BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

In how many different ways can a group of 9 people be

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
Legendary Member
Posts: 503
Joined: Thu Jul 20, 2017 9:03 am
Thanked: 86 times
Followed by:15 members
GMAT Score:770

by ErikaPrepScholar » Sat Mar 31, 2018 11:19 am
Let's think through the scenario. First we pick 3 people out of 9. It doesn't matter what order we pick the 3 people in, since they'll all end up in the same group anyway. Since order doesn't matter, we have a combination.

So for this first step, we have: $$\frac{9!}{3!\left(9-3\right)!}=\frac{9!}{3!6!}=\frac{9\cdot8\cdot7}{3\cdot2\cdot1}=3\cdot4\cdot7=84$$

So we have 84 possibilities for how we pick the first group.

Then we need to pick the next group. We pick 3 people out of the remaining 6. Again, it doesn't matter what order we pick them in, so we have a combination:

$$\frac{6!}{3!\left(6-3\right)!}=\frac{6!}{3!3!}=\frac{^{6\cdot5\cdot4}}{3\cdot2\cdot1}=2\cdot5\cdot2=20$$

So for each of the 84 possibilities for how we pick the first group, we have 20 possibilities for how we pick the second group. That's 84*20 = 1680 possibilities in total.

We're left with 3 people. There's only one option at this point, which is to pick all 3 of the remaining people for the last group. So there are 1680 possibilities in total for how we can pick the 3 groups.

HOWEVER, before we pick C, we need to recognize that based on this scenario, we don't care which order the groups themselves are chosen in. This means we need to account for duplicates where the same groups of people are chosen in different orders. (Ex: if the groups are chosen 1.ABC, 2.DEF, 3.GHI, it is the same three groups as when they are chosen 1. GHI, 2.ABC, 3.DEF.) 3 groups can be ordered in 3! ways. So to get rid of any duplicates, we divide our total number of possibilities by 3!.

$$\frac{1680}{3!}=\ 280$$

So the correct answer is A.

Note: this final element of the problem (discounting group ordering) makes this problem particularly challenging. There are other equations by which we can get to the solution faster, but they're so unlikely to come up on a real GMAT exam that they aren't worth memorizing. Using logic and basic combinatorics as we did above is your best bet.
Image

Erika John - Content Manager/Lead Instructor
https://gmat.prepscholar.com/gmat/s/

Get tutoring from me or another PrepScholar GMAT expert: https://gmat.prepscholar.com/gmat/s/tutoring/

Learn about our exclusive savings for BTG members (up to 25% off) and our 5 day free trial

Check out our PrepScholar GMAT YouTube channel, and read our expert guides on the PrepScholar GMAT blog
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8086
Joined: Sat Apr 25, 2015 10:56 am
Location: Los Angeles, CA
Thanked: 43 times
Followed by:29 members

by Scott@TargetTestPrep » Mon May 27, 2019 4:25 am
M7MBA wrote:In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people?

A. 280
B. 1,260
C. 1,680
D. 2,520
E. 3,360

The OA is A.
The first group of 3 can be chosen in 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84 ways.

The second group of 3 can be chosen in 6C3 = (6 x 5 x 4)/(3 x 2) = 5 x 4 = 20 ways.

The third group of 3 can be chosen in 3C3 = 1 way.

Therefore, the 3 groups can be chosen 84 x 20 x 1 = 1680 ways. However, since the order of the 3 groups doesn't matter, we have to divide 1680 by 3!. Hence, the number of ways 9 people can be divided into 3 groups is 1680/3! = 1680/6 = 280.

Answer: A

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Post new topic Post Reply
• Page 1 of 1