Find all School-related info fast with the new School-Specific MBA Forum

It is currently 30 Jun 2016, 09:23
GMAT Club Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How many 5 person committees chosen at random from a group c

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
Joined: 02 Jul 2013
Posts: 30
Concentration: Technology, Other
GMAT Date: 01-17-2014
GPA: 3.84
Followers: 0

Kudos [?]: 4 [0], given: 91

How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 03 Dec 2013, 09:41
2
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

72% (03:15) correct 28% (02:45) wrong based on 90 sessions

HideShow timer Statistics

How many 5 person committees chosen at random from a group consisting of 5 men, 5 women, and 5 children contain at least 1 woman?

A. 700
B. 1221
C. 1434
D. 2751
E. 3011


Well i know this may look easy. but for me it wasn't.
[Reveal] Spoiler: OA
Expert Post
2 KUDOS received
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 3187
Followers: 1063

Kudos [?]: 4621 [2] , given: 52

Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 03 Dec 2013, 14:45
2
This post received
KUDOS
Expert's post
mumbijoh wrote:
How many 5 person committees chosen at random from a group consisting of 5 men, 5 women, and 5 children contain at least 1 woman?
a) 700
b) 1221
c) 1434
d) 2751
e) 3011
Well i know this may look easy. but for me it wasn't.

Dear mumbijoh
I'm happy to help. :-) This is NOT an easy question. You may find this blog helpful.
http://magoosh.com/gmat/2013/difficult- ... -problems/

Let's first think about this conceptually. The two number of 5-person committees we could choose from 15 people is 15C5. Put that on hold. That's the total number.

Of that total number, all of them would include at least one woman EXCEPT the ones that are all men & children. How many 5-person groups can we form from these 10 people (5 men, 5 children)? That would be 10C5. Those would be the only groups that do NOT have at least one woman.

Thus, subtract this second number from the first. Answer = (15C5) - (10C5)

That's conceptually how we get the answer. Now, calculating that without a calculator is a more of a calculation task than the GMAT would typically make folks do. For this reason, I don't think this is a very good GMAT question. If the calculations are very hairy, the GMAT would typically leave all the answer choices in symbolic form, as I found above.

With a calculator, I will tell you:
15C5 = 3003
10C5 = 252
Answer = 3003 -252 = 2751
Once again, there is absolutely no way the GMAT would expect you to do that calculation on your own, with no calculator.

Does all this make sense?
Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Intern
Intern
avatar
Joined: 02 Jul 2013
Posts: 30
Concentration: Technology, Other
GMAT Date: 01-17-2014
GPA: 3.84
Followers: 0

Kudos [?]: 4 [0], given: 91

Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 03 Dec 2013, 22:41
Hi Mike,
Thank you for the explanation.I was getting lost with choosing the 5 from men and children I thought that I would have to permutate 5 in each case since you could choose to have all 5 men or all 5 women.
But from your explanation I have gathered that how those two are picked is not in question so long as women are still not part of the group.
Thanks a lot!
Intern
Intern
avatar
Status: Hulk
Joined: 17 Jul 2013
Posts: 11
Location: Singapore
Concentration: Entrepreneurship, Finance
Schools: Insead '14 (M)
GMAT 1: 600 Q44 V30
GMAT 2: 650 Q45 V34
GMAT 3: 690 Q49 V34
Followers: 0

Kudos [?]: 39 [0], given: 10

Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 05 Dec 2013, 22:47
I added all the possible combinations of a 5 person committee with at least one woman:

5C1*10C4 + 5C2*10C3 + 5C3*10C2 + 5C4*10C1 + 5C5*10C0
Expert Post
Magoosh GMAT Instructor
User avatar
Joined: 28 Dec 2011
Posts: 3187
Followers: 1063

Kudos [?]: 4621 [0], given: 52

Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 06 Dec 2013, 11:02
Expert's post
xhimi wrote:
I added all the possible combinations of a 5 person committee with at least one woman:

5C1*10C4 + 5C2*10C3 + 5C3*10C2 + 5C4*10C1 + 5C5*10C0

Dear xhimi,
Your approach gives the correct answer, but in a way, you fell into the trap of the question. Yes, everything about this expression would yield the correct answer, but without a calculator, this calculation would take quite a long time, longer than the 90 seconds or so you have for a GMAT math problem.

The words "at least" is an important hint that a shortcut is possible. Here's a blog that talks about this in probability:
http://magoosh.com/gmat/2012/gmat-math- ... -question/
It's similar in counting problems --- when the problem say "at least", and especially when it says "at least one" (the most typical case), then it's a trap to get drawn into calculating all the separate cases for one and greater. Instead, it's usually very easy to calculate the total (here, 15C5) and simply subtract the one included in the case for zero. You see, the opposite of "at least one" is "none", and that is almost always the key to solution.

Does all this make sense?
Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Image

Image

Intern
Intern
avatar
Status: Hulk
Joined: 17 Jul 2013
Posts: 11
Location: Singapore
Concentration: Entrepreneurship, Finance
Schools: Insead '14 (M)
GMAT 1: 600 Q44 V30
GMAT 2: 650 Q45 V34
GMAT 3: 690 Q49 V34
Followers: 0

Kudos [?]: 39 [0], given: 10

Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 07 Dec 2013, 01:53
mikemcgarry wrote:
xhimi wrote:
I added all the possible combinations of a 5 person committee with at least one woman:

5C1*10C4 + 5C2*10C3 + 5C3*10C2 + 5C4*10C1 + 5C5*10C0

Dear xhimi,
Your approach gives the correct answer, but in a way, you fell into the trap of the question. Yes, everything about this expression would yield the correct answer, but without a calculator, this calculation would take quite a long time, longer than the 90 seconds or so you have for a GMAT math problem.

The words "at least" is an important hint that a shortcut is possible. Here's a blog that talks about this in probability:

It's similar in counting problems --- when the problem say "at least", and especially when it says "at least one" (the most typical case), then it's a trap to get drawn into calculating all the separate cases for one and greater. Instead, it's usually very easy to calculate the total (here, 15C5) and simply subtract the one included in the case for zero. You see, the opposite of "at least one" is "none", and that is almost always the key to solution.

Does all this make sense?
Mike :-)



makes totally sense. i'm just not very secure in probability yet. my approach indeed took me more than 3 minutes. but your explanation and blog is very helpful, thanks! I'm getting there :-D
Intern
Intern
User avatar
Joined: 19 Mar 2013
Posts: 23
Followers: 0

Kudos [?]: 2 [0], given: 27

Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 12 Dec 2013, 10:35
Number of all possible committees of 15 people minus number of committees without women = number of committees with at least one woman
15C5-10C5= answer
i didn't go through all the calculations though, doesn't seem to be potential gmat question
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 10254
Followers: 483

Kudos [?]: 124 [0], given: 0

Premium Member
Re: How many 5 person committees chosen at random from a group c [#permalink]

Show Tags

New post 15 Jun 2016, 16:32
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.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Re: How many 5 person committees chosen at random from a group c   [#permalink] 15 Jun 2016, 16:32
    Similar topics Author Replies Last post
Similar
Topics:
1 A committee of 5 people is to be selected from 8 people. How many diff AbdurRakib 1 11 Jun 2016, 10:27
Experts publish their posts in the topic How many different committees can be formed from a group of two women Bunuel 4 10 Jun 2016, 01:01
3 Experts publish their posts in the topic What is the probability that a 4 person committee chosen at random fro reto 7 12 Jul 2015, 08:07
2 Experts publish their posts in the topic From a total of 5 boys and 4 girls, how many 4-person committees can Bunuel 5 26 Mar 2015, 04:21
6 Experts publish their posts in the topic A three-person committee must be chosen from a group of 7 anujkch 11 26 May 2012, 11:58
Display posts from previous: Sort by

How many 5 person committees chosen at random from a group c

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.