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From a group of 6 men & 5 women, 4 persons are to be selected to form a committee such that at least 2 men are in committee. In how many ways it can be done?  

(A) 150        (B) 250    (C) 265    (D) 115

Ans:C)265

To find the number of ways to select 4 persons from a group of 6 men and 5 women to form a committee such that at least 2 men are in the committee, we can use combinations.

We need to consider the following cases:

Selecting 2 men and 2 women
Selecting 3 men and 1 woman
Selecting 4 men

Case 1: Selecting 2 men and 2 women Number of ways to select 2 men from 6 men = 6C2 Number of ways to select 2 women from 5 women = 5C2 Total ways for this case = 6C2 * 5C2

Case 2: Selecting 3 men and 1 woman Number of ways to select 3 men from 6 men = 6C3 Number of ways to select 1 woman from 5 women = 5C1 Total ways for this case = 6C3 * 5C1

Case 3: Selecting 4 men Number of ways to select 4 men from 6 men = 6C4

Total number of ways to form the committee = Total ways for Case 1 + Total ways for Case 2 + Total ways for Case 3 Total ways = (6C2 * 5C2) + (6C3 * 5C1) + 6C4

Calculating the values: 6C2 = 15 5C2 = 10 6C3 = 20 5C1 = 5 6C4 = 15

Substitute these values into the formula: Total ways = (15 * 10) + (20 * 5) + 15 Total ways = 150 + 100 + 15 Total ways = 265

Therefore, there are 265 ways to select 4 persons from the group of 6 men and 5 women to form a committee such that at least 2 men are in the committee.

Comments

Abhilash Thampy said…
The probability of All the members are mens=6c4=6*5*4*3/(1*2*3*4)=360/24=15
The probability of 3 members are men's and one woman=6c3*5c1=(6*5*4/(1*2*3))*5=20*5=100
Probability of 2 men and 2 women=6c2*5c2=15*10=150
Total probability =15+100+150=265
December 22, 2021 at 5:49 AM
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