Question

Of the 12 members of a high school drama club, 8 seniors. The club plans to establish an 8-member committee. If exactly 6 members of the committee must be seniors, calculate the total number of possible committees.
[a] 21
[b] 34
[c] 168
[d] 336
[e] 495

Answer 1

Hint: First, we select six members out from the eight seniors and then from the remaining members, i.e. two seniors and four juniors, we select two juniors and hence form a committee of 8 members. The committee will have exactly six members as seniors and exactly two members as juniors.
Use the fundamental principle of counting. Use the fact that the number of ways of selecting r items out from n given items is given by $^{n}{{C}_{r}}$. Use $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Hence find the total number of ways of forming an eight-member committee with the given restrictions.

Complete step-by-step answer:
While forming a committee of 8 members in which we have exactly six seniors, we need to select six seniors and two juniors out of the given lot of 8 seniors and 4 juniors.
Selection of six seniors is equivalent to selection of 6 items out of given 8 items. We know that selection of r items out of given n items is given by $^{n}{{C}_{r}}$. Hence the number of ways of selecting six seniors is $^{8}{{C}_{6}}$.
Selection of two juniors is equivalent to selection of 2 items from a given lot of 4 items which can be done in $^{4}{{C}_{2}}$ ways.
Hence the total number of ways of forming a committee of 8 members in which there are exactly 6 senior member is $^{8}{{C}_{6}}{{\times }^{4}}{{C}_{2}}=\dfrac{8!}{\left( 8-6 \right)!6!}\times \dfrac{4!}{\left( 4-2 \right)!2!}=\dfrac{8\times 7\times 6!}{2!6!}\times \dfrac{4\times 3\times 2!}{2!2!}=\dfrac{56}{2}\times \dfrac{12}{2}=168$.
Hence the total number of possible committees is 168.
Hence option [c] is correct.

Note: [1] Students often get confused when to add the individual number of ways and when to multiply. It can be remembered as follows:
[i] If we need to perform both the tasks then we multiply their individual number of ways.
[ii] If we need to perform either of the tasks then we add their individual number of ways.
In the above problem, we needed to perform both the tasks, i.e. selection of 6 seniors and selection of 2 juniors.
Hence in finding the total number of ways we multiplied their individual number of ways.