Question 445117
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That depends on how you are using the word "Combination."  In the strict mathematical sense it means how many different ways to choose groups where the order of elements in an individual group doesn't matter.  Furthermore, elements are typically not repeated.


So, if ROY is a different combination than YRO or ORY, or if RRB is a valid choice, then you are using combination in a non-standard way.  In such case, there are 7 ways to chose the first of the three times 7 ways to chose the second of the three times 7 ways to pick the third element, for a total of 343 different arrangements.


If order matters but you can't duplicate, then there are 7 ways to pick the first element, 6 ways to pick the second element, and 5 ways to pick the 3rd element for a total of 210 different arrangements.


However, the number of combinations (where order doesn't matter) of n things taken k at a time is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(n\cr k\right\)\ =\ \frac{n!}{k!(n\,-\,k)!]


So for 7 things taken 3 at a time:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(7\cr 3\right\)\ =\ \frac{7!}{3!(7\,-\,3)!}\ =\ \frac{7!}{3!4!}\ =\ \frac{7\,\cdot\,6\,\cdot\,5}{3\,\cdot\,2}\ =\ 35]


Notice that this result is the previous result divided by 6, which, by the way is the number of ways that three different things can be arranged, for example  ROY, RYO, YOR, YRO, OYR, ORY.


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