be selected from the group if at least 2 boys must be included?

 Apr 13, 2022

Best Answer 

 #2
avatar+1384 
+2

There can be: 

 

0 girls + 7 boys: \({4 \choose 0} \times {8 \choose 7} = 8\)

1 girl + 6 boys: \({4 \choose 1} \times {8 \choose 6} = 112\)

2 girls + 5 boys: \({4 \choose 2} \times {8 \choose 5} = 336\)

3 girls + 4 boys: \({4 \choose 3} \times {8 \choose 4} = 280\)

4 girls + 3 boys: \({4 \choose 4} \times {8 \choose 3} = 56\)

 

Adding everything up, there can be \(\color{brown}\boxed{792}\) committees.

 Apr 13, 2022
 #1
avatar+122 
-1

2, 3 or 4 girls must be selected

 

If 2 are selected we have  C(4,2) * C(8, 5)

If 3 are selected, we have C(4,3) * C(8, 4)

If 4 are selected, we have C(4,4) * C(8, 3)

 

So we have

 

C(4,2)*C(8,5)  + C(4,3) * C(8,4)  + C(4,4)* C(8,3)   =  672 ways

 Apr 13, 2022
 #2
avatar+1384 
+2
Best Answer

There can be: 

 

0 girls + 7 boys: \({4 \choose 0} \times {8 \choose 7} = 8\)

1 girl + 6 boys: \({4 \choose 1} \times {8 \choose 6} = 112\)

2 girls + 5 boys: \({4 \choose 2} \times {8 \choose 5} = 336\)

3 girls + 4 boys: \({4 \choose 3} \times {8 \choose 4} = 280\)

4 girls + 3 boys: \({4 \choose 4} \times {8 \choose 3} = 56\)

 

Adding everything up, there can be \(\color{brown}\boxed{792}\) committees.

BuilderBoi Apr 13, 2022

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