# Question 9 Exercise 6.3

Solutions of Question 9 of Exercise 6.3 of Unit 06: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

An $8$-persons committee is to be formed from a group of $6$ women and $7$ men. In how many ways can the committee be chosen if (i) committee must contain four men and four women?

Total men are $7$ and total women are $6.$

Therefore, Total number of persons $=7+6=13$

Committee consist of 8 persons four men and four women

If committee contain exactly four men and four women.

Total number of different ways that four men to be selected are: ${ }^7 C_4$.

Total number of different ways that four women to be selected are ${ }^6 C_4$.

By fundamental principle of counting the total number of different committees that will exactly contain four men and four women are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4)}\\\ &= 525\end{align}

An $8$-persons committee is to be formed from a group of $6$ women and $7$ men. In how many ways can the committee be chosen if there must be at least two men?

Total men are $7$ and total women are $6$.

Therefore, Total number of persons $=7+6=13$

There must be at least two men The number of committees that will contain at least two men may contain $3,4,5,6$ or 7 men.

If committee contains two men and then it will obviously contain $6$ women.

so in this case the total number of committees are: \begin{align}{ }^7 C_2 \cdot{ }^6 C_6&=\dfrac{7 !}{(7-2) ! 2 !} \cdot \dfrac{6 !}{(6-6) ! 6 !}\\ &=21\end{align}

If committee contains $3$ men then it will contain $5$ women,

in this case the total number of committees are: \begin{align}{ }^7 C_3 \cdot{ }^6 C_5&=\dfrac{7 !}{(7-3) ! 3 !} \cdot \dfrac{6 !}{(6-5) ! 5 !}\\\ &=210\end{align} If committee contain $4$ men then it will contain $4$ women,

in this case total number of committees are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4) ! 4 !}\\\ &=525\end{align} If committee contains $5$ men then it will contain $3$ women, in this case total number of committees are: \begin{align}{ }^7 \mathrm{C}_5 \cdot{ }^6 \mathrm{C}_3&=\dfrac{7 !}{(7-5) ! 5 !} \dfrac{6 !}{(6-3) ! 3 !}\\ &=420 \end{align} If committee contains $6$ men then it will contain $2$ women,

in this case total number of committees are: \begin{align}{ }^7 C_6 \cdot{ }^6 C_2&=\dfrac{7 !}{(7-6) ! 6 !} \cdot \dfrac{6 !}{(6-2) ! 2 !}\\ &=105\end{align} If committee contain $7$ men then it will contain $1$ women,

in this case total number of committees are: \begin{align}{ }^7 C_7 \cdot{ }^6 C_1&=\dfrac{7 !}{(7-7) ! 7 !} \dfrac{6 !}{(6-1) ! 1 !}\\ &=6\end{align} Thus the total number of committees that will contain at least $2$ men are: $$21+210+525+420+105+6=1,287$$

An $8$-persons committee is to be formed from a group of $6$ women and $7$ men. In how many ways can the committee be chosen if there must be at least two women?

Total men are $7$ and total women are $6.$

Therefore, Total number of persons $=7+6=13$

There must be at least two women.

The number of committees that will contain at least two women may contain $3,4,5$, or $6$ women.

If committee contain two women and then obviously it will contain $6$ men.

So, in this case the total number of commitlees are: \begin{align}{ }^6 \mathrm{C}_2 \cdot{ }^7 \mathrm{C}_6& =\dfrac{6 !}{(6-2) ! 2 !} \cdot \dfrac{7 !}{(7-6) ! 6 !}\\ &=105 \end{align} If committee contain $3$ women then obviously it will contain $5$ men,

so in this case the total number of committees are: \begin{align}{ }^6 \mathrm{C}_3 .{ }^7 \mathrm{C}_5& =\dfrac{6 !}{(6-3) ! 3 !} \cdot \dfrac{7 !}{(7-5) ! 5 !}\\ &=420 \end{align} If committee contain $4$ women then obviously it will contain $4$ men.

in this case the total number of committees are: $$=525$$ If committee contain $5$ women then it will contain $3$ men.

In this case the total number of committees are: \begin{align} { }^6 C_5 \cdot{ }^7 C_3&=\dfrac{6 !}{(6-5) ! 5 !} \cdot \dfrac{7 !}{(7-3) ! 3 !} \\ & =6.35=210 \end{align} If committee contain $6$ women and then it will contain $2$ men,

so in this case the total number of committees are:

\begin{align}{ }^6 C_6 \cdot{ }^7 C_2&=\dfrac{6 !}{(6-6) ! 6 !} \cdot \dfrac{7 !}{(7-2) ! 2 !}\\ &=1.21=21\end{align} Hence the total number of committees containing at least two women are: $$105+420+525+210+21=1281$$

An $8$-persons committee is to be formed from a group of $6$ women and $7$ men. In how many ways can the committee be chosen if there must be more women than men?

Total men are $7$ and total women are $6$

Therefore, Total number of persons $=7+6=13$

There must be more women than men.

If committee contains $5$ women then it will contain $3$ men,

in this case total number of committees are: \begin{align}{ }^7 C_5 \cdot{ }^6 C_3&=\dfrac{7 !}{(7-5) ! 5 !} \dfrac{6 !}{(6-3) ! 3 !}\\ &=420 \end{align} If committee contains $6$ women then it will contain $2$ men,

in this case total number of committees are: \begin{align}{ }^7 C_6 \cdot{ }^6 C_2&=\dfrac{7 !}{(7-6) ! 6 !} \cdot \dfrac{6 !}{(6-2) ! 2 !}\\ &=105 \end{align} If committee contain $7$ women then it will contain $1$ men,

in this case total number of committees are: \begin{align}{ }^7 C_7 \cdot{ }^6 C_1&=\dfrac{7 !}{(7-7) ! 7 !} \dfrac{6 !}{(6-1) ! 1 !}\\ &=6\end{align} Thus the total number of committees that will contain more men than women are: $$420+105+6=531$$