# Selection from identical objects - distinction between number of ways and number of outcomes

The number of ways in which $2$ oranges can be selected from $5$ identical oranges is given by $5 \choose 2$ . In another interpretation, since the oranges are identical, the number of ways is $1$. Is the following explanation correct ?

When we say the number of ways in which $2$ oranges can be selected from $5$ identical oranges is $5 \choose 2$, we are referring to number of ways in which the oranges can be taken in groups of $2$. The oranges look identical, but they themselves are distinct. Let's say we assign each orange a number. Then the problem becomes one of choosing two numbers from the set $\{1,2,3,4,5\}$. The answer is $5 \choose 2$.

In another interpretation, the answer is $1$. No matter which two oranges we pick, they look identical. But the answer $1$ refers to number of distinct outcomes. In the first situation(# of ways is $5 \choose 2$), we were more concerned with all possible variations in $2$-groupings but the order within the grouping did not matter. Here, we are more concerned with all possible variations in appearance.

If those oranges were distinct, then both the answers (# of ways and # of outcomes) would be $5 \choose 2$

• With $5$ identical oranges, there is $1$ distinct outcome, $5\choose2$ pairs to reach that outcome, and $2{5\choose2}$ ways to select those pairs. – user137794 May 3 '14 at 6:00
• There is no such thing as $5$ identical oranges. If they are identical, there is only one of them. Also, if we really mean a multi-set that contains one orange with multiplicity $5$, then selecting $2$ of them makes no sense, since they are the same. In how many ways can you choose $2$ roots of the polynomial $x^5$, and what would that mean? – Marc van Leeuwen Apr 23 '18 at 14:13

When a combinatorialist talks about counting selections from a set of indistinguishable objects, she means counting orbits under the action of a symmetric group permuting those objects. In the example of the question, take $\mathbf S_5$ acting on a set of $5$ distinct objects, there are $6$ orbits on its power-set, one of which is the orbit of $2$-element subsets. So if you must consider this type of problem, there is only one way to select $2$ from a set of $5$ indistinguishable elements (but there are $6$ ways to select some elements).
This is a pretty standard combinatorial argument, where enumerating all objects gives $N$, but each object will be counted $d$ times, and the number of objects is $N/d$. You're just considering the trivial case, where you find $N$, but all objects are identical, thus $d = N$.
A convenient way is to use generating functions. Suppose you have objects of types $1,2,\dots,n$, with an infinite number of objects of each type. Any two objects of the same type are identical, and the order of the objects in the draw doesn't matter. The number of ways (I will use "the number of ways" only as a synonym for "the number of outcomes") to choose $k$ objects is $$[x^k] (1 + x + x^2 + \dots)^n = [x^k] (1 - x)^{-n} = {-n \choose k} (-1)^k = {n + k - 1 \choose k},$$ which is the multiset coefficient. The idea is that choosing $j$ objects of type $i$ corresponds to taking $x^j$ from the $i$th factor. But if you have only $m_i$ objects of type $i$, then the number of ways becomes $$[x^k](1 + x + \ldots + x^{m_1}) \cdots (1 + x + \ldots + x^{m_n}) = [x^k](x - 1)^{-n} \prod_{i=1}^n (x^{m_i + 1} - 1).$$ If each $m_i = 1$, this gives the binomial coefficient. For $n = 1, m_1 = 5$, we get $[0 \leqslant k \leqslant 5]$.
Your interpretation "the number of ways $1$" is not correct since in probability there are two kind of random experiences:
1. those who neglect the order and hence the position of your oranges does not matter that is no matter if you pick up the orange 1 then orange 2 the most important is that you will get two specific distincts oranges and then you have to use $\left(_2^5\right)$ since the order don't matter: its interpretation combination of $2$ among $5$ "distincts: because every single orange represent only itself as an orange but similar since they belong to the same kind"
2. those who does not neglect the order that is "the order matter". In such cases you have to use arrangement instead of combination. Its interpretation the oranges holds numbers saying from $1$ to $5$ and you pick the first then you pick the second. It differs also when you pick up and drop the orange back to repeat the experience again or you hold the first orange while picking up another.