4.4.2 Permutation Short Questions (Question 5 – 8)

Question 5:
A committee that consists of 6 members is to be selected from 5 teachers and 4 students. Find the number of different committees that can be formed if
(a) there is no restriction,
(b) the number of teachers must exceed the number of students.

Solution:
(a)
Total number of committees = 5 + 4 = 9
6 members to be selected from 9 committees with no restriction
${}^{\mathrm{=\; 9}}{C}_6=84$

(b)
$\begin{array}{l}\text{If the number of teachers must exceed the}\\ \text{number of students, the combination}\\ \text{= 4 teachers 2 students + 5 teachers 1 student}\\ ={}^5{C}_4\times {}^4{C}_2+{}^5{C}_5\times {}^4{C}_1\\ =30+4\\ =34\end{array}$

Question 6:
Six members of a committee of a school are to be selected from 6 male teachers, 4 female teachers and a male principal. Find the number of different committees that can be formed if
(a) the principal is the chairman of the committee,
(b) there are exactly 2 females in the committee,
(c) there are not more than 4 males in the committee.

Solution:
(a)
If the principal is the chairman of the committee, the remaining number of committee is 5 members.
Hence, the number of different committees that can be formed from the remaining 6 male teachers and 4 female teachers
$\begin{array}{l}={}^{10}{C}_5\\ =252\end{array}$

Question 7:
A school prefect committee that consists of 6 persons is to be chosen from 6 Malays, 5 Chinese and 4 Indians. Calculate the number of different committees that can be formed if the number of Malays, Chinese and Indians must be equal.

Solution:
Number of different committees that can be formed for 2 Malays, 2 Chinese and 2 Indians
$\begin{array}{l}={}^6{C}_2\times {}^5{C}_2\times {}^4{C}_2\\ =900\end{array}$

Question 8:
There are 10 different flavour candies in a plastic bag.
Find
(a) the number of ways 3 candies can be chosen from the plastic bag.
(b) the number of ways at least 8 candies can be chosen from the plastic bag.

Solution:
(a)
Number of ways choosing 3 candies out of 10 candies
$\begin{array}{l}={}^{10}{C}_3\\ =120\end{array}$

(b)
Number of ways choosing 8 candies =   $={}^{10}{C}_8$
Number of ways choosing 9 candies = ${}^{10}{C}_9$
Number of ways choosing 10 candies = ${}^{10}{C}_{10}$

Hence, number of ways of choosing at least 8 candies
$\begin{array}{l}={}^{10}{C}_8{\text{+}}^{10}{C}_9+{}^{10}{C}_{10}\\ =56\end{array}$