Question

Four sophomores, five juniors and six seniors have volunteered to serve on a committee consisting of five people. In how many ways can the membership of the committee be selected under the following restrictions respectively?

(a) Any of the volunteers can serve on the committee.

(b) Only seniors can serve on the committee.

(c) At least one senior must serve on the committee.

(d) At least two juniors and at least two seniors must serve on the committee.

(e) At least one person from each class must serve on the committee, but no more than two members of the same class can serve on the committee.

Answer 1

(a) Since any of the volunteers can serve on the committee, hence the number of ways of selecting the committee membership is the number of ways of selecting any 5 students from the collection of all students.
The total number of students = 4+5+6 =15.

Thus, the required number of ways is ¹⁵C₅.


(b) Since only seniors can serve on the committee, hence the number of ways of selecting the committee membership
is the number of ways of selecting 5 students from the collection of 6 senior students.

Thus, the required number of ways is ⁶C₅ = 6.


(c) Let us calculate the number of ways of selecting the committee membership such that no seniors serve as the committee
members. This is equal to the number of ways of selecting 5 students out of the collection of 4 sophomores and 5 juniors.
Thus, the number of ways of selecting the committee such that no seniors are in the committee is ⁹C₅ .

Thus, the required number of ways of selecting the committee so that at least one senior serves on the
committee is ¹⁵C₅ - ⁹C₅ .


(d) If at least two juniors and at least two seniors serve on the committee of 5 members, then the following mutually exclusive cases hold:
CASE 1: 2 Juniors(out of 5 Juniors), 2 Seniors(out of 6 Seniors) and 1 Sophomore(out of 4 Sophomores) are in the Committee.
CASE 2: 3 Juniors(out of 5 Juniors) and 2 Seniors(out of 6 Seniors) are in the Committee.
CASE 3: 2 Juniors(out of 5 Juniors) and 3 Seniors(out of 6 Seniors) are in the Committee.

The number of ways of selecting the committee in CASE 1 is ⁵C₂ * ⁶C₂ * ⁴C₁ = 600 (Multiplication(AND) Rule).
The number of ways of selecting the committee in CASE 2 is ⁵C₃ * ⁶C₂ = 150 (Multiplication(AND) Rule).
The number of ways of selecting the committee in CASE 3 is ⁵C₂ * ⁶C₃ = 200 (Multiplication(AND) Rule).

Thus, the required number of ways is 600+150+200 = 950 (Addition(OR) Rule).


(e) If at least one person from each class must serve on the committee, but no more than two members of the same class can serve on the committee, then the following mutually exclusive cases arise:

CASE 1: 1 Sophomore, 2 Juniors and 2 Seniors are in the Committee.
CASE 2: 2 Sophomores, 2 Juniors and 1 Senior are in the Committee.
CASE 3: 2 Sophomores, 1 Junior and 2 Seniors are in the Committee.

  The number of ways of selecting the committee in CASE 1 is ⁴C₁ * ⁵C₂ * ⁶C₂ = 600 (Multiplication(AND) Rule).
  The number of ways of selecting the committee in CASE 2 is ⁴C₂ * ⁵C₂ * ⁶C₁ = 360 (Multiplication(AND) Rule).
  The number of ways of selecting the committee in CASE 3 is ⁴C₂ * ⁵C₁ * ⁶C₂ = 450 (Multiplication(AND) Rule).

Thus, the required number of ways is 600+360+450 = 1410 (Addition(OR) Rule).