Probability - Cominations and Integer Valued Vectors

AsianMan
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This problem comes from Sheldon Ross's book "A First Course in Probability (6th ed)."

There are 5 hotels in a certain town. If 3 people check into hotels in a day, what is the probability that they each check into a different hotel?


Attempt at a solution:

There are 5C3 = 10 different combinations of hotels where each individual person picks a different hotel.

I also decided that there were 7C4 = 35 possible ways for 3 individuals to choose from the 5 hotels, if more than 1 can stay in the same hotel. I got this answer because there are (n+r-1)C(r-1) distinct nonnegative integer-valued vectors (x1,x2,...,xr) satisfying x1 + x2 + ... + xr = n, where n = 3 and r = 5.

Therefore, I got 10/35 as my answer, but the answer is actually .48 (rounded?)

Interestingly, I got very close this answer mistakenly at first by dividing 5C3 by 7C2.
 
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Your'e making it too complicated

\frac {P(5,3)}{5^{3}}
 
The first person arrives and checks into any hotel. The second person arrives and checks into a hotel. What is the probability that person checks into a different hotel? The third person arrives. What is the probability this person checks into yet a different hotel? The probability that they check into three different hotels is the product of those two probabilities.. This is exactly the same as Random Variable gives- although, Random Variable, it would be better not to just "give" answers. Especially in the "coursework and homework sections".
 
Wow, I'm embarrassed.

Thanks guys!

BTW, is there any way to do it the way that I was doing it?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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