We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
∴ Required number
of ways |
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) |
|
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) |
|
| = (6 x 4) + |
❨ |
6 x 5 |
x |
4 x 3 |
❩ |
+ |
❨ |
6 x 5 x 4 |
x 4 |
❩ |
+ |
❨ |
6 x 5 |
❩ |
| 2 x 1 |
2 x 1 |
3 x 2 x 1 |
2 x 1 |
|
|
= (24 + 90 + 80 + 15) |
|
= 209. |
