Anisha and Dequon are 2 of the 10 people in a group. In how many diff

Question

Anisha and Dequon are 2 of the 10 people in a group. In how many different ways can this group of 10 people be divided into a group of 7 people and a group of 3 people if Anisha and Dequon are to be in the same group?
A) 48
B) 56
C) 64
D) 80
E) 128

u1983 · Accepted Answer

According to the Q-stem there could be 2 possible cases for the groups:

Case 01 :  Group 01 : A, D & another 5 people & Group 02: Other 3 people 
A & D already assigned to Group 01 . Thus remaining 5 people can be chosen from 8 people in 8C5 = 56 ways
Then for the last 3 people of the other group , 3 people can be chosen from 3 people in 3C3=1 ways
Hence total Case 01 possibilities : 56*1=56

Case 02 :  Group 01 : A, D & another 1 person & Group 02: Other 7 people 
A & D already assigned to Group 01 . Thus remaining 1 person can be chosen from 8 people in 8C1 = 8 ways
Then for the last 7 people of the other group , 7 people can be chosen from 7 people in 7C7=1 ways
Hence total Case 02 possibilities : 8*1=8

Hence total number of possibilities : 56+8 = 64.......... I would go for  option C .

varundixitmro2512 · Answer

Considering 2 will join group of 7 , then 8c5 (5 people needed from 8 remaining) and rest for the group of three
Or 2 will join group of 3, then 8c1(1 people needed from 8 remaining)

Add these two=64

niks18 · Answer

Case 1: if the two join the group of 7, then we need to divide 8 remaining people into two groups of 5 and 3

This can be done in  \frac{8!}{5!*3!}  ways  =56

Case 2: if the two join the group of 3, then we need to divide 8 remaining people into two groups of 7 and 1

This can be done in  \frac{8!}{7!*1!}  ways  = 8

Hence total number of ways  = 56+8=64

Option  C

-------------------------------------------

Why we are dividing here:

You have 8! ways to arrange 8 different persons, after that we can divide each of these persons into 2 groups of strength 5 and 3 for the first case (or 7 & 1 for the second case). But we do not care about the order of each group. The number of ways to form a group with 5 people is 5!. The same for group of 3 people, so we must eliminate all of these duplicates.

So we will have 8!/5!*3! ways to form these different groups of 5 and 3 people in the first case. Similarly for the second case.

GMATBusters · Answer

This question is based on the distribution of objects into groups:
To more about division into groups :
Refer this  Division into groups- combination

zishu912 · Answer

10 people need to be divided in 2 group of 7 and 3 people

Now suppose that Anisha and Dequon are already included in the group of 7 people so for rest of the 5 people will be selected from 8 in 8C5 = 56 ways and rest 3 will be selected in only one way

Next Anisha and Dequon are included in a group of 3 people, so from 8 we need to select only one and it could be done in 8C1 = 8 ways and for the rest 7 there is only one way to select

Hence the total selection = 56 + 8 = 64

GYANENDRA88 · Answer

alternate method
Select 7 (or 3) people from 10 and then subtract those selections in which only one of them is selected
i.e. 10C7 - 2(8C2)= 64

achloes · Answer

Bunuel   KarishmaB   gmatophobia  how would you approach this question? Thanks in advance!

gmatophobia · Answer

achloes  - The constraint in this question is around A and D. A and D are to be in the same group. So if we consider them as a single unit say 'group AD' and then distribute the users into groups of 7 and groups of 3, wherever group AD is placed A and D will remain together.

Case 1  : Group AD is a part of the group that has 7 members

As group AD is already a part of this group, we need 5 more members in this group from the 8 available members.

Number of ways of selecting 5 members from 8 members =  ^8C_5 = ^8C_3 = 56

The group having 3 members can be selected in 1 way.

Case 2  : Group AD is a part of the group that has 3 members

As group AD is already a part of this group, we need 1 more member in this group from the 8 available members.

Number of ways of selecting 1 member from 8 members =  ^8C_1 = 8

The group having 7 members can be selected in 1 way.

Total = 8 + 56 = 64

Option C

KarishmaB · Answer

As shown by gmatophobia above, I would also use 8C1 + 8C3

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Anisha and Dequon are 2 of the 10 people in a group. In how many diff

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Anisha and Dequon are 2 of the 10 people in a group. In how many diff

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards