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Eight persons in an organization including Andrew and Bob were to be d

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Bunuel
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Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   14 Jan 2020, 01:45
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Competition Mode Question



Eight persons in an organization including Andrew and Bob were to be divided in two groups of 4 members each. The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed?

A. 1/4
B. 1/70
C. 3/14
D. 1/10
E. 11/14
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C
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   14 Jan 2020, 07:17
total possible groups l 8c4 ; 70
group containing andrew and bob ; 6c2 ;
P ; 15/70 ; 3/14
IMO C

Eight persons in an organization including Andrew and Bob were to be divided in two groups of 4 members each. The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed?
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   14 Jan 2020, 09:51
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Number of ways in which 2 groups of 4 that can be formed = \(8c_4\) = \(\frac{8*7*6*5}{4!} = 70\)

Favorable outcome = Andrew & Bob to be in same team of 4
--> Favorable ways = selecting 2 members from remaining 6 = \(6c_2 = 6*5/2! = 15\)

--> Required Probability = \(\frac{15}{70}\) = \(\frac{3}{14}\)

Option C
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   14 Jan 2020, 14:22
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There are 8C4 ways to choose the first of the two groups and only 1 way to choose the second group after the first is chosen so: 140 ways to create 2 groups of 4 members each.

Andrew and Bob can be both in the first group or both in the second group.

If both are in the first group, we anchor them in and choose 2 people from the remaining 6 to complete the first group. This can be done in 15 ways. The other scenario, that is when they are both in the second group, is the mirror image of the former scenario. So there are 15+15=30 ways for this to happen.

30/140 or 3/14.

Answer is C
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   14 Jan 2020, 14:56
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The total number of groups which can be formed = 8C4 = 70 groups (this includes the second group and sequence is not important here)

The total number of groups containing both Andrew and Bob (A, B, _, _) = 6C2 = 15

Fraction of total groups = 15/70 = 3/14

FINAL ANSWER IS (C)

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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   14 Jan 2020, 22:53
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Total number of employees = 8;
To form a group of 4 employees that contain Andrew and Bob out of 8, we only need to add 2 of the remaining 6 employees to Andrew and Bob. This can be achieved in 6C2 ways = 15 ways.

The total number of possible ways to form groups of 4 out of 8 = 8C4 = 70 ways.

The number of groups containing Andrew and Bob as a fraction of the total possible groups = 15/70 = 3/14.

The answer is C.
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   15 Jan 2020, 10:19
Im confused

by using formula by theory given by chetan2u https://gmatclub.com/forum/topic215915.html#p1667366

Formula-


If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= \(\frac{(xa)!}{x!(a!)^x}\)

so total cases are 35, and no of groups in which A and B are together is 6C2 =15
my answer is 15/35.

Can someone clarify.
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   15 Jan 2020, 10:22
SREYNANS wrote:
Im confused

by using formula by theory given by chetan2u https://gmatclub.com/forum/topic215915.html#p1667366

Formula-


If you divide 'xa' items in 'x' groups of 'a' items each, the number of ways= \(\frac{(xa)!}{x!(a!)^x}\)

so total cases are 35, and no of groups in which A and B are together is 6C2 =15
my answer is 15/35.

Can someone clarify.
Regards
Sreynans


Ok got it, the no.of groups would be double the no. of ways. so 15/70 =3/14
Ans C
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   21 Jan 2020, 09:36
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Bunuel wrote:

Competition Mode Question



Eight persons in an organization including Andrew and Bob were to be divided in two groups of 4 members each. The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed?

A. 1/4
B. 1/70
C. 3/14
D. 1/10
E. 11/14


The total number of 4-person groups is:

8C4 * 4C4 =(8 x 7 x 6 x 5) / (4 x 3 x 2) * 1 = 70

Now let’s determine the total number of 4-person groups containing both Andrew and Bob.
Since Bob and Andrew are guaranteed to be chosen (2C2 = 1), we have 6 remaining individuals to fill the 2 remaining slots:

2C2 * 6C2 = 1 * (6 x 5)/2 = 15

Therefore, the probability is 15/70 = 3/14.

Answer: C
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   30 Jan 2020, 22:02
Question: The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed

Bolded 1: The total number of groups containing both Andrew and Bob= 8C4= 70

Bolded 2: total number of groups which can be formed = 6C2=15
6 represents the remaining employees after removing Andrew & Bob
2 represents the vacancies left from an initial 4 which was taken by Andrew & Bob

Answer= 15/70=3/14 (C)
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   20 Oct 2020, 10:31
Can you please explain the logic of how you get 6C2?

Archit3110 wrote:
total possible groups l 8c4 ; 70
group containing andrew and bob ; 6c2 ;
P ; 15/70 ; 3/14
IMO C

Eight persons in an organization including Andrew and Bob were to be divided in two groups of 4 members each. The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed?
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   20 Oct 2020, 10:42
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It's a given condition that A&B both have to be part of the selection so in that case we combine them say X which can be chosen in 2c2 ways
then we are left with 6 members & 2 members are to be chosen. 6c2
2c2:1
1*6c2
Tiburcio1987




Tiburcio1987 wrote:
Can you please explain the logic of how you get 6C2?

Archit3110 wrote:
total possible groups l 8c4 ; 70
group containing andrew and bob ; 6c2 ;
P ; 15/70 ; 3/14
IMO C

Eight persons in an organization including Andrew and Bob were to be divided in two groups of 4 members each. The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed?


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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   20 Oct 2020, 13:33
Can anyone please explain why the total no: of cases are 8C4 and not 8C4/2

Say A, B, C, D, E, F , G, H are 8 persons

Two groups of ABCD and EFGH will be same as EFGH and ABCD.

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Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   Updated on: 24 Apr 2022, 06:21
Expert Reply
Bunuel wrote:

Competition Mode Question



Eight persons in an organization including Andrew and Bob were to be divided in two groups of 4 members each. The total number of groups containing both Andrew and Bob is what fraction of the total number of groups which can be formed?

A. 1/4
B. 1/70
C. 3/14
D. 1/10
E. 11/14


Since the 8 people are divided into two groups of 4, Andrew must be selected for one of the two groups.
If we select one of the two groups:
Since Andrew has an equal chance of being included in either group, P(Andrew is included in the selected group) = 1/2
Since there are 3 spots left in Andrew's group -- and 7 remaining people who could occupy these 3 spots -- P(Bob is one of the 3 people selected to join Andrew) = 3/7
Thus:
P(Andrew and Bob are in the same group) = 1/2 * 3/7 = 3/14

Since the probability that Andrew and Bob are in the same group = 3/14, they must be included together in 3/14 of all possible 4-member groups.

Show Spoiler
C

Originally posted by GMATGuruNY on 23 Apr 2022, 06:50.
Last edited by GMATGuruNY on 24 Apr 2022, 06:21, edited 2 times in total.
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink] New post   24 Apr 2022, 03:48
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Case 1:
How many pairs can be formed from four people A, B, C and D?
Answer: 4C2 \(= \frac{4*3}{2*1} = 6\)
The approach above counts the following options:
AB, AC, AD, BC, BD, CD

Case 2:
How many ways can four people be divided into two pairs?
Answer: \(\frac{4C2}{2!} = \frac{6}{2} = 3\)
The approach above counts the following options:
AB and CD
AC and BD
AD and BC

Akshat1994 wrote:
Can anyone please explain why the total no: of cases are 8C4 and not 8C4/2

Say A, B, C, D, E, F , G, H are 8 persons

Two groups of ABCD and EFGH will be same as EFGH and ABCD.

Posted from my mobile device


How many groups of 4 can be formed from 8 people?
This question is similar to Case 1 above.
Thus:
Number of possible 4-person groups = 8C4
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Re: Eight persons in an organization including Andrew and Bob were to be d [#permalink]
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