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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.

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up vote 3 down vote accepted

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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+1 though if your denominator is going to be $52!$ then you may need an extra $4! \times 48!$ in the numerator. – Henry Dec 2 '12 at 23:06
Thanks, so the answer is $1-\frac{48!4!\binom{49}{4}}{52!}\approx21.73\%$ ? – Belgi Dec 2 '12 at 23:08
I'll correct to include Henry's comment – Belgi Dec 2 '12 at 23:09
I think your answer is for at least one pair of Aces together. You did not need to subtract from $1$ – Henry Dec 2 '12 at 23:14
@Henry - I just looked at the answer and your comment again, I agree with you. thanks! – Belgi Dec 2 '12 at 23:52

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