Question

A group consists of 10 persons of
whom 6 are men and 4 women. In
how many ways can a committee
of 5 persons be selected so that
the men are in majority?​

Answer 1

Given:

A group consists of 10 persons of  whom 6 are men and 4 women.

To Find:

No. of ways can a committee  of 5 persons be selected so that  the men are in majority?​

Step-by-step explanation:

It is given that in a group consists of 10 persons of  whom 6 are men and 4 women. we have to make a committee  of 5 persons so that  the men are in majority.

We can select it in 2 ways:

1) 4 men and 1 women

2)3 men and 2 women

                   $^4C_1\times ^6C_4+^4C_2\times ^6C_3\\\\\Rightarrow\frac{4!}{3!} \times\frac{6!}{2!4!} +\frac{4!}{2!2!} \times\frac{6!}{3!3!} \\\\\Rightarrow 4\times15+6\times40\\\Rightarrow300$

Hence, total no. ways are 300

Answer 2

There are 186 ways in which a committee of 5 persons be selected so that the men are in majority

Given:

• A group consists of 10 persons 6  men and 4 women.
• A committee of 5 persons be selected

To Find:

• Number of ways so that men are in majority

Solution:

Step 1:

A committee of 5 persons can have Men and women

(M , W)  = (5 , 0) , (4 , 1) , (3 , 2) , (2 , 3) , ( 1 , 4)  

{ note (0,5) not possible as there are only 4 women)

Step 2:

Identify the cases where Men > Woman

(M , W)  = (5 , 0) , (4 , 1) , (3 , 2)

Step 3:

Number of Ways of selection

⁶C₅.⁴C₀ + ⁶C₄.⁴C₁ + ⁶C₃ .⁴C₂

= 6(1) + 15(4) + (20)(6)

= 6 + 60 + 120

= 186

Hence, there are 186 ways in which a committee of 5 persons be selected so that the men are in majority.