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.

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.

Yes, if Hilary Clinton runs as an independent, she will win.

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 !