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Re: Three men (out of 7) and 3 women (out of 6)
[#permalink]
27 Jun 2017, 11:54
1
Vrushali wrote:
Three men (out of 7) and 3 women (out of 6) will be chosen to serve on a committee. In how many ways can the committee be formed?
Take the task of seating the 6 people and break it into stages.
Stage 1: Select 3 men to be on the committee Since the order in which we select the men does not matter, we can use combinations. We can select 3 men from 7 men in 7C3 ways (35 ways) So, we can complete stage 1 in 35 ways
Stage 2: Select 3 women to be on the committee Since the order in which we select the women does not matter, we can use combinations. We can select 3 women from 6 women in 6C3 ways (20 ways) So, we can complete stage 2 in 20 ways
By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 6-person committee) in (35)(20) ways (= 700 ways)
Re: Three men (out of 7) and 3 women (out of 6)
[#permalink]
06 Sep 2022, 20:17
Three men (out of 7) and 3 women (out of 6) will be chosen to serve on a committee. In how many ways can the committee be formed?
Choosing 3 men out of 7, we require a combination of 3 out of 7 which is \( = \frac{7!}{(3! * (7 - 3)!)} = 35\)
Choosing 3 women out of 6, we require combination of 3 out of 6 which is \( = \frac{6!}{(3! * (6 - 3)!)} = 20\)
Now, we apply the fundamental counting principle which says that if there are n ways of doing something, and m ways of doing another thing after that, then there are \(n*m\) ways to perform both of these actions.
Therefore, there are \(35 * 20 = 700\) ways of forming the committee.
gmatclubot
Re: Three men (out of 7) and 3 women (out of 6) [#permalink]