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

^{= 9} C_{6} = 84

$^{= 9} C_{6} = 84$

(b)

\begin{matrix} If the number of teachers must exceed the number of students, the combination = 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{matrix}

$\begin{array}{l} If the number of teachers must exceed the \\ number of students, the combination \\ = 4 teachers 2 students + 5 teachers 1 student \\ =^{5} C_{4} \times^{4} C_{2} +^{5} C_{5} \times^{4} C_{1} \\ = 30 + 4 \\ = 34 \end{array}$

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{matrix} =^{10} C_{5} = 252 \end{matrix}

$\begin{array}{l} =^{10} C_{5} \\ = 252 \end{array}$

(b)

\begin{matrix} Exactly 2 females in the committee =^{4} C_{2} \times^{7} C_{4} = 210 \end{matrix}

$\begin{array}{l} Exactly 2 females in the committee \\ =^{4} C_{2} \times^{7} C_{4} \\ = 210 \end{array}$

(c)

\begin{matrix} There are not more than 4 males in the committee = 4 males 2 females + 3 males 3 females + 2 males 4 females =^{7} C_{4} \times^{4} C_{2} +^{7} C_{3} \times^{4} C_{3} +^{7} C_{2} \times^{4} C_{4} = 210 + 140 + 21 = 371 \end{matrix}

$\begin{array}{l} There are not more than 4 males in the committee \\ = 4 males 2 females + 3 males 3 females + 2 males 4 females \\ =^{7} C_{4} \times^{4} C_{2} +^{7} C_{3} \times^{4} C_{3} +^{7} C_{2} \times^{4} C_{4} \\ = 210 + 140 + 21 \\ = 371 \end{array}$

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{matrix} =^{6} C_{2} \times^{5} C_{2} \times^{4} C_{2} = 900 \end{matrix}

$\begin{array}{l} =^{6} C_{2} \times^{5} C_{2} \times^{4} C_{2} \\ = 900 \end{array}$

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{matrix} =^{10} C_{3} = 120 \end{matrix}

$\begin{array}{l} =^{10} C_{3} \\ = 120 \end{array}$

(b)
Number of ways choosing 8 candies =

=^{10} C_{8}

$=^{10} C_{8}$
Number of ways choosing 9 candies =

^{10} C_{9}

$^{10} C_{9}$
Number of ways choosing 10 candies =

^{10} C_{10}

$^{10} C_{10}$

Hence, number of ways of choosing at least 8 candies

\begin{matrix} =^{10} C_{8} +^{10} C_{9} +^{10} C_{10} = 56 \end{matrix}

$\begin{array}{l} =^{10} C_{8} +^{10} C_{9} +^{10} C_{10} \\ = 56 \end{array}$

4.4.2 Permutation Short Questions (Question 5 – 8)

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