Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register
 

Important Permutations & Combinations Concept

This topic has 5 member replies
Bookmark
Previous Topic Next Topic
aneesh.kg GMAT Destroyer!
Joined
16 Apr 2012
Posted:
382 messages
Followed by:
23 members
Thanked:
174 times
Important Permutations & Combinations Concept Post Wed May 02, 2012 8:36 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    I want to discuss a very important concept in Permutations & Combinations. This concept lies at the heart of this topic.

    Please Read carefully.

    Question: There are five people named A,B,C,D and E. In how many ways can a team of 2 people be formed from these 5 people?

    Solution 1:
    2 people can be selected out 5 people in 5C2 ways.
    So, the number of teams = 5C2 = 10

    Solution 2:
    Let's select one person from the 5 people. The number of ways of doing this is 5C1.
    From the remaining 4 people (i.e. excluding the person already selected), let's selected one more person. The number of ways of this is 4C1.
    So, total number of ways = 5C1*4C1 = 20

    What did just happen? I solved the question with two methods, both of which seem correct and yet the answers from the two methods are different. Infact, the first one is just half of the other. Can someone explain which one is correct and what is the problem with the other one?

    Please don't come up with an explanation unless you are convinced of your own explanation.

    Think!

    _________________
    Aneesh Bangia
    GMAT Math Coach
    aneesh.bangia@gmail.com

    GMATPad:
    Facebook Page: http://www.facebook.com/GMATPad



    Last edited by aneesh.kg on Thu May 03, 2012 2:19 am; edited 2 times in total

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    mathbyvemuri Really wants to Beat The GMAT! Default Avatar
    Joined
    26 Apr 2012
    Posted:
    142 messages
    Thanked:
    28 times
    Post Thu May 03, 2012 1:39 am
    Aneesh it's a good one.
    The answer is 5C2, ie., 10 only.
    In the first method we get the following combinations: AB,AC,AD,AE,BC,BD,BE,CD,CE,DE
    But,in the second method we get some additional combinations which are just repeats but in a reverse order: AB,AC,AD,AE,BA,BC,BD,BE,CA,CB,CD,CE,DA,DB,DC,DE,EA,EB,EC,ED
    The underlined ones are repeats in reverse order. These are redundant as it makes no difference to have BA as a team when you already have AB at hand.
    Kudos to your post, which is conceptual.

    gmatNooB8787 Rising GMAT Star Default Avatar
    Joined
    16 Oct 2011
    Posted:
    31 messages
    Followed by:
    2 members
    Thanked:
    4 times
    Post Thu May 03, 2012 1:39 pm
    I understand your reply "mathbyvemuri" , but this set thing becomes obvious only when we start searching for actual values in the sample set, but not while calculating.

    mathbyvemuri Really wants to Beat The GMAT! Default Avatar
    Joined
    26 Apr 2012
    Posted:
    142 messages
    Thanked:
    28 times
    Post Fri May 04, 2012 5:59 pm
    "gmatNooB8787", I did not get your point, pl elaborate on that

    aneesh.kg GMAT Destroyer!
    Joined
    16 Apr 2012
    Posted:
    382 messages
    Followed by:
    23 members
    Thanked:
    174 times
    Post Mon May 07, 2012 10:06 pm
    I don't see too many attempts on this but I hope a few people did think about this concept.

    Before answering this, let me revise a few concepts of Permutations & Combinations for you pertaining to this problem.
    (i) Arrangement of n different objects in n places is by n! ways
    (ii) The number of ways of selecting r objects from n different objects is nCr.
    (iii) AND means Multiplication.

    The number of teams of 2 people from 5 people = The number of ways of selecting 2 people from 5 people = 5C2

    So, Solution 1 is correct. I am sure you would've figured this much.

    Solution 2 is wrong. But what is the mistake?
    The mistake is in selecting objects one by one from the same pool.
    Say we select A out of the 5 people and then select B.
    In a separate way of selection, say we select B first, and A later.
    So, we got (A,B) as well as (B,A) - which is the same team.
    But, we wanted this team just once. Notice the question just wants teams of people and not their arrangement also.

    We have arranged the two people of each team in 2! ways also, which is unnecessary. We have (C,D) as well as a (D,C), a (B,E) as well as a (E,B), and so on. No wonder the wrong answer is 2! times more than the correct answer of 10.

    To get the correct answer, you can now divide 5C1*4C1 by 2! to nullify the redundant cases.
    (5C1*4C1)/2! = 10

    Do you notice now that 5C2 (or rather nCr) is such an awesome formula?
    5C2 is inbuilt with a 2! at its denominator.
    5C2 = (5C1*4C1)/2!
    The formula makes sure that arrangements are nullified and only unique set of teams are counted. No worries!

    I have seen a lot of students make this mistake. So, it's important that you understand this well.

    Method3 in this solution:
    http://www.beatthegmat.com/probability-of-cars-t111282.html#469279
    is based on this concept.

    @mathybyvemuri: you were correct. thanks for your input.

    _________________
    Aneesh Bangia
    GMAT Math Coach
    aneesh.bangia@gmail.com

    GMATPad:
    Facebook Page: http://www.facebook.com/GMATPad

    Thanked by: sushmita0327, imhimanshu, eaakbari
    sunman Really wants to Beat The GMAT!
    Joined
    17 Feb 2011
    Posted:
    162 messages
    Followed by:
    7 members
    Thanked:
    12 times
    Test Date:
    Sept 2012
    Target GMAT Score:
    750+
    GMAT Score:
    750
    Post Tue May 08, 2012 3:38 am
    aneesh.kg wrote:
    I want to discuss a very important concept in Permutations & Combinations. This concept lies at the heart of this topic.

    Please Read carefully.

    Question: There are five people named A,B,C,D and E. In how many ways can a team of 2 people be formed from these 5 people?

    Solution 1:
    2 people can be selected out 5 people in 5C2 ways.
    So, the number of teams = 5C2 = 10

    Solution 2:
    Let's select one person from the 5 people. The number of ways of doing this is 5C1.
    From the remaining 4 people (i.e. excluding the person already selected), let's selected one more person. The number of ways of this is 4C1.
    So, total number of ways = 5C1*4C1 = 20

    What did just happen? I solved the question with two methods, both of which seem correct and yet the answers from the two methods are different. Infact, the first one is just half of the other. Can someone explain which one is correct and what is the problem with the other one?

    Please don't come up with an explanation unless you are convinced of your own explanation.

    Think!
    Solution 1 is correct.

    Solution 2 is wrong because in a team of two, order doesn't matter. DE is the same exact team as ED, so the answer is:

    (5! x 4!)/(2!) = 10.

    _________________
    "Never doubt that a small group of thoughtful, committed citizens can change the world. Indeed, it's the only thing that ever has" - Margaret Mead

    Best Conversation Starters
    1 varun289 38 topics
    2 killerdrummer 22 topics
    3 sana.noor 20 topics
    4 sanaa.rizwan 14 topics
    5 guerrero 14 topics
    See More Top Beat The GMAT Members...

    Most Active Experts
    1 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    		 204 posts
    2 image description GMATGuruNY

    The Princeton Review Teacher

    		 136 posts
    3 image description Jim@StratusPrep

    Stratus Prep

    		 100 posts
    4 image description Anju@Gurome

    Gurome

    		 74 posts
    5 image description Jon@Admissionado

    Admissionado

    		 51 posts
    See More Top Beat The GMAT Experts