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(A) To form a committee of $5$ members,we consider the two given cases:Case $1$: The committee has all $4$ professors and $1$ research associate.The n

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$12$

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(A) To form a committee of $5$ members,we consider the two given cases:Case $1$: The committee has all $4$ professors and $1$ research associate.The number of ways to choose $4$ professors out of $4$ is ${}^{4}C_{4} = 1$.The number of ways to choose $1$ research associate out of $6$ is ${}^{6}C_{1} = 6$.Total ways for Case $1 = 1 	imes 6 = 6$.Case $2$: The committee has all $3$ trainees and $2$ professors.The number of ways to choose $3$ trainees out of $3$ is ${}^{3}C_{3} = 1$.The number of ways to choose $2$ professors out of $4$ is ${}^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$.Total ways for Case $2 = 1 	imes 6 = 6$.Since these cases are mutually exclusive,the total number of ways is the sum of the ways from both cases:Total ways $= 6 + 6 = 12$.

$A$ committee of $5$ members is to be formed out of $3$ trainees,$4$ professors,and $6$ research associates. In how many different ways can this be done if the committee should have all $4$ professors and $1$ research associate $OR$ all $3$ trainees and $2$ professors?

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