Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 25 Oct 2005
Posts: 12

Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
16 Nov 2005, 02:01
Question Stats:
81% (01:35) correct 19% (02:18) wrong based on 580 sessions
HideShow timer Statistics
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 employees consist of 3 women and one man? A. 22 B. 35 C. 56 D. 70 E. 105
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 64243

Re: GMAT Prep: Probabilities
[#permalink]
Show Tags
27 Dec 2009, 16:54
GMAT TIGER wrote: dk94588 wrote: This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.
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 employees consist of 3 women and one man?
a) 22 b) 35 c) 56 d) 70 e) 105 Select 3 out of 5 = 5c3 = 5 Select 1 out of 7 = 7c1 = 7 Total = 5c3 x 7c1 = 5x7 = 35 Just a little correction 5C3=10 not 5. So, 5C3*7C1=10*7=70. (5C3 # of selections of 3 women out of 5 and 7C1 # of selections of 1 man out of 7) Answer: D
_________________




Manager
Joined: 04 Feb 2007
Posts: 64

Re: GMAT Prep: Probabilities
[#permalink]
Show Tags
21 Feb 2010, 18:51
dk94588 wrote: This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.
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 employees consist of 3 women and one man?
a) 22 b) 35 c) 56 d) 70 e) 105 The wording of this question is pure evil. First it gives you information regarding 4 permanent employees, then it asks you concerning groups of temporary employees. The information of the permanent employees is irrelevant to this problem. Make sure to understand exactly what the question is asking before attacking it.




Director
Joined: 13 Nov 2003
Posts: 592
Location: BULGARIA

Select three women out of five in 5C3=10 ways.Select one man in 7 ways or 10x7=70 different groups of 3 women and 1 man.



Director
Joined: 06 Jun 2004
Posts: 678
Location: CA

Re: PS: employees
[#permalink]
Show Tags
16 Nov 2005, 11:13
themagiccarpet wrote: Of the 12 temporary employees in a certain company, 4 will be hired as permanent emploees. 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
Possible groups = Possible groups of 3 women * possible groups of 1 men
Since order does not matter, its a combination problem.
=[5!/(3!2!)} * [7!/(6!1!)]
=70



SVP
Joined: 29 Aug 2007
Posts: 1733

Re: GMAT Prep: Probabilities
[#permalink]
Show Tags
27 Dec 2009, 12:56
dk94588 wrote: This is a section I have had a lot of trouble with, and I was hoping someone could show me how to do this.
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 employees consist of 3 women and one man?
a) 22 b) 35 c) 56 d) 70 e) 105 Select 3 out of 5 = 5c3 = 5 Select 1 out of 7 = 7c1 = 7 Total = 5c3 x 7c1 = 5x7 = 35



Manager
Joined: 17 Oct 2010
Posts: 69

Re: GMAT Prep: Probabilities
[#permalink]
Show Tags
08 Jun 2012, 00:10
another approach :
one scenario : [WWWM]
now we have 5 ways for the first women , 4 ways for the 2nd women and , 3 ways for the third women.
and 7 ways to select one man
so 5*4*3*7 = 420
now the three women are same so combinations with win them are irrelevant the 3 women can we arranged in 3! ways
hence 420/3!= 420/6= 70



Intern
Joined: 25 Jun 2012
Posts: 7

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
09 Jul 2012, 20:20
You need to decide if the problem at hand asks for permutation (order matters) or combination (order not important) or hybrid of the two.
In your first problem you have two separate slots for males and females. The problem doesn't care about the order of the selected employees so each slot contains a combination and you multiply them: 5C3 * 7C1, to get the total number of possibilities.
In the probability problems of coin or die tosses, you always calculate the probability of a particular outcome having a fixed order = permutation. The reason for that is that the formula for multiplying probabilities is defined only for events in a logical sequence: first even A, then event B whose probability may depend on A, etc. If the problem doesn't care about the order at the end, which is usually the case, it is asking for a the probability of a COMBINATION, so you have to find how many of the permutations merge into the desired combination and sum up their probabilities because they are exclusive (cannot happen at the same time).
Note that in coin or die tosses you always pretend they are tossed in a sequence one after another (otherwise you can't apply the formula for multiplying probabilities) even when they are actually tossed at the same time.



Manager
Joined: 13 May 2010
Posts: 95

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
10 Jul 2012, 01:11
Thanks for replying, tutorphd. What do we do we have to deal with conditional probability?
Let's say you have this question 
A certain junior class has 1000 students and a certain senior class has 800 students. Among these students, there are 60 siblings pairs each consisting of 1 junior and 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?
A, 3/40000, B 1/3600, C. 9/2000, D, 1/60, E, 1/15
In this question
Why won't we add the probability of two cases  (Case I  Junior first then Senior) + (Case II Senior First then Junior)
= 60/1000 * 1/800 + 60/800 * 1/1000
What is wrong with this ?
Answer is A
Wouldn't JS , SJ be two muutually exlusive possibilities, and we should apply the probabiliy rule P(A or B) = P(A) +P(B)  P(A and B)
Given this rule shouldn't we be doing the above proposed solution, however the answer I get is double of the official answer?
I kind of understand the point that you made in your post above, as a request can I please ask you to give me an example of the points that you made in your previous post. I will really appreciate if you can give me some examples to explain your previous explaination. I am doing fine in combinatorics at this point but this concept of multiplying with number of things in the end is messing me up often. Please help.



Intern
Joined: 25 Jun 2012
Posts: 7

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
10 Jul 2012, 04:52
The general formula for multiplying probabilities is P(A and B) = P(A)*P(BA) and uses the conditional probability P(BA). For independent events, the conditional probability P(BA)=P(B) and the formula simplifies to P(A and B) = P(A)*P(B), it is still called "conditional probability formula". The ordering in this formula is arbitrary logical ordering (first event A, then event B) obtained by tracing a probability tree. It has nothing to do with ordering in time with which you are mixing it up.
In the problems with coin or die tosses, you can label each coin, say 'coin A', 'coin B' ... and trace a probability tree by first considering the outcome for coin A, then coin B, coin C... You DON'T add that to another probability tree by first considering the outcome of coin B, then A, then C ... You use a SINGLE probability tree = single arbitrary logical ordering which has nothing to do with which coin was tossed first or second in time. That is why problems in which you toss coins/dice at the same time are equivalent to problems where you toss then in sequence, because the order is arbitrary logical ordering BY LABELING THEM and tracing a SINGLE probability tree, not by ordering them in time.
In the sibling pair problem, the labels are already there 'junior' and 'senior'. You trace the probability tree either as outcomes for junior first and senior second, or the the other way around BUT NOT BOTH. You are getting a double result because you are trying to add the results of two probability trees, not two 'exclusive events'.
In the coin toss problems, you are tracing a single tree with an arbitrary logical ordering you have chosen, say outcome of 'coin A' first, outcome of 'coin B' second etc. The exclusive events/outcomes are separate branches in that same tree, say {A,B}={H, T} or {T,H}, not part of another tree.



Director
Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 964
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

Re: Of the 12 temporary employees in a certain company,
[#permalink]
Show Tags
02 Jan 2013, 20:44
kiyo0610 wrote: 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 Number of ways to select 3 women out of 5 is \(5C3\) and the number of ways to select 1 man out of 7 is \(7C1\). Therefore number of possible groups of 4 temporary employees consist of 3 women and 1 man are \(7C1 * 5C3 = 70\). There are 70 possible groups. +1D
_________________



Manager
Joined: 03 Jan 2015
Posts: 72

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
21 Feb 2016, 10:10
Combinations will do.
5 out of the 12 employees are women, 3 of whom will be chosen and 2 whom will not. Therefore: The women = \(\frac{5!}{3!2!} = 5*2 = 10\) 7 out of the 12 employees are men, 1 of whom will be chosen en 6 whom will not. Therefore: The men = \(\frac{7!}{6!1!} = 7\)
\(7 * 10 = 70\) possible groups consisting of 3 women en 1 man.



GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4879
Location: Canada

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
21 Feb 2016, 13:56
Quote: 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 employees consist of 3 women and one man?
A. 22 B. 35 C. 56 D. 70 E. 105 Take the task of selecting the employees and break it into stages. Stage 1: Select the 3 women The order in which we select the women does not matter, so we can use combinations. We can select 3 women from 5 women in 5C3 ways (= 10 ways) Aside: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: http://www.gmatprepnow.com/module/gmatcounting?id=789Stage 2: Select the 1 man There are 7 men, so we can complete this stage in 7 ways. By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus select the permanent employees) in (10)(7) ways ([spoiler]= 70 ways[/spoiler]) Answer: D Note: the FCP can be used to solve the majority of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmatcounting?id=775Cheers, Brent
_________________
Test confidently with gmatprepnow.com



Manager
Joined: 12 Feb 2011
Posts: 75

Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
22 Feb 2016, 16:22
GMATPrepNow wrote: Take the task of selecting the employees and break it into stages. Stage 1: Select the 3 women The order in which we select the women does not matter, so we can use combinations. We can select 3 women from 5 women in 5C3 ways (= 10 ways) Aside: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: http://www.gmatprepnow.com/module/gmatcounting?id=789Stage 2: Select the 1 man There are 7 men, so we can complete this stage in 7 ways. By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus select the permanent employees) in (10)(7) ways ([spoiler]= 70 ways[/spoiler]) Answer: D Note: the FCP can be used to solve the majority of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmatcounting?id=775Cheers, Brent Question  Why aren't we multiplying 70 with 4 to count possibilities to arrange 3w and 1m in the group, i.e. mwww, wmww, wwmw, and wwwm? I guess I am confusing/missing something here. Thanks!



GMAT Club Legend
Joined: 11 Sep 2015
Posts: 4879
Location: Canada

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
22 Feb 2016, 18:34
Dienekes wrote: Question  Why aren't we multiplying 70 with 4 to count possibilities to arrange 3w and 1m in the group, i.e. mwww, wmww, wwmw, and wwwm? I guess I am confusing/missing something here. Thanks! Those 4 cases are all the same. For example, the group consisting of Ann, Bob, Bill, and Bart is the same as the group consisting of Bob, Bart, Ann, and Bill Cheers, Brent
_________________
Test confidently with gmatprepnow.com



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10594
Location: United States (CA)

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
14 Mar 2019, 06:13
themagiccarpet wrote: 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 employees consist of 3 women and one man?
A. 22 B. 35 C. 56 D. 70 E. 105 We are asked to find the number of ways of choosing 3 women from a group of 5 women and 1 man from a group of 7 men. Let’s first find the number of ways one can choose 3 women from 5. Since the order of how the 3 women are chosen doesn’t matter, we use combinations: 5C3 = 5!/(3! x 2!) = (5 x 4 x 3)/3! = 60/6 = 10 Similarly, the number of ways one can choose 1 man from a group of 7 men is: 7C1 = 7!/(1! x 6!) = 7 Finally, the number of ways we can choose 3 women from 5 women and 1 man from 7 men is: 5C3 x 7C1 = 10 x 7 = 70 Answer: D
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



NonHuman User
Joined: 09 Sep 2013
Posts: 15023

Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
Show Tags
08 Oct 2019, 06:05
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Of the 12 temporary employees in a certain company, 4 will
[#permalink]
08 Oct 2019, 06:05




