Question

A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at least five women have to be included in a committee? In how many of these committees
(a) the women are in majority?
(b) the men are in majority"?

Answer 1

Hint: Proceed the solution of this question by making pairs of women and men considering all the possibilities such that, there should be 12 members in the committee on keeping at least 5 women. Then using the commutation formula, we can select them according to formed pairs.

Complete step-by-step solution -
Given that there are 9 women and 8 men. A committee of 12 is to be formed including at least 5 women. This can be done in the following ways.
= (5 women and 7 men) + (6 women and 6 men) , + (7 women and 5 men) + (8 women and 4 men) + (9 women and 3 men)
This can also have written in short form-
As 5W and 7M
 or 6W and 6M
Or 7W and 5M
Or 8W and 4M
Or 9W and 3M
So, Total number of ways of forming committee =\[ \Rightarrow {}^9{{\text{C}}_5} \times {}^8{{\text{C}}_7} + {}^9{{\text{C}}_6} \times {}^8{{\text{C}}_6} + {}^9{{\text{C}}_7} \times {}^8{{\text{C}}_5} + {}^9{{\text{C}}_8} \times {}^8{{\text{C}}_4} + {}^9{{\text{C}}_9} \times {}^8{{\text{C}}_3}\]
We know that
$ \Rightarrow {\text{ }}{}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{\text{n}}!}}{{{\text{r}}!\left( {{\text{n}} - {\text{r}}} \right)!}}$
On further solving above expression
$ \Rightarrow \dfrac{{9!}}{{5!\left( {9 - 5} \right)!}} \times \dfrac{{8!}}{{7!\left( {8 - 7} \right)!}} + \dfrac{{9!}}{{6!\left( {9 - 6} \right)!}} \times \dfrac{{8!}}{{6!\left( {8 - 6} \right)!}} + \dfrac{{9!}}{{7!\left( {9 - 7} \right)!}} \times \dfrac{{8!}}{{5!\left( {8 - 5} \right)!}}{\text{ + }}$
$\dfrac{{9!}}{{8!\left( {9 - 8} \right)!}} \times \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} + \dfrac{{9!}}{{9!\left( {9 - 9} \right)!}} \times \dfrac{{8!}}{{3!\left( {8 - 3} \right)!}}$
On cancelling common terms from Numerator and denominator
$ \Rightarrow \dfrac{{9!}}{{5!\left( 4 \right)!}} \times \dfrac{{8!}}{{7!\left( 1 \right)!}} + \dfrac{{9!}}{{6!\left( 3 \right)!}} \times \dfrac{{8!}}{{6!\left( 2 \right)!}} + \dfrac{{9!}}{{7!\left( 2 \right)!}} \times \dfrac{{8!}}{{5!\left( 3 \right)!}}{\text{ + }}$$\dfrac{{9!}}{{8!\left( 1 \right)!}} \times \dfrac{{8!}}{{4!\left( 4 \right)!}} + \dfrac{{9!}}{{9!\left( 0 \right)!}} \times \dfrac{{8!}}{{3!\left( 5 \right)!}}$
$ \Rightarrow \dfrac{{9 \times 8 \times 7 \times 6}}{{\left( {4 \times 3 \times 2 \times 1} \right)}} \times \dfrac{8}{{\left( 1 \right)}} + \dfrac{{9 \times 8 \times 7}}{{\left( {3 \times 2 \times 1} \right)}} \times \dfrac{{8 \times 7}}{{\left( 2 \right)}} + \dfrac{{9 \times 8}}{{\left( 2 \right)}} \times \dfrac{{8 \times 7 \times 6}}{{\left( {3 \times 2 \times 1} \right)}}{\text{ }}$
$ + \dfrac{9}{{\left( 1 \right)}} \times \dfrac{{8 \times 7 \times 6 \times 5}}{{\left( {4 \times 3 \times 2 \times 1} \right)}} + \dfrac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}$
on doing multiplication

⇒ 126 x 8 + 84 x 28 + 36 x 56 + 9 x 70 + 56 =
⇒1008+2352+2016+630+56 = 6062 ways.
(a) The women are in majority in when we select (7 women and 5 men) + (8 women and 4 men) + (9 women and 3 men)


\[ \Rightarrow {}^9{{\text{C}}_7} \times {}^8{{\text{C}}_5} + {}^9{{\text{C}}_8} \times {}^8{{\text{C}}_4} + {}^9{{\text{C}}_9} \times {}^8{{\text{C}}_3}\]
⇒2016+630+56 =2702 ways


 (b) The men are in majority in when we select (5 women and 7 men)

\[ \Rightarrow {}^9{{\text{C}}_5} \times {}^8{{\text{C}}_7}\]
⇒ 1008 ways.

Note: In this particular question, we should know that, in mathematics a permutation is the choice of r things from a set of n things without replacement and where the order matters.
$ \Rightarrow {}^{\text{n}}{{\text{P}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}}$
while a combination is the choice of r things from a set of n things without replacement and where order doesn't matter.
${\text{ }} \Rightarrow {}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{{}^{\text{n}}{{\text{C}}_{\text{r}}}}}{{{\text{n!}}}} = \dfrac{{{\text{n}}!}}{{{\text{r}}!\left( {{\text{n}} - {\text{r}}} \right)!}}$.
  So in the above question, selection of women and men doesn’t matter so we select them by combination.