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I want to arrange n red and 5 orange balls in a way so all the orange balls are adjacent to at least one other orange one.

I'm thinking that there either there will be a group of 5 orange balls, a group of 3 or a group of 2, so adjacency could take place. Where to from here? or should i take a different approach ?

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Your analysis is fine as far as it goes. Now consider the two cases. If the $5$ orange balls form a single group, there is one arrangement for every possible placement of that group in a line of $n$ red balls; how many places are there to put it?

If the orange balls are split $2$ and $3$ or $3$ and $2$, you have to pick $2$ of the positions in the line of $n$ red balls, and then you can insert the two groups of orange balls in $2$ ways, either $3$ and then $2$, or $2$ and then $3$, so you’ll get something of the form $2\binom{?}2$. Can you finish it from there?

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    hmmm so (n + 1) for 5 and (n+2) for the other cases .. does the 1 represent the 5 balls as one unit, and the 2 for one unit being 3 and the other unit being 2?2012-09-21
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    There is a problem in the second analysis if you choose your two positions if your line of $n$ red balls next to each other ; in that case you are double counting, because permuting the $2$ and $3$ gives you the same arrangement.2012-09-21
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    @Beginnernato: You get $n+1$ in the first case simply because there are $n+1$ ‘slots’ in a string of $n$ red balls: $n-1$ between adjacent balls, plus one at each end. When you have two groups of orange balls, you have to pick two of these $n+1$ slots, which you can do in $\binom{n+1}2$ ways; there’s no $n+2$ here.2012-09-21
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    @Patrick: No, it doesn’t. That would happen only if you put the two groups into the *same* slot.2012-09-21
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    @Brian : I guess I understood you wrong then. Actually I read you a little be more carefully and we have the same answer but I give a different way to get there.2012-09-21
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    @Patrick: You do not have a different answer. Read mine again.2012-09-21
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    So n+1 + 2((n+1) choose 2)) ? If we use (n+1) no overcounting would take place right?2012-09-21
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    @Beginnernato: That’s right. We’re counting separately the cases in which the $5$ orange balls are all together and the cases in which they’re split into two groups. Now you can simplify the expression and get $n^2+2n+1$, or if you prefer, $(n+1)^2$.2012-09-21
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    @BrianM.Scott Could you help me with http://math.stackexchange.com/questions/200931/probability-with-cards ?2012-09-23
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    @Beginnernato: It’ll be a few minutes before I get to it, but I will.2012-09-23
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    @BrianM.Scott Thanks a lot Brian2012-09-23
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I think the easiest way to think about it is this : The case where you have $5$ adjacent orange balls is actually just the case where you have $2$ orange-$3$ orange balls, but that those two groups are now sticked together. So replace the original problem by this : You have $n$ red balls, one $2$-group of orange balls and one $3$-group of orange balls, i.e. $3$ types of objects. Therefore there are $\begin{pmatrix} n + 2 \\ n,1,1 \end{pmatrix} = (n+2)(n+1)$ ways to arrange those. But when the orange balls are in a group of $5$, you're double counting, because it is the same thing if you place the group of $2$-orange balls to the left or to the right of the group of $3$-orange balls when they are next to each other, so in this count you remove once the number of cases where the $5$ orange balls are together, which gives you a final count of $$ \begin{pmatrix} n+2 \\ n,1,1 \end{pmatrix} - (n+1) = (n+2)(n+1) - (n+1) = (n+1)^2 = n^2 + 2n + 1. $$