From a group of 8 women and 6 men, a committee consisting of 3 men and 3 womenis to be formed. How many different committees are possible if 2 of the men refuse toserve together? How many different committees are possible if 1 man and 1 womanrefuse to serve together?

a) 896 ways

c) 910 ways

Step-by-step explanation:

The question is a combination problem since it has to do with selection. In combination, if r object is selected from a pool of n objects, this can be done in nCr ways.

nCr = n!/(n-r)!r!

If from a group of 8 women and 6 men, a committee consisting of 3 men and 3 women is to be formed, we need to first calculate the total committee formed without any condition.

8C3 × 6C3

= 8!/5!3! × 6!/3!3!

= 8×7×6×5!/5!×6 × 6×5×4×3!/3!×6

= 56×20

= 1120ways

Now if;

a) two of the men refuse to serve together.

Since two men refuse to serve together, we will select 1 man from remaining 4 men since two has already been selected then select 3 from 8 women as shown.

4C1×8C3

= 4×56

= 224ways

Number of ways this can be done will be 1120-224 = 896ways

b) 1 man and 1 woman refuse to serve together

We need to choose 2 men from remaining 5 and 2 women from the rest of the men which is 7 as shown:

= (8-1)C(3-1) × (6-1)C(3-1)

= 7C2 × 5C2

= 7!/5!2! × 5!/3!2!

= 7×6×5!/5!×2 × 5×4×3!/3!×2

= 21×10

= 210ways

The number of ways this can be done will be 1120-210 = 910ways

391.6 is the correct answer

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