Question 628135
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I think what you are looking at is *[tex \LARGE {{n}\choose{r\,-\,1}}] (read: "n choose r - 1") which is the number of combinations of *[tex \LARGE n] things taken *[tex \LARGE r\ -\ 1] at a time.  Such as, how many ways can I choose from 10 different books on a shelf if I take them 3 at a time and I don't care what order the three are in when I get them in my hand?


In general, *[tex \LARGE {{n}\choose{k}}] is calculated by *[tex \LARGE \frac{n!}{k!(n\,-\,k)!].  Hence, your *[tex \LARGE \frac{6!}{3!(6\,-\,3)!}\ =\ 20] is exactly correct given *[tex \LARGE n\ =\ 6] and *[tex \LARGE r\ =\ 4].


Here's the logic:  Let's say you have 6 things and want to choose 3 of them.  There are 6 ways to choose the first one, then since you didn't replace the first one you chose, there are 5 ways to choose the second one for each one of the 6 ways to choose the first one.  Then there  are 4 ways to choose the third one for each of the 30 ways to pick the first two, which works out to *[tex \LARGE 6\ \times\ 5\ \times\ 4] which comes from *[tex \LARGE \frac{6!}{(6-3)!}\ =\ \frac{6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2}{3\ \times\ 2}].  But that number is too large by a factor of the number of ways to arrange the three things in your hand, namely *[tex \LARGE 3!], and that is where the other denominator factor comes from.


You might want to compare this to the number of permutations of *[tex \LARGE n] things taken *[tex \LARGE k] at a time.  With permutations, order matters.  Such as you have 20 people in your club and you want to know how many different ways you can select a President, Secretary, and Treasurer.  Here order matters because Suzy being the president is a different outcome than Suzy being the Secretary, for example.  Permutations are calculated *[tex \LARGE _nP_r\ =\ \frac{n!}{(n\,-\,r)!}].  See the difference?


By the way, this is also the *[tex \LARGE k]th coefficient (counting from zero) of the binomial expansion of *[tex \LARGE \left(x\ +\ y\right)^n] and the *[tex \LARGE k]th element of the *[tex \LARGE n]th row of Pascal's Triangle.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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