Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees are women, how many of the possible groups of 4 temporary employees consist of 3 women and 1 man?

A)22
B)35
C)56
D)70
E)105

Please help & explain. Greatly appreciated.

women: 5!/3!(5-3)!=5!/3!2!=10
men: 7!/1!(7-1)!=7!/1!6!=7
the possible groups of 4 temporary employees=w*m=10*7=70
thus, answer D

can you confirm the OA? thanks

Great job! Yes, the OA is D. 70.

For some reason I also felt the need to do 12!/(4!)(8!). Perhaps I'm confusing that with probability.

Can you explain how you setup the factorial for the women and men groups? This is my understanding:

5! / 3! (5-3)!

5! represents 5 women out of 12 employees (W W W W W M M M M M M M )
3! represents 3 women out of 8 temp employees (W W W M M M M M)
(5-3)! .....do you have to include this because there is overlap or double-count in the 5! and 3!?

Please explain!

Thanks.

3 things:

(1) yes, that fraction could arise in the context of probability. specifically, if you want the probability that a RANDOMLY selected group of four people DOES consist of 3w 1m, then you'd take the answer to this problem (70) and divide it by the total number of ways of selecting any four of the 12 people, which is the cited fraction (12!)/(4!8!).

(2) in the context of this problem - which is pure combinatorics, not probability - there's no need to calculate the number of ways of selecting four people in general, because your choice is restricted to 3w 1m from the beginning.

(3) whatever you do, DON'T think of the general number of ways of selecting 4 people (12!/4!8!) as something that only arises in the context of probability. instead, just make sure that you can articulate clearly what that fraction represents: the total number of ways of selecting any four people out of the whole group.

How can I know that the problem is combination problem or a probability problem and whats the difference between them.I m really confuses .please help.

A probability question would use words such as "What is the chance that...?" or "What is the probability that..."

A combinatorics question would use words such as "How many ways can ____ happen...?" or "How many of ____ can be created?"

Confusion/solution overlap arises from the fact that you can do many probability questions by answering the "how many?" question first.

For example, if this question had been about probability, the probability of selecting a group of 4 temporary employees that consists of 3 women and 1 man is:

Prob of 3W1M = # of groups with 3W1M/# of groups total, ignoring gender.

Ron's point is just that if you have to answer "how many?" don't do more math than is necessary. Since this is not about probability, we don't care about # of groups total, we only care about the # of groups with the gender restricted to 3W1M.

_________________
Emily Sledge
Instructor
ManhattanGMAT

we seem to ignore the effect of the statement in bold; can someone please explain in what circumstance we take the expression in bold into consideration.
Thanks

The statement in bold does not have any other implication except to indicate that a group of 4 has to be formed out of available 12 persons.

12 C 4: Total No. of ways in which this can be done.

Then a specific condition is given regarding how that group of 4 persons has to be formed. (3 Women + 1 Men). It is also indicated that Out of 12 employees 5 are women. Thus 7 employees are males.

Favorable ways in which this can be done: (5C3 x 7C1)

Probability can be obtained.

we are actually not ignoring the boldface statement; note the following text in the problem:
how many of the possible groups of 4 temporary employees consist of...

the words i've marked in purple refer to the same four people who are described in the boldface statement. there is no need to mention again that these people are going to be hired as permanent employees, since there are no other groups of four people considered in the problem.

_________________
Being well-dressed gives a feeling of inward tranquillity [that] religion is powerless to bestow.
C.F. Forbes

Permutations and Combinations is not my strong suit. So here's a laconic solution for anyone who might get some help from it:

7c1 x 5c3. Solve this and you get 70 as the answer.

Thanks everyone!

_________________
Jamie Nelson
ManhattanGMAT Instructor

I have a problem that I encounter often while solving probability and combinatorics problems -

In this question - I did 5C3 * 7C1 * 4!/!3 = 280 thinking that we also need to account for the number of ways you can pick these people let's say

WWWM or WMWW or WWMW or MWWW

The reason for this confusion is that I have seen a lot in probability questions in which we need to consider the number of possible ways an event can happen and multiply that number to get the final answer such as -



For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads 'at least' 4 times?

The answer to this question is
5[((0.6)^4)(0.4)] + (0.6)^5

P(atleast 4 heads) = P(4 heads) + P( 5 heads)

Now while solving this question when we calculate P(atleast 4 heads)-

P(4heads) = (0.6)^4)(0.4) * 4 - Here we multiply by four because there are four ways to get four heads and one tail

Thinking on the similar lines, I utilized the same logic for the question in discussion and multiplied by the number of ways in which we can pick the required group.

I am not clear with respect to when to multiply with the number of ways a group can be formed/ number of ways we can get four heads etc etc VERSUS when not to multiply by the number of ways---- in order to get the final answer.

the best general advice i can give you here is that when you are making successive choices (as you did with your coin example), you need to divide off by the number of permutations you can get, because you will get the same final outcome multiple times (eg. HHHHT HHHTH etc.). when using a combinatorics approach (selecting all items at once rather than sequentially), you will not have this problem and thus will not need to worry about overcounting..

_________________
Tim Sanders
Manhattan GMAT Instructor

Thanks,Tim.

One more question -

In the given question, if this question was about calculating the probability of picking 3 Women and 1 Men then this should be

Approach 1: P(picking three women)*P(picking one man) = 3/12*1/11

OR

Should it be -

Approach 2: [P(three women) * P(one man)] + P[(one man)* P(three women)]

Should it matter while calculating the Prob. whether women gets picked up first or man gets picked up first? When we use conditional probability...shouldn't that matter who gets picked up first since the conditional probability of the first event will affect the probability of the other event that comes later. Do we need to account for both the possibilities as considered in approach 2?