An office has 6 employees; there are 5 female employees and 1 male emp

\(C^2_5*C^1_1=10\): there is only ONE male employee, so # of ways to choose 1 out of 1 is \(C^1_1=1\), # of ways to choose 2 other members from 5 women is \(C^2_5\).

Hi chetan2u EMPOWERgmatRichC GMATPrepNow VeritasKarishma Bunuel

Can you point out why did I falter solving this question using Fundamental counting principle?

I have to fill 3 places:
Place 1: I like to use restrictions first. Can be filled by the only male, 1 male to choose from for 1 slot, or 1C1 = 1
Place 2: 5 females to choose from , for second slot , 5C1 ie 5
Place 3: 4 female to choose from , for third slot . 4C1 ie 4

Total no of arrangements = 4*5 = 20

After choosing 1 male, you have got into ARRANGEMENTS, 5*4, while the question is about SELECTION, 5C2.

SO, 5*4 contains 2 different arrangements where a,b is different from b,a. Thus, divide your answer by 2---20/2=10

Hi adkikani,

We're told that an office has 6 employees (5 female employees and 1 male employee). We're asked for the number of ways that a 3-person committee can be created if the committee must include the male employee. This question can be treated as either a Permutation OR a Combination (depending on how you want to do the math). In addition, since the number of options is relatively small, you could 'brute force' the solution by writing out all of the possibilities.

Since we're forming a 'group', the order of the 3 people does NOT matter. Thus, using the Combination Formula would make sense. The one male employee would take that one 'spot', leaving the 5 female employees for the remaining 2 spots.... 5c2 = (5!)/(2!)(3!) = (5)(4)/(2)(1) = 10 possible groups.

After placing the one male employee, you could do the rest of the calculation as a Permutation. However, you have to realize that a Permutation will create "duplicate" groups - and we will have to remove those duplicates.

5 women for the 1st spot
4 women for the 2nd spot
(1)(5)(4) = 20.... however, each group in there has been counted twice (since the pair A,B is the same as the pair B,A). Thus, we have to divide this total by 2... 20/2 = 10 groups.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

We are told that the total employees in an office is 6 of which 5 are females and 1 is male.

We are asked to make a committee of 3 and it should include a male.

Now from the total employees, there is only 1 male and as in committee selection the order does not matter as there is no designation here, we can select any one spot for the male. So the number of ways to choose for a male is 1C1 = 1.

For the remaining two spots, we got to choose 2 from 5 females, the number of ways females can be chosen is 5C2=10.
Total number of ways committee can be made is 10*1=10.

In order to ensure that the 3-person committee includes the male employee, we'll start by placing the male employee on the committee.
Our task now is to choose 2 more people (from the 5 female employees) to join the male on the committee.

Since the order in which we select the two people does not matter, we can use combinations.
We can select 2 employees from 5 employees in 5C2 ways (= 10 ways)

Answer: A

Cheers,
Brent

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