# Combination question involving apples and oranges

Suppose that I have n apples and n oranges, I want to arrange the apples and oranges in a row such that no apples will be next to another apple and no oranges will be next to another orange.

attempt 1:

suppose apples and oranges are interchangeable (no order)

then we have $(n)*(n)*(n-1)*(n-1)*(n-2)...*(n-n+1)*(n-n+1)$ ways

attempt 2:

suppose apples and oranges are not interchangeable (ordered)

then we have ${\binom{n}{1} * \binom{n}{1}* \binom{n}{2}* \binom{n}{2}*...* \binom{n}{n-1}*\binom{n}{n-1}}$ ways, and this is equal to $\sum_{i=1}^{n-1}\binom{n}{i}$

Please inform me whether I've done this correctly and give pointers, thank you.

Note: suppose that each apple and orange has a unique number on them i.e. apple 1, apple 2....orange 1, orange 2

• Since you have the same number $n$ of apples and oranges, they must alternate. Thus if they're interchangeable, there are only two ways to order them. – Dave Sep 6 '14 at 3:37
• My interpretation is that all apples are identical, and all oranges are identical, so the answer is very simple. – André Nicolas Sep 6 '14 at 3:38

If the apples and oranges are individuals, perhaps because each has a student number, then there are only $2$ basic patterns allowed, $AOAOAO\dots AO$ and $OAOAOA\dots OA$.
In either case, the $n$ apples can be placed in the $A$ slots in $n!$ possible orders, and for each order the $n$ oranges can be placed in the $O$ slots in $n!$ ways, for a total of $2(n!)^2$.
But I think that unless we are told explicitly that the apples are distinct from each other, as are the oranges, the natural interpretation is that they are not, giving answer $2$.
Remark: Your first attempt yielded $(n!)^2$. That is close to right under the "distinct" hypothesis, except that it does not take into account that there are $2$ basic allowed patterns.
I have not understood the reasoning that may underlie the second attempt. The product you get is not equal to $\sum_1^{n-1}\binom{n}{i}$.