+0

# help counting

0
35
2

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

#2
+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
+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
+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.

BuilderBoi Apr 13, 2022