Question

A society has 6 men and 5 women as its members. In how many ways can a committee of 5 members be selected with
a majority of women?
a.180
b.181
с.285
d. 281

Answer 1

$\huge \mathcal \blue{answer}$

in a committee of 5 members selected from 6 men and 5 women consisting 3 men and 2 women is 281 ways.

Answer 2

Answer:

Number of ways a committee of 5 members can be selected with

a majority of women  from 6 men and 5 women is 181.

Step-by-step explanation:

A society has 6 men and 5 women as its members.

To select 5 members with a majority of women, possible ways are :

• 4 women and 1 men

• 3 women and 2 men
• 5 women only

For 4 women and 1 men :

The number of ways of selecting r people from a group of n people is $nC_{r}$

Formula for $nC_{r}$ is

⇒$nC_{r}$=$\frac{n!}{r!.(r- n)!}$ .............................................................................(i)

so, no. of ways to select 1 men from 6 men is $\frac{6!}{1! . 5!}$= 6

no. of ways to select 4 women from 5 women is = $\frac{5!}{4! . 1! }$ = 5

Hence , no. of ways to select 4 women and 1 men from 6 men and 5 women is : $6C_{1}$ × $5C_{4}$ = 6 × 5 = 30 .........................................................(ii)

For 3 women and 2 men :

no. of ways to select 3 women and 2 men from 5 women and 6 men is :

$5C_{3}$ × $6C_{2}$ = $\frac{5!}{3! . 2!}$  × $\frac{6!}{2! . 4!}$ = $\frac{5 . 4 }{2 . 1}$ ×$\frac{6 . 5}{2 . 1}$ = 6 × 5 × 5 = 150 ................................(iii)

For 5 women only :

no. of ways to select 5 women from 5 women and 6 men is : 1 ........(iv)

Hence, number of ways a committee of 5 members be selected with

a majority of women  from 6 men and 5 women is :

30 + 150 + 1 = 181 [ By adding (ii), (iii) and (iv)]