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# If 4 people are selected from a group of 6 married couples

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Manager
Joined: 29 Nov 2011
Posts: 114

Kudos [?]: 24 [0], given: 367

Re: If 4 people are selected from a group of 6 married couples [#permalink]

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04 Apr 2016, 23:05
HI Bunnel

Number of ways to select...Atleast one married couple is

6C1- Choose 1 married couple for 2 seats
10C2- Chose 2 members from the remaining 5 couples for the remaining 2 seats

6C1*10C2= 270
No of ways to select 4 people from 12=12C4

P= 270/495

Answer=1- the prob of the above
=5/11

What am I missing here?

Kudos [?]: 24 [0], given: 367

Manager
Joined: 01 Nov 2016
Posts: 71

Kudos [?]: 9 [0], given: 70

Concentration: Technology, Operations
Re: If 4 people are selected from a group of 6 married couples [#permalink]

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27 Mar 2017, 20:51
What is the chance that four chosen have nobody married to each other? Well we want to find $$\frac{chance of success}{number of outcomes}$$. Without using combitronics:

Chance of success is 12*10*8*6
Total outcomes are 12*11*9*8

$$\frac{12*10*8*6}{12*11*10*9}=\frac{8*6}{11*9}=\frac{16}{33}$$

Kudos [?]: 9 [0], given: 70

Manager
Joined: 12 Nov 2016
Posts: 124

Kudos [?]: 12 [0], given: 71

Location: Kazakhstan
Concentration: Entrepreneurship, Finance
GMAT 1: 620 Q36 V39
GMAT 2: 650 Q47 V33
GPA: 3.2
Re: If 4 people are selected from a group of 6 married couples [#permalink]

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16 Dec 2017, 07:51
metallicafan wrote:
+1 D

A faster way to solve it:

$$\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}$$

Isn't it a short and great approach? is there anything wrong with it? Why for problems like this is combinatorics a preferred approach?

Kudos [?]: 12 [0], given: 71

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 810

Kudos [?]: 379 [1], given: 42

Location: India
GPA: 3.82
Re: If 4 people are selected from a group of 6 married couples [#permalink]

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16 Dec 2017, 08:41
1
KUDOS
Erjan_S wrote:
metallicafan wrote:
+1 D

A faster way to solve it:

$$\frac{12}{12} * \frac{10}{11} * \frac{8}{10} * \frac{6}{9} = \frac{48}{99} = \frac{16}{33}$$

Isn't it a short and great approach? is there anything wrong with it? Why for problems like this is combinatorics a preferred approach?

Hi Erjan_S

There is nothing wrong with the approach. As the numbers are simple so you can simply count the possible and total outcomes and then keep on removing the pairs. There could be multiple ways to solve counting problems. P & C is preferred because you get the value through direct formula. in case of simple numbers you can manually count the number of possible ways or selections, but when the numbers or situation become a bit complex then you will have to rely upon P & C.

Kudos [?]: 379 [1], given: 42

Re: If 4 people are selected from a group of 6 married couples   [#permalink] 16 Dec 2017, 08:41

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# If 4 people are selected from a group of 6 married couples

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