Question

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

• A. 3375
• B. 4375
• C. 3875
• D. 4250

Answer 1

Answer: (B)

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
$={}^5{C}_1\times {}^5{C}_1\times {}^5{C}_4+{}^5{C}_1\times {}^5{C}_2\times {}^5{C}_3+{}^5{C}_1\times {}^5{C}_3$
$\times {}^5{C}_2$
$+{}^5{C}_1\times {}^5{C}_4\times {}^5{C}_1+{}^5{C}_2\times {}^5{C}_1\times {}^5{C}_3+{}^5{C}_2\times {}^5{C}_2$
$\times {}^5{C}_2+{}^5{C}_2\times {}^5{C}_3\times {}^5{C}_1+{}^5{C}_3\times {}^5{C}_1\times {}^5{C}_2+{}^5{C}_3$
$\times {}^5{C}_2\times {}^5{C}_1+{}^5{C}_4\times {}^5{C}_1\times {}^5{C}_1$
$=5\times 5\times 5+5\times 10\times 10+5\times 10\times 10+5\times 5\times 5$
$+10\times 5\times 10+10\times 10\times 10+10\times 10\times 5+10\times 5$
$\times 10+10\times 10\times 5+5\times 5\times 5$
$=125+500+500+125+500+1000+500+500$
$+500+125$
$=4375$