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Consider the number of way a committee of 3 can be selected

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

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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.
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Re: combination problem justification [#permalink]

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New post 20 Apr 2010, 11:28
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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 \(C^1_5\)).
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Re: combination problem justification [#permalink]

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New post 20 Apr 2010, 12:27
When you say C(1,5) do you mean 1x1xC(1,5)?

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?


Essentially the same thing as at least 1 problem?
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Re: combination problem justification [#permalink]

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New post 20 Apr 2010, 12:42
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ksharma12 wrote:
When you say C(1,5) do you mean 1x1xC(1,5)?

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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New post 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

Hope this is correct?
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Re: combination problem justification [#permalink]

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New post 10 Sep 2012, 03:11
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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 :o), 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!...
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Re: combination problem justification [#permalink]

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New post 10 Sep 2012, 06:27
Thanks for the reassurance
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Re: Consider the number of way a committee of 3 can be selected [#permalink]

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

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

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New post 23 Jan 2014, 03:25
m3equals333 wrote:
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!


Your approach is correct.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Consider the number of way a committee of 3 can be selected   [#permalink] 23 Jan 2014, 03:25
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