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(B) To form a committee of $6$ members with at least one member from each of the $3$ countries (Indians,Americans,Australians),we consider the possibl

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$4375$

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(B) To form a committee of $6$ members with at least one member from each of the $3$ countries (Indians,Americans,Australians),we consider the possible distributions of members $(n_I, n_A, n_{Au})$ such that $n_I + n_A + n_{Au} = 6$ and $n_I, n_A, n_{Au} \ge 1$.The possible partitions of $6$ into $3$ parts are:$1. (4, 1, 1)$ and its permutations: $(4, 1, 1), (1, 4, 1), (1, 1, 4)$. There are $3$ such cases.Number of ways $= 3 	imes \binom{5}{4} 	imes \binom{5}{1} 	imes \binom{5}{1} = 3 	imes 5 	imes 5 	imes 5 = 375$.$2. (3, 2, 1)$ and its permutations: $(3, 2, 1), (3, 1, 2), (2, 3, 1), (2, 1, 3), (1, 3, 2), (1, 2, 3)$. There are $3! = 6$ such cases.Number of ways $= 6 	imes \binom{5}{3} 	imes \binom{5}{2} 	imes \binom{5}{1} = 6 	imes 10 	imes 10 	imes 5 = 3000$.$3. (2, 2, 2)$. There is $1$ such case.Number of ways $= 1 	imes \binom{5}{2} 	imes \binom{5}{2} 	imes \binom{5}{2} = 10 	imes 10 	imes 10 = 1000$.Total ways $= 375 + 3000 + 1000 = 4375$.

The number of ways of forming a committee of $6$ members out of $5$ Indians,$5$ Americans,and $5$ Australians such that there will be at least one member from each country in the committee is:

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