|
Author |
Message |
|
Manager
Joined: 22 Jul 2008
Posts: 156
Followers: 1
Kudos [?]:
5
[0], given: 0
|
Re: PS - Combinatorics [#permalink]
 26 Aug 2008, 14:30
4 C 3 * 2 C 1 * 2 C 1 * 2 C 1 = 4*2*2*2 = 32
|
|
|
|
|
|
|
Intern
Joined: 10 Oct 2006
Posts: 3
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
23 Aug 2009, 12:34
xALIx wrote: Well put!! bmwhype2 wrote: Mishari wrote: Total possibilities excluding consideration for the restrictions :
8C3 = 8!/[3!x5!] = 8x7x6/3x2 = 56
The exluded possibilities due to the restriction (i.e. no sibiling in the 3-members committee ):
6 possible people, we choose 1 of them.
6C1=6
corresponding possibilities for each pair (the 3rd member) = 6
we can do this with each pair of the twins.
total pairs of siblings = 4
(Unfavorable pairs)(6C1)
So it is 4x6 = 24
ANSWER = total possibilities - restricted possibilities = 56 - 24 = 32
ANSWER: C excellent answer. small elaboration above. I like this explanation but i am hung up on "6 possible people, we choose 1 of them" if there are 8 people total and you want to form 3 person committees from 4 pairs of siblings who cannot be together in a committee and if you have you have 6 possible people (i'm assuming you've picked 1 person and the person's sibling is excluded) shouldn't we be choosing 2 people out of the 6 people that are left to form the 3 person committee? Please go really slowly, i am far from a math whiz.
|
|
|
|
|
|
Intern
Joined: 10 Oct 2006
Posts: 3
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
23 Aug 2009, 12:47
Nevermind i got it: The operating phrase here "excluded possibilities!"
in otherwords what cannot be.
The first two people are indeed siblings of whom you ignore to grab someone else from the the remaining six to form a committee that cannot be.
subtracting the the committees that cannot be from the sum of those that can and cannot be leaves you with those that can be.
I hate that it take me so stinking long to figure these out!
Thanks
|
|
|
|
|
|
Intern
Joined: 02 Nov 2009
Posts: 21
Followers: 0
Kudos [?]:
2
[0], given: 9
|
Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 07:38
1. All possbilities
C3/8= 8!/3!*(8-3)!=56
2. eliminate the pairs with siblings:
AaB
AaC
AaD
Aab
Aac
Aad
...24 pairs
3. 56-24=32
Answer: C
|
|
|
|
|
|
Intern
Joined: 29 Dec 2009
Posts: 1
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 09:53
Hi Guys, I have reverse question here if somebody can help. I found 3 ways of calculating the answer: 1. 4C3 (All Boys)+4C3 (All Girls)+4C2x2C1(2 Boys and 1 girl)+2C1x4C2(2 Girl and 1 Boy) 2. (Total ways of making the committe)-( ways for selecting one pair)x(Selecting final committe member ) i.e [8C3-4C1x6C1] 3. This is tricky and I don't know why this is wrong  (Ways of Selecting First Committe Member)x( Ways of selecting 2nd memeber)x(Ways of selecting 3rd member) i.e [8C1x6C1x4C1] I need help proving 3rd way of calculation is wrong. Plz help Thanks, RRH
|
|
|
|
|
|
Intern
Joined: 15 Nov 2009
Posts: 31
Location: Moscow, Russia
Followers: 0
Kudos [?]:
8
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 10:17
First member can chosen from any of 8
Second - from the 6 left (we exclude the sibling).
The third from the remaining 4.
Finally, don't forget about repetitions like abc, acb...l number of which is 3!=6.
So, (8*6*4)/6=32.
Ans. is C.
|
|
|
|
|
|
Intern
Joined: 22 Dec 2009
Posts: 41
Followers: 0
Kudos [?]:
1
[0], given: 13
|
Re: PS - Combinatorics (m02q05) [#permalink]
09 Jan 2010, 21:37
Mishari,
Total possibilities excluding consideration for the restrictions :
8C3 = 8!/[3!x5!] = 8x7x6/3x2 = 56
How come [3!x5!] yields 3x2?
Isnt it suppose to be 3!(1・2・3)×5!(1・2・3・4・5)??
|
|
|
|
|
|
Intern
Joined: 10 Jan 2010
Posts: 18
Followers: 0
Kudos [?]:
1
[0], given: 22
|
Re: PS - Combinatorics (m02q05) [#permalink]
12 Jan 2010, 08:14
I think that 4C2 should be 4C1 because we are only constraining the second seat on the committee from being a sibling before considering the third seat. 4C1*6C1 = 24.
|
|
|
|
|
|
Intern
Joined: 20 Dec 2010
Posts: 6
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
12 Jan 2011, 08:33
OA C = 32
Total Combinations - Not accepted
(8!/(5!3!)) - 4 x (6!/(5!1!))
Supose you pick two siblings, there are 6 candidates for the last post. This can occur four times as there are four pairs of brothers.
|
|
|
|
|
|
Intern
Joined: 09 Jan 2011
Posts: 1
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
12 Jan 2011, 10:32
The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways
Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. Hence to get the no of combination divide the no of ways by 3!
The committee can be formed in = 192/6 = 32 ways
Sanjay
|
|
|
|
|
|
Manager
Joined: 27 Jan 2010
Posts: 150
Concentration: Strategy, Other
GMAT 1: Q V
GMAT 2: Q V
WE: Business Development (Consulting)
Followers: 7
Kudos [?]:
16
[0], given: 53
|
Re: PS - Combinatorics (m02q05) [#permalink]
04 Jun 2011, 18:17
This is how I did it -
Consider 4 siblings:
B1G1 B2G2 B3G3 B4G4
From any one pair of siblings, choose 1 person = 2C1 (From a brother & sister pair choose 1)
We are selecting 3 people, so repeat this 3 times from 3 different pairs of siblings.
So we get = 2C1 * 2C1 * 2C1 = 8
Now we have 4 pairs and we need to select any 3 pairs (From which to select 3 individual people - this part has been done in the above step) = 4C3 = 4
So we now get 8*4 = 32.
|
|
|
|
|
|
Senior Manager
Joined: 24 Mar 2011
Posts: 479
Location: Texas
Followers: 4
Kudos [?]:
33
[0], given: 20
|
Re: PS - Combinatorics (m02q05) [#permalink]
04 Jun 2011, 23:36
4C1 * 3C2 + 4C1 * 3C2 + 4C3 + 4C3 = 32
1B, 2G
1G, 2B
3B
3G
|
|
|
|
|
|
Senior Manager
Joined: 29 Jan 2011
Posts: 401
Followers: 0
Kudos [?]:
8
[0], given: 87
|
Mishari wrote: Total possibilities excluding consideration for the restrictions :
8C3 = 8!/[3!x5!] = 8x7x6/3x2 = 56
The exluded possibilities due to the restriction (i.e. no sibiling in the 3-members committee ):
total pairs of sibilings = 4
corresponding possibilities for each pair (the 3rd member) = 6
So it is 4x6 = 24
ANSWER = total possibilities - restricted possibilities = 56 - 24 = 32
ANSWER: C How do you get corresponding possibilities for each pair (the 3rd member) = 6 ??? I dont understand
|
|
|
|
|
|
Manager
Status: Bell the GMAT!!!
Affiliations: Aidha
Joined: 16 Aug 2011
Posts: 192
Location: Singapore
Concentration: Finance, General Management
GMAT 1: 680 Q46 V37
GMAT 2: 620 Q49 V27
GMAT 3: 700 Q49 V36
WE: Other (Other)
Followers: 5
Kudos [?]:
22
[0], given: 43
|
Re: PS - Combinatorics (m02q05) [#permalink]
04 Oct 2011, 20:56
Bunuel wrote: The third way is also valid, but you should divide 8C1x6C1x4C1 by 3! to get rid of duplications.
With 8C1x6C1x4C1 you can have ABC members as well as BCA members, which is basically the same group.
Bunnel, how you came to the conclusion that 8C1x6C1x4C1 should be divided by 3! to get rid of duplications. Why not 4! or 2!. Please explain.
_________________
If my post did a dance in your mind, send me the steps through kudos :)
My MBA journey at http://mbadilemma.wordpress.com/
|
|
|
|
|
|
Intern
Joined: 03 Aug 2011
Posts: 4
Followers: 0
Kudos [?]:
1
[0], given: 0
|
Re: PS - Combinatorics (m02q05) [#permalink]
20 Oct 2011, 18:14
Let the 4 groups be A B C D.
Following are the 4 ways to select 3 people
ABC, BCD, ACD,ABD. Since there are 2 ways to select a person from a group we have 8 different possibilities for each pairing. Since there are 4 pairings 8 x4=32 the final answer
|
|
|
|
|
|
Senior Manager
Joined: 23 Oct 2010
Posts: 335
Location: Azerbaijan
Followers: 6
Kudos [?]:
68
[0], given: 67
|
Re: PS - Combinatorics (m02q05) [#permalink]
18 Jan 2012, 00:54
8*6*4/3!=32 u have to divide (8*6*4) by 3!, since all there are double counted. I mean first u chose any of 8, then any of 6. that any of 8 may consist any of 6. same with any of 4.
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11611
Followers: 1801
Kudos [?]:
9597
[0], given: 829
|
Re: PS - Combinatorics (m02q05) [#permalink]
18 Jan 2012, 04:30
|
|
|
|
|
|
Manager
Joined: 10 Jul 2010
Posts: 197
Followers: 1
Kudos [?]:
15
[0], given: 12
|
Re: PS - Combinatorics (m02q05) [#permalink]
02 Feb 2012, 05:33
Sanjay76 wrote: The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways
Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. Hence to get the no of combination divide the no of ways by 3!
The committee can be formed in = 192/6 = 32 ways
Sanjay Thanks Sanjay, this was the easiest post to comprehend and helped me understand why they 3! was needed
|
|
|
|
|
|
Intern
Joined: 25 Dec 2012
Posts: 38
Followers: 0
Kudos [?]:
0
[0], given: 6
|
Re: PS - Combinatorics (m02q05) [#permalink]
16 Jan 2013, 06:11
The three pair of sibling can be selected in 4C3 ways.
And among the three siblings, you can select one from each in 2 ways = 4C3 * 2 * 2 * 2 = 32
|
|
|
|
|
|
|
Re: PS - Combinatorics (m02q05)
[#permalink]
16 Jan 2013, 06:11
|
|
|
|
|
|
|
|
|
|
|
close
