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

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Math Expert
Joined: 02 Sep 2009
Posts: 59561
Alice and David are two of the ten people in a group. In how many diff  [#permalink]

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15 Aug 2018, 00:51
00:00

Difficulty:

35% (medium)

Question Stats:

71% (02:15) correct 29% (02:19) wrong based on 59 sessions

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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
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Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

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15 Aug 2018, 02:04
Bunuel wrote:
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

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

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Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

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15 Aug 2018, 02:17
64

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

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Joined: 04 Jan 2015
Posts: 3158
Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

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16 Aug 2018, 02:14
1

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.

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Re: Alice and David are two of the ten people in a group. In how many diff  [#permalink]

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16 Aug 2018, 02:38
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

$$=120-56 = 64$$
Re: Alice and David are two of the ten people in a group. In how many diff   [#permalink] 16 Aug 2018, 02:38
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