gmatclub combinatorics

Expert replies

bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Wed May 18, 2011 11:49 pm

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

C

Cheers !!

Quant 47-Striving for 50
Verbal 34-Striving for 40

My gmat journey :
https://www.beatthegmat.com/710-bblast-s ... 90735.html
My take on the GMAT RC :
https://www.beatthegmat.com/ways-to-bbla ... 90808.html
How to prepare before your MBA:
https://www.youtube.com/watch?v=upz46D7 ... TWBZF14TKW_

Like this post

Source: Beat The GMAT — Problem Solving | Wed May 18, 2011 11:49 pm

kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Thu May 19, 2011 12:03 am

You could reason that we need to choose 3 of the 8 people but exclude the groups consisting of two siblings and a third person:

8C3 - 4 x 6 = (8 x 7 x 6)/3 x 2 x 1 - 24 = 32

Alternatively, imagine for a moment that the order of choice mattered: We need 3 unrelated people 8 x x 4. Since the order of choice does not matter, we need to divide by 3!

Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid

Like this post

dv2020: Master | Next Rank: 500 Posts; Posts: 112; Joined: Sun Nov 08, 2009 9:39 pm; Location: Delhi; Thanked: 2 times

by dv2020 » Thu May 19, 2011 12:04 am

There are three SLOTs to be filled none of the siblings can be on the committee and here order doesn't matter between the three selected members so we will have to divide by n!

Hence,

First place can be taken by any of the 8 members, next taken by remaining 6(one selected member and sibling gone) & the third slot taken by (8-2 set of siblings gone now=4)

8X6X4/3!=8X4=32

Like this post

bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Thu May 19, 2011 12:42 am

kevin i did the easy part

wrote "8C3- ", but then could not figure out what to subtract. Why subtract 24 from it ?

DV2020
i got the part 8*6*4 but why divide by 3! ? is this because its a combo nd not permutation ?

Like this post

dv2020: Master | Next Rank: 500 Posts; Posts: 112; Joined: Sun Nov 08, 2009 9:39 pm; Location: Delhi; Thanked: 2 times

by dv2020 » Thu May 19, 2011 1:04 am

The idea is to figure out whether order matters or not...generally we use combinations formula when order doesn't matter and permutations when order does matter...if you can figure out that the order doesn't matter dividing by 3! is the same as the combinations formula

For Eg if brothers are B1 B2 B3 B4 & sisters S1 S2 S3 S4
so let us select a committee
B1B2S3
the five other orders within these 3 are not counted separately here but all six orders count as one
B1S3B2
B2B1S3
B2S3B1
S3B1B2
S3B2B1

So the order doesn't matter. hence we divide by 3!

to be ultra clear about using this method i recommend take some time and watch Ron Purewal's combinatoric study hall...I was also always confused about which formula Permutation or combinations which one to use but i found the answers after going though this study hall

Like this post

kevincanspain: GMAT Instructor; Posts: 613; Joined: Thu Mar 22, 2007 6:17 amFollowed by: 64 members; Location: madrid; Thanked: 171 times; GMAT Score:790

by kevincanspain » Thu May 19, 2011 1:46 am

kevincanspain wrote:You could reason that we need to choose 3 of the 8 people but exclude the groups consisting of two siblings and a third person:

8C3 - 4 x 6 = (8 x 7 x 6)/3 x 2 x 1 - 24 = 32

Alternatively, imagine for a moment that the order of choice mattered: We need 3 unrelated people 8 x 6 x 4. Since the order of choice does not matter, we need to divide by 3!

You could also reason that 3 of the 4 sibling pairs will have a member on the committee, and that each of these 3 pairs selected can 'donate' either of its two members:

4C3 x 2 x 2 x 2

Note that 4C3 =4C1 = 4

Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu May 19, 2011 2:56 am

bblast wrote:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

C

Good committees = Total possible committees - Bad committees.

Total possible committees:
Total number of combinations of 3 that can be formed from 8 choices = 8C3 = 56.

Bad committee = a sibling pair combined with one other person:
Number of sibling pairs = 4.
Once a sibling pair has been chosen from the 8 people, the number of people left that can combined with the sibling pair = 8-2 = 6.
Multiplying, we get 4*6 = 24 bad committees.

Thus, good committees = total - bad = 56-24 = 32.

The correct answer is C.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

559 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review