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I have the following HW question:

Recall that the number of solutions in $\mathbb{N}$ (including $0$) for $\sum_{j=1}^{m}a_{j}=n$ is $\binom{n+m-1}{n}$

Also Recall that the number of solutions in $\mathbb{N}\backslash\{0\}$ for $\sum_{j=1}^{m}a_{j}=n$ is $\binom{n-1}{m-1}$

Use these to solve the following: What is the probability that in a deck of $52$ cards no two aces are next to each other ?

My efforts:

First I arrange the $52-4$ cards (there are $48!$ ways of doing so), now I would like to take my $4$ aces and count the number of ways of putting them into the deck of the $48$ cards in a way that at least two are next to each other (or maybe find the complement).

I know exactly how to use the reminder: I see it this way - before the first ace there is some number $a_{1}$ of cards, after the first ace and before the second one there are $a_{2}$ cards...and after the last ace there are $a_{5}$ cards and $a_{1},a_{5}\geq0$ and $a_{2},a_{3},a_{4}>0$ because no two aces are next to each other.

Here I am stuck and I could use some help.

1 Answers 1

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Hint: "No two Aces next to each other" is easier. Look at the $48$ non-Aces. These determine $49$ gaps, including the end gaps. We need to choose $4$ of these.

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    @Henry - I just looked at the answer and your comment again, I agree with you. thanks!2012-12-02