Question 976201
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No, because there are 5 ways to choose who gets the large prize, then there are 4 ways to choose who gets the first $100 prize, then 3 ways to choose who gets the second $100 prize, then 2 ways to choose who gets the  first $50 prize, and finally 1 way to choose who gets the last $50 prize (so 5 times 4 times 3 times 2 times 1), all divided by 2 ways to arrange 2 $100 prizes times 2 ways to arrange 2 $50 prizes.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{5!}{2!\cdot 2!}\ =\ \frac{5\ \times\ 4\ \times\ 3}{2}\ = 30]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \