Combinations question

[Imranp]

So basically I am trying to figure out the # of possible combinations of a 3-person committee from a total of 13 members.

 

I used 13_C_3 and that gave me 286 which I think is correct. I am having trouble with the follow-up questions.

 

First one is, how many of the 286 committees would a particular person, let's say John, be on ? I manually calculated (using a spreadsheet) that number to be 66. Is this correct ? Is there a mathematical expression I can use to calculate that ?

 

Second, how many of the 286 committees would two particular people, let's say John OR Vanessa, be on ? I could expand me spreadsheet and by tedious, visual counting figure this out, but is there a mathematical expression I can use to calculate that ? I am thinking that I need to calculate the number of committees they are on individually (66 each ?) and then subtract the # of committees they are both on (11 committees ??) so that gives me 66+66-11 = 121.. does this make sense ?

 

 

Thanks !

[timo]

What is "13_C_3" and why do you use it to solve the first question? The number of committees that John is on in the number of combination that you can make out of John and two of the other twelve people. And "yes", your method to evaluate the last question seems okay. Can you explain why you substracted the number of committees that they are both on (which is correct - I am just checking if you understood what you did)?

Edited November 19, 2011 by timo

[Imranp]

By 13_C_3 I mean "3 choose 13", using n!/k!(n-k)!

 

Yes, but is there a mathematical expression to figure out how many committees John is on ? I had to count to get the answer, 66. Is this correct ? Is there a formula to calculate this ? Thanks.

[timo]

By 13_C_3 I mean "3 choose 13", using n!/k!(n-k)!

I know it as "13 over 3", but whatever. Why do you use it?

 

Yes, but is there a mathematical expression to figure out how many committees John is on?

Yes. And if you can answer my previous question and then reread my previous post the expression should be obvious.

I had to count to get the answer, 66. Is this correct ?

Yes.

 

Is there a formula to calculate this ?

Isn't that the same question as above?

 

Thanks.

You're welcome. You should try to answer my question to you, though. I don't ask them for fun or because I wouldn't know a correct answer to them myself. I'm asking them because you need to be able to answer them to understand the solution to your questions. I do know that students are often interested in having the answer, only, and do not really bother understanding them. But I also know that understanding the answers is actually the important part of homework. Edited November 19, 2011 by timo

[Imranp]

I thought I did answer your question.

 

You asked: What is "13_C_3" and why do you use it to solve the first question

I answered: By 13_C_3 I mean "3 choose 13", using n!/k!(n-k)!

Are we missing anything here ?

 

You use 13C3 to choose combinations for 3 from a pool of 13. I am not sure what' you are asking. Please rephrase, elaborate.

[timo]

Well, "3 choose 13" is not exactly a proper sentence. "You use 13C3 to choose combinations for 3 from a pool of 13" comes closer to a proper sentence (still sounds odd to me, but I am not a native speaker, anyways). How would you then calculate the number of possibilities to choose two people out of a set of twelve?

Btw.: No need to rush the answer. I'll go to bed soon, anyways.

Edited November 19, 2011 by timo

[DrRocket]

Well, "3 choose 13" is not exactly a proper sentence. "You use 13C3 to choose combinations for 3 from a pool of 13" comes closer to a proper sentence (still sounds odd to me, but I am not a native speaker, anyways). How would you then calculate the number of possibilities to choose two people out of a set of twelve?

Btw.: No need to rush the answer. I'll go to bed soon, anyways.

 

"13 choose 3" is the usual terminology in the mathematics community, and it is reflected in LaTex as well -- {13 \choose 3 }= \frac {13!}{3!10!} gives you

 

[math] {13 \choose 3 }= \frac {13!}{3!10!}[/math] .

 

and in general for "n choose m" you have

 

[math] {n \choose m} = \frac {n!}{m!(n-m)!}[/math]

[Xittenn]

 

Second, how many of the 286 committees would two particular people, let's say John OR Vanessa, be on ? I could expand me spreadsheet and by tedious, visual counting figure this out, but is there a mathematical expression I can use to calculate that ? I am thinking that I need to calculate the number of committees they are on individually (66 each ?) and then subtract the # of committees they are both on (11 committees ??) so that gives me 66+66-11 = 121.. does this make sense ?

 

 

Typical approaches to individual portions of your questions . . . . break this down into cases . . . .

 

Case 1: How many committees is John on?

 

We know that there are 13 possible committee members including John. Assume that John has been selected and that we need to select 2 more from the remaining.

 

 

Case 2: How many committees is Vanessa on?

 

We know that there are 13 possible committee members including Vanessa. Assume that Vanessa has been selected and that we need to select 2 more from the remaining.

 

 

Case 3: How many committees are they both on?

 

We know that there are 13 possible committee members including John and Vanessa. Assume that John and Vanessa have been selected and that we need to select 1 more from the remaining.

 

Now how many are one or the other on but not both?? Combinatorics and Permutations often involve a good deal of dissection to come to the proper answer. Be prepared to break the problems down into individual cases to get the figures required to combine the results appropriately and come to an answer. Spread sheets are really not the proper way to come to a solution, even though it is an approach that can be taken.