(i) no girl
|
Total No. |
No. to be chosen |
No. of ways to choose |
Girls |
4 |
0 |
4C0 |
Boys |
7 |
5 |
7C5 |
Total number of ways = 4C0 × 7C5
\(=\frac{4!}{0!(4-0)!}\times\frac{7!}{5!(7-5)}\)
\(=\frac{4!}{4!}\times \frac{7!}{5!\space2!}\)
\(=1\times\frac{7\times6}{2}\)
= 21
(ii) at least 3 girls? Since, the team has to consist of at least 3 girls, the team can consist of
(a) 3 girls and 2 boys
(b) 4 girls and 1 boy
(a) 3 girls and 2 boys
|
Total No. |
No. to be chosen |
No. of ways to chosen |
Girls |
4 |
3 |
4C3 |
Boys |
7 |
2 |
7C2 |
Number of ways selecting = 7C2 × 4C3
=\(\frac{7!}{2!\space5!}\times\frac{4!}{3!\space1!}\)
\(=\frac{7\times6}{2\times1}\times4\)
= 84
(b) 4 girls and 1 boy
|
Total No. |
No. to be chosen |
No. of ways to choose |
Girls |
4 |
4 |
4C4 |
Boys |
7 |
1 |
7C1 |
Number if ways selecting = 7C1 × 4C4
\(=\frac{7!}{1!\space6!}\times\frac{4!}{4!\space0\!}\)
\(=7\times1\)
= 7
∴ Total number of ways = 84 + 7 = 91
(iii) at least one girl and one boy?
A group giving at least one boy and one girl will consist of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girl
Number of ways of selecting 1 boy and 4 girls = 7C1 × 4C4 = 7
Number of ways of selecting 2 boys and 3 girls = 7C1 × 4C3 = 84
Number of ways of selecting 3 boys and 2 girls = 7C2 × 4C2 = 210
Number of ways of selecting 4 boys and 1 girl = 7C4 × 4C1 = 140
Hence, total number of ways = 7 + 84 + 210 + 140 = 441 ways.