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Consider the number of way a committee of 3 can be selected [#permalink]

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20 Apr 2010, 11:09

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Consider the number of way a committee of 3 can be selected from 7 people- A, B, C, D, E, F, G if order does not matter, and C and E cannot be chosen together.

Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:

Order does not matter,

and C and E cannot be chosen together.

What is the justification?

\(C^3_7-C^1_5=30\)

\(C^3_7\) - # of ways we can choose any 3 out of 7 (without restriction). \(C^1_5\) - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is \(C^1_5\)).
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Because only 1 way to choose C and 1 Way to choose E?

Therefore you minus that from the complete total ways to pick 3 from 7?

Yes. You can write this as \(C^1_1*C^1_1*C^1_5=5\) (one way to choose C and one way to choose choose E) OR \(C^2_2*C^1^5=5\) (one way to choose C and E, from C and E), which is basically \(C^1_5=5\).

ksharma12 wrote:

Essentially the same thing as at least 1 problem?

If you mean in a way: total-opposite, then yes.
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Re: combination problem justification [#permalink]

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09 Sep 2012, 23:11

Bunuel wrote:

ksharma12 wrote:

Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:

Order does not matter,

and C and E cannot be chosen together.

What is the justification?

\(C^3_7-C^1_5=30\)

\(C^3_7\) - # of ways we can choose any 3 out of 7 (without restriction). \(C^1_5\) - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is \(C^1_5\)).

just for discussion sake and to clear the concept I was wondering what would happen if the order mattered in the above scenario. Eva ..Bunuel and others in the community request you to please contribute .

lets assume ABCDEFG is a 7 letter word so ABC is different from CAB meaning order matters so if I wanted to form 3 letter words such that CE should not be together then , what would be the number of ways this could be done ?

I tried to a certain extent please verify if this is correct or not.

\(C^7_3 * 3! - C^1_1 *C^1_1*C^5_1*3!\)=180

\(C^7_3\) = # of ways to select 3 out of 7 3! = # of ways to arrange them among themselves as order matters

\(C^1_1\)= choosing c \(C^1_1\) = choosing e \(C^5_1\)= # of ways to select one from remaining 5 , after C and E have been chosen in the community. \(3!\) = # of to arrange the 3 letter committee containing both C and E

Re: combination problem justification [#permalink]

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10 Sep 2012, 03:11

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This post received KUDOS

stne wrote:

Bunuel wrote:

ksharma12 wrote:

Consider the number of way a committee of 3 can be selected from 7 people- A,B,C,D,E,F,G if:

Order does not matter,

and C and E cannot be chosen together.

What is the justification?

\(C^3_7-C^1_5=30\)

\(C^3_7\) - # of ways we can choose any 3 out of 7 (without restriction). \(C^1_5\) - # of groups with C and E together (if C and E are in chosen group, then third member can be any out of 5 left, so total # of groups is \(C^1_5\)).

just for discussion sake and to clear the concept I was wondering what would happen if the order mattered in the above scenario. Eva ..Bunuel and others in the community request you to please contribute .

lets assume ABCDEFG is a 7 letter word so ABC is different from CAB meaning order matters so if I wanted to form 3 letter words such that CE should not be together then , what would be the number of ways this could be done ?

I tried to a certain extent please verify if this is correct or not.

\(C^7_3 * 3! - C^1_1 *C^1_1*C^5_1*3!\)=180

\(C^7_3\) = # of ways to select 3 out of 7 3! = # of ways to arrange them among themselves as order matters

\(C^1_1\)= choosing c \(C^1_1\) = choosing e \(C^5_1\)= # of ways to select one from remaining 5 , after C and E have been chosen in the community. \(3!\) = # of to arrange the 3 letter committee containing both C and E

Hope this is correct?

Yes, this is correct.

I just have my own preferences to count...so, for example, choose 3 out of 7 when order matters I write directly 7*6*5, meaning I directly take into account the order (why write the formula with the factorials for nCk and then multiply by k!, reduce...). First choice 7 options, second choice 6, third 5. And I am not even writing the 1C1 factors. There is nothing wrong with it, but I know that C and E must be chosen, then I need just one extra person (letter ), so I can choose 1 out of 5, for which again I am not writing the 5C1, and then having C, E and * (somebody), I have to consider all the permutations of the three, so 3!...
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Consider the number of way a committee of 3 can be selected [#permalink]

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22 Jan 2014, 15:31

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Re: Consider the number of way a committee of 3 can be selected [#permalink]

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22 Jan 2014, 16:26

Bunuel, can we also think of it as 6C3 + 6C3 - 3C5?

I am trying to calculate combinations each C and E with other 5 separately, adding them together, then subtracting out one of the double counted 3 team combinations of the other 5 members. Not sure if this is a proper approach? Thanks!
_________________

Bunuel, can we also think of it as 6C3 + 6C3 - 3C5?

I am trying to calculate combinations each C and E with other 5 separately, adding them together, then subtracting out one of the double counted 3 team combinations of the other 5 members. Not sure if this is a proper approach? Thanks!

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