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.
Please check my solution
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
Best Answer
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 $\sum_1^{n-1}\binom{n}{i}$.