Question

From a group of 5 women and 4 men, 3 persons are to be selected to form a committee so that at least 2 women are there are on the committee. in how many ways can it be done?

Answer 1

5c2×4c1
(5×4÷2)×4=40 +5c3×4c0=(5×4×3)÷(3×2)=10
so its 50

i suppose

Answer 2

Answer:

From a group of 5 women and 4 men, if 3 persons are to be selected to form a committee so that at least 2 women are there are on the committee then we can do it in 50 ways.

Step-by-step explanation:

• The group has 5 women and 4 men .

• The number of ways to choose a committee with 2 women and 1 men is $5C_{2}$ × $4C_{1}$ = $\frac{5!}{3! 2!}$ × $\frac{4!}{3! 1!}$ = (5×2) × (4) = 40.

• The number of ways to choose a committee with 3 women and no men is  $5C_{3}$ × $4C_{0}$ = $\frac{5!}{3! 2!}$ × 1 = (5 × 2) × 1 = 10.

• So, the total number of way = 40 + 10 = 50.
• Hence we can conclude that from a group of 5 women and 4 men, if 3 persons are to be selected to form a committee so that at least 2 women are there are on the committee, we can do it in 50 ways.

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