Alice and David are two of the ten people in a group. In how many diff

Question

Alice and David are two of the ten 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 Alice and David are to be in the same group?

A. 48

B. 56

C. 64

D. 80

E. 128

A. 48
B. 56
C. 64
D. 80
E. 128

GMATinsight · Answer

There are two scenarios

Case 1  : When Alice and David are in group of 7 : Total ways to make such groups of 7 and 3 = 8C5 (To select 5 more out of 8 to make group of 7 which includes Alice and David already)

This way the other group of 3 is automatically formed, 8C5 = 56

Case 2  : When Alice and David are in group of 3 : Total ways to make such groups of 3 and 7 = 8C1 (To select 1 more out of 8 to make group of 3 which includes Alice and David already)

This way the other group of 7 is automatically formed, 8C1 = 8

Total Possible ways = 56+8 = 64

Answer: Option C

rohanramgude10 · Answer

64

1*1*8c5+1*1*8c1

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EgmatQuantExpert · Answer

Solution 
  Given:  
 • There are 10 people in a group
• Alice and David are two among the ten people

To find:  
 • In how many ways can these 10 people be divided into a group of 7 and a group of 3, such that Alice and David are in the same group

Approach and Working:   
There are two possible scenarios,
 • Scenario 1: Alice and David are in the group of 7 people
• Scenario 2: Alice and David are in the group of 3 people

In Scenario 1, the total number of ways will be  ^8C_5 
 • Since, Alice and David are two among the 7 people, the other 5 people can be selected from the remaining 8 people in 8C5 ways. 
• Later the remaining 3 will form another group.

In Scenario 2, the total number of ways will be  ^8C_1 
 • Since, Alice and David are two among the 3 people, the other person can be selected from the remaining 8 people in 8C1 ways. 
• Later the remaining 7 will form another group.

Therefore, the total number of ways =  ^8C_5 + ^8C_1 = 8 + 56 = 64

Hence, the correct answer is option C.

Answer: C

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-AT-e1533292152363.png

Princ · Answer

OA:C

No of ways, when Alice and David are in different groups  : 2 * C(8,6) * C(2,2) = 2 *\frac{8!}{6!*2!}*1=56

Total number to form two teams one of  7  members and another  3  members, out of total  10  members, without any restrictions

= C(10,7)*C(3,3)=\frac{10!}{7!*3!}*1=120

Numbers of ways when Alice and David are in the same team = Total number of ways, without restrictions - No of cases when they are in different teams

=120-56 = 64

kevincan · Answer

Another way to do this is to focus on the group of three people. How many unique compositions are there? consider two disjoint cases: one in which the pair are included, 8c1 = 8 and the other in which the pair is not included, 8c3 =56

So we get 64 in total

So we get 64 in total

luisdicampo · Answer

Deconstructing the Question

We have a group of 10 people divided into a group of 7 and a group of 3. 
Alice and David must be in the same group.

Key idea: count by cases:
either both are in the group of 7 or both are in the group of 3.

Step-by-step

Case 1: Alice and David are in the group of 7.
We already placed 2 people, so we choose 5 more from the remaining 8.

\frac{8!}{5!3!} = 56

Case 2: Alice and David are in the group of 3.
We already placed 2 people, so we choose 1 more from the remaining 8.

\frac{8!}{1!7!} = 8

Total ways:

56 + 8 = 64

Answer: 64

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