Short Question 5 – 8 – user's Blog!

Question 4:
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

(b)

\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 5:
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 6:
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} +^{10} C_{9} +^{10} C_{10} \\ = 56 \end{array}