Question 584053
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It depends on whether order matters.  That is, is Q1, Q2, Q3, Q4 the same or different from Q2, Q1, Q3, Q4?


If order matters, that is if the two examples given above are actually considered different selections, then you want *[tex \LARGE _{10}P_4] where *[tex \LARGE _nP_r] is calculated by *[tex \LARGE \frac{n!}{(n\ -\ r)!}].  The idea being, in your example, that there are 10 ways to pick the first question and for each of these 10 ways there are 9 ways to pick the second question, so 10 times 9, and then for those 90 ways there are 8 ways to pick the third question, and so on.  The math is thus:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10\ \times\ 9\ \times\ 8\ \times\ 7\ \times\ 6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2}{6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2}\ =\ 10\ \times\ 9\ \times\ 8\ \times\ 7]


On the other hand, if order doesn't matter, using the above arithmetic overcounts by a factor of the number of ways you can arrange 4 things, namely 4!.  Hence, if order doesn't matter you need another factor in the denominator: You want *[tex \LARGE _{10}C_4] where *[tex \LARGE _nC_r] is calculated by *[tex \LARGE \frac{n!}{r!(n\ -\ r)!}].


And the arithmetic for your problem becomes:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{10\ \times\ 9\ \times\ 8\ \times\ 7\ \times\ 6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2}{(4\ \times\ 3\ \times\ 2)(6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2)}\ =\ \frac{10\ \times\ 9\ \times\ 8\ \times\ 7}{4\ \times\ 3\ \times\ 2}\ =\ 5\ \times\ 3\ \times\ 2\ \times\ 7]


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