Question 455916
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In this problem, order matters.  That is to say that Bob on June 1, Sue on June 2, Joanne on June 3, and Jasper on June 4 is a different outcome than, say, Bob on June 4, Sue on June 3, Joanne on June 2, and Jasper on June 1.  Since order matters, you need the number of permutations of 30 things taken 4 at a time.  How do we know it is 30?  Because June has 30 days.


The number of permutations of *[tex \LARGE n] things taken *[tex \LARGE k] at a time is given by:


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


in your case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{30!}{26!}\ =\ 30\ \times\ 29\ \times\ 28\ \times\ 27]


You can do your own arithmetic.


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