Q. Number of ways of forming a committee of $6$ members out of $5$ Indians, $5$ Americans and $5$ Australians such that there will be atleast one member from each country in the committee is
AP EAMCETAP EAMCET 2019
Solution:
Required cases are as follow:
Indians
Americans
Australians
1
1
4
1
2
3
1
3
2
1
4
1
2
1
3
2
2
2
2
3
1
3
1
2
3
2
1
4
1
1
$\therefore $ Required number of committee
$={ }^{5} C_{1} \times{ }^{5} C_{1} \times{ }^{5} C_{4}+{ }^{5} C_{1} \times{ }^{5} C_{2} \times{ }^{5} C_{3}+{ }^{5} C_{1} \times{ }^{5} C_{3} $
$ \times{ }^{5} C_{2} $
$+{ }^{5} C_{1} \times{ }^{5} C_{4} \times{ }^{5} C_{1}+{ }^{5} C_{2} \times{ }^{5} C_{1} \times{ }^{5} C_{3}+{ }^{5} C_{2} \times{ }^{5} C_{2} $
$ \times{ }^{5} C_{2}+{ }^{5} C_{2} \times{ }^{5} C_{3} \times{ }^{5} C_{1}+{ }^{5} C_{3} \times{ }^{5} C_{1} \times{ }^{5} C_{2}+{ }^{5} C_{3} $
$ \times{ }^{5} C_{2} \times{ }^{5} C_{1}+{ }^{5} C_{4} \times{ }^{5} C_{1} \times{ }^{5} C_{1} $
$= 5 \times 5 \times 5+5 \times 10 \times 10+5 \times 10 \times 10+5 \times 5 \times 5 $
$+10 \times 5 \times 10+10 \times 10 \times 10+10 \times 10 \times 5+10 \times 5$
$ \times 10+10 \times 10 \times 5+5 \times 5 \times 5 $
$= 125+500+500+125+500+1000+500+500$
$+500+125 $
$= 4375$
| Indians | Americans | Australians |
|---|---|---|
| 1 | 1 | 4 |
| 1 | 2 | 3 |
| 1 | 3 | 2 |
| 1 | 4 | 1 |
| 2 | 1 | 3 |
| 2 | 2 | 2 |
| 2 | 3 | 1 |
| 3 | 1 | 2 |
| 3 | 2 | 1 |
| 4 | 1 | 1 |