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• ### $450 Tuition Credit & Official CAT Packs FREE February 15, 2019 February 15, 2019 10:00 PM EST 11:00 PM PST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # Ten telegenic contestants with a variety of personality  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Manager Joined: 22 Jan 2012 Posts: 78 Location: India Concentration: General Management, Technology GPA: 3.3 WE: Engineering (Consulting) Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 19 Mar 2012, 01:04 2 5 00:00 Difficulty: 25% (medium) Question Stats: 78% (01:47) correct 22% (01:53) wrong based on 207 sessions ### HideShow timer Statistics Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible? A. 120 B. 126 C. 252 D. 1200 E. 1260 _________________ Press +1 Kudos rather than saying thanks which is more helpful infact.. Ill be posting good questions as many as I can... Towards Success ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 19 Mar 2012, 02:33 5 6 iwillcrackgmat wrote: Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible? A. 120 B. 126 C. 252 D. 1200 E. 1260 Selecting 5 contestants for tribe A: $$C^5_{10}=252$$. The rest 5 contestants will automatically form tribe B. Answer: C. Notice that if we were asked about different ways of splitting 10 people into 2 equal groups of 5 then the answer would be 252/2=126, since we wouldn't have group #1 and group #2 in that case. Questions about the same concept to practice: in-how-many-different-ways-can-a-group-of-8-people-be-125985.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html how-many-ways-are-there-to-split-a-group-of-6-boys-into-two-105381.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-104807.html in-how-many-different-ways-can-a-group-of-9-people-be-101722.html in-how-many-different-ways-can-a-group-of-8-people-be-99053.html nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html in-how-many-different-ways-can-a-group-of-8-people-be-85707.html Hope it helps. _________________ ##### General Discussion Manager Joined: 22 Jan 2012 Posts: 78 Location: India Concentration: General Management, Technology GPA: 3.3 WE: Engineering (Consulting) Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 19 Mar 2012, 01:40 1 This question appears like a pitfall for me... Someone confident of the ans pls write Ur explanation.. My ans is 10C5 = 252 _________________ Press +1 Kudos rather than saying thanks which is more helpful infact.. Ill be posting good questions as many as I can... Towards Success Manager Joined: 22 Jan 2012 Posts: 78 Location: India Concentration: General Management, Technology GPA: 3.3 WE: Engineering (Consulting) Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 19 Mar 2012, 02:38 Thats really useful thanks for the explanation. _________________ Press +1 Kudos rather than saying thanks which is more helpful infact.. Ill be posting good questions as many as I can... Towards Success Manager Joined: 07 Dec 2011 Posts: 81 Location: India Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 19 Mar 2012, 13:26 Bunuel wrote: iwillcrackgmat wrote: Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible? A. 120 B. 126 C. 252 D. 1200 E. 1260 Selecting 5 contestants for tribe A: $$C^5_{10}=252$$. The rest 5 contestants will automatically form tribe B. Answer: C. Notice that if we were asked about different ways of splitting 10 people into 2 equal groups of 5 then the answer would be 252/2=126, since we wouldn't have group #1 and group #2 in that case. Questions about the same concept to practice: in-how-many-different-ways-can-a-group-of-8-people-be-125985.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html how-many-ways-are-there-to-split-a-group-of-6-boys-into-two-105381.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-104807.html in-how-many-different-ways-can-a-group-of-9-people-be-101722.html in-how-many-different-ways-can-a-group-of-8-people-be-99053.html nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html in-how-many-different-ways-can-a-group-of-8-people-be-85707.html Hope it helps. I got C as well but could you explain what you mean by the additional statement you made?? And while I'm at it, A BIG THANK YOU for your contributions in this forum. It's amazing the amount of help you provide to anyone who is willing to look for it. Intern Joined: 18 Mar 2012 Posts: 7 Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 28 Mar 2012, 01:16 thanks for the question and thanks for the explanatin as well Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 28 Mar 2012, 12:49 1 karun0109 wrote: Bunuel wrote: iwillcrackgmat wrote: Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible? A. 120 B. 126 C. 252 D. 1200 E. 1260 Selecting 5 contestants for tribe A: $$C^5_{10}=252$$. The rest 5 contestants will automatically form tribe B. Answer: C. Notice that if we were asked about different ways of splitting 10 people into 2 equal groups of 5 then the answer would be 252/2=126, since we wouldn't have group #1 and group #2 in that case. Questions about the same concept to practice: in-how-many-different-ways-can-a-group-of-8-people-be-125985.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-106277.html how-many-ways-are-there-to-split-a-group-of-6-boys-into-two-105381.html a-group-of-8-friends-want-to-play-doubles-tennis-how-many-104807.html in-how-many-different-ways-can-a-group-of-9-people-be-101722.html in-how-many-different-ways-can-a-group-of-8-people-be-99053.html nine-dogs-are-split-into-3-groups-to-pull-one-of-three-88685.html in-how-many-different-ways-can-a-group-of-8-people-be-85707.html Hope it helps. I got C as well but could you explain what you mean by the additional statement you made?? The difference between the case when we have group #1 and #2 and the case when we don't, is explained for example here: in-how-many-different-ways-can-a-group-of-8-people-be-125985.html as well as in other links presented. _________________ Math Expert Joined: 02 Sep 2009 Posts: 52905 Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 30 Jun 2013, 23:58 Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html _________________ MBA Section Director Affiliations: GMAT Club Joined: 21 Feb 2012 Posts: 6043 City: Pune Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 01 Jul 2013, 20:00 Do not have any alternate explanation. Yours is perfect. 10C5 AND 5C5 --------> 10!/5! X 1 = 252 _________________ Starts from Feb 4th: MBA Video Series, Video answers to specific components and questions about the MBA application. Manager Joined: 21 Oct 2013 Posts: 185 Location: Germany GMAT 1: 660 Q45 V36 GPA: 3.51 Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 08 Nov 2013, 03:55 Narenn wrote: Do not have any alternate explanation. Yours is perfect. 10C5 AND 5C5 --------> 10!/5! X 1 = 252 Hey Guys, since I haven't had probabilty in high school, I don't get anything here. With the C and 5! and so on. Could anyone explain a little further! Would be great! EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13536 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Ten telegenic contestants with a variety of personality [#permalink] ### Show Tags 24 Jan 2015, 11:00 Hi All, The last post in this thread (from unceldolan) points out that each of the explanations takes it for granted that the reader knows the Combination Formula. If that's the case, then the other explanations should suffice. If you DON'T know the Combination Formula, then here's what it is and how to use it. Any time a prompt asks for "groups" or "combinations" of things, then the order of the things DOES NOT MATTER. For example, if a 2-person team consists of A and B, then A,B is the same as B,A --> thus, order does NOT matter. Mathematically though, you're allowed to count this team TWICE - A,B and B,A are the same team, so it should only be counted ONCE. The Combination Formula removes all of the "duplicates", leaving you with the unique combinations for whatever situation you're working with. The Combination Formula itself is: N!/[K!(N-K)!] N = the total number of items/people K = the size of the subgroup Here, we have 10 people and we're asked to form 2 groups of 5. For the first group, N = 10 and K = 5.... 10!/[5!5!] = (10)(9)(8)(7)(6)/(5)(4)(3)(2)(1) = 256 unique groups of 5 people Once you have formed that group of 5, then the remaining 5 form the other group (so there's no more math to do) Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: Ten telegenic contestants with a variety of personality  [#permalink]

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11 Apr 2016, 04:33
iwillcrackgmat wrote:
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

I got C as well. But l'm confused in one thing - why are we not diving 252 by 2 to avoid repetition here. Could someone please clarify?
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Re: Ten telegenic contestants with a variety of personality  [#permalink]

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25 Apr 2016, 02:51
1
iwillcrackgmat wrote:
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

I got C as well. But l'm confused in one thing - why are we not diving 252 by 2 to avoid repetition here. Could someone please clarify?

Hi,

I would say the reason is that the tribes are named as A and B..
so say c,d,e,f,g are in A and h,i,j,k,l are in B..
this will be different to c,d,e,f,g are in B and h,i,j,k,l are in A

But as you correctly mentioned if it were just two groups with no difference, we would divide the numbers by 2..
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentage-increase-decrease-what-should-be-the-denominator-287528.html

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Re: Ten telegenic contestants with a variety of personality  [#permalink]

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07 Feb 2019, 10:14
Top Contributor
iwillcrackgmat wrote:
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

Let's take the task of creating the teams and break it into stages.

Stage 1: Select two 5 contestants to be in tribe A
Since the order in which we select the contestants does not matter, we can use combinations.
We can select 5 contestants from 10 contestants in 10C5 ways
10C5 = (10)(9)(8)(7)(6)/(5)(4)(3)(2)(1) = 252
So, we can complete stage 1 in 252 ways

Stage 2: Place the remaining 5 people in tribe B
There's only 1 way to place all 5 remaining people in tribe B
So we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create 2 tribes of 5 contestants each) in (252)(1) ways (= 252 ways)

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: Ten telegenic contestants with a variety of personality  [#permalink]

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10 Feb 2019, 07:32
iwillcrackgmat wrote:
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

The number of ways to select the first tribe is 10C5:

(10!)/[(10 - 5)! x 5!]

= (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2 x 1)

= 3 x 2 x 7 x 6 = 252 ways

The next tribe can be selected in 5C5 = 1 way. So the total number of ways the two tribes can be selected is 252 x 1 = 252.

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Re: Ten telegenic contestants with a variety of personality  [#permalink]

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10 Feb 2019, 19:33
iwillcrackgmat wrote:
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

The number of ways to select the first tribe is 10C5:

10C5 = (10!)/(10 - 5)! x 5!

= (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2 x 1)

= 3 x 2 x 7 x 6 = 252

The next tribe can be selected in 5C5 = 1 way.

So there are 252 x 1 = 252 ways to select the two tribes.

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Re: Ten telegenic contestants with a variety of personality   [#permalink] 10 Feb 2019, 19:33
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