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?
+2668 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.
+64 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
+2668 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.
