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

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Eight persons in an organization including Andrew and Bob were to be d  [#permalink]

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New post 14 Jan 2020, 00:45
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A
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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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Re: Eight persons in an organization including Andrew and Bob were to be d  [#permalink]

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New post 14 Jan 2020, 06:17
1
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]

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New post 14 Jan 2020, 08:51
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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]

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New post 14 Jan 2020, 13:22
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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]

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New post 14 Jan 2020, 13:56
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1
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]

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New post 14 Jan 2020, 21:53
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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]

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New post 15 Jan 2020, 09:19
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.
Regards
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Re: Eight persons in an organization including Andrew and Bob were to be d  [#permalink]

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New post 15 Jan 2020, 09:22
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]

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New post 21 Jan 2020, 08:36
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]

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New post 30 Jan 2020, 21:02
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] 30 Jan 2020, 21:02
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