Combinations Question

[TGE]

Please see the problem below:

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

(A) 16
(B) 24
(C) 26
(D) 30
(E) 32

OA: E

Any help would be greatly appreciated. Thanks.

[cramya]

Couples of ways:

First method

We need to choose 3 people

First person we hve 8 choices

For the 2nd person(out of 7 remaining) we have 6 choices (since one of them would make a couple with the first person already chosen)


For the 3rd person(out of 6remaining) we have 4 choices (since two of them would make a couple with the first person/second person already chosen) = 4

Since we have permuted above where the order is prevalent we need to divide by 3!

8*6*4 / 3! = 32


Second way:

8C3 is the number of ways we can choose 3 members from 8 members = 56

We can find the number of arrangements where the couple are together and then subtract this number from 56 to get to the number of arrangements without any couples

Lets say the couples are AB CD EF GH

Lets fix AB in a commitee.

We have 6 choices for the 3rd person. Since there are 4 married couples each couple pair will have 6 different committees including them

6*4 = 24

Number of committes withiut any couples = 56-24 = 32


Hope this helps!


Regards,
CR

[TGE]

Pure genius. Thanks, cramya.

[apple100]

Couples of ways:

First method

We need to choose 3 people

First person we hve 8 choices

For the 2nd person(out of 7 remaining) we have 6 choices (since one of them would make a couple with the first person already chosen)


For the 3rd person(out of 6remaining) we have 4 choices (since two of them would make a couple with the first person/second person already chosen) = 4

Since we have permuted above where the order is prevalent we need to divide by 3!

8*6*4 / 3! = 32


Second way:

8C3 is the number of ways we can choose 3 members from 8 members = 56

We can find the number of arrangements where the couple are together and then subtract this number from 56 to get to the number of arrangements without any couples

Lets say the couples are AB CD EF GH

Lets fix AB in a commitee.

We have 6 choices for the 3rd person. Since there are 4 married couples each couple pair will have 6 different committees including them

6*4 = 24

Number of committes withiut any couples = 56-24 = 32


Hope this helps!


Regards,
CR

For the first method, can you explain why you divided by 3!

8*6*4 / 3! = 32

[TGE]

Apple100,

You divide by 3! since we don't care about the order, and there are 3! different ways (i.e., permutations) to arrange the 3 people. So you have to divide the number of permutations (i.e., 8 x 6 x 4) by 3! so that you count the arrangements of any 3 people only once.

Hope this helps. Anyone - please feel free to jump if I missed anything.

[cramya]

U can analyze it coceptually or go by the definition

In a problem where the order of the elements is important and where the number of elements selected(R) is smaller than the list of items(N)

Total number of permutations (NpR)= N ! / (N-R) !

In a problem where the order of the elements is NOT important and where the number of elements selected(R) is smaller than the list of items(N)

Total number of combinations (NcR) = N ! / R! (N-R) !

In the problem above the order was not important since a commitee of ABC, CBA,CAB is all the same (i.e with three members A,B,C order not important)

Since there were 3 items selected (R) I divided by 3! (R!) (TO ELIMINATE ALL THE DUPLICATE COUNTS)

With ABC only 1 combination count possible but 6 different permutation count possible

8*6*4 would be 6 times the number of counts that would be applicable when the order is not important so I divide by 3! = 6



Hope this helps!


Regards,
CR