We learnt , about counting arrangements of objects, where each arrangement is treated as distinct from a different arrangement of the same objects.
For example the arrangement $\unicode{0x201C} bac \unicode{0x201D}$ is distinct from the arrangement $\unicode{0x201C} cba \unicode{0x201D}$ of the letters $a$, $b$ and $c$.
In contrast, to the above, the word $combination$ means combining (or choosing) some number of things from a given set of things, where order does not matter. For example the selection $\{a,\ b\}$ from the set $\{a,\ b,\ c,\ d\}$ is same as selecting $\{b,\ a\}$.
Let us say we are given the problem of selecting $3$ types of fruits for a fruit salad from a basket containing $5$ different types of fruits, namely $Apples,\ Bananas,\ Cherries,\ Dragon\ fruits$ and $Elderberries$, in short $\{a,\ b,\ c,\ d,\ e\}$
As we learnt , we can form an arrangement of $3$ fruits from a set of $5$ fruits in $\xaperm{5}{3}$ ways, that is:
$\dfrac{5!}{(5 - 3)!} = \dfrac{5!}{2!}$
Now, we already know that these arrangements with contain the arrangements:
Since in case of selections, order is not important, all of the above $6$ selections are identical.
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Similarly, for any other choice of $3$ fruits we will again have $3! = 6$ possible arrangements, that should be treated as the same.
Therefore, the actual number of distinct selections will be:
$\xaperm{5}{2} \div 3! = \dfrac{5!}{2!\ 3!}$
For the general case of choosing $r$ items from a set of $n$ distinct objects, the total number of ways is:
$\dfrac{n!}{r!\ (n - r)!}$ ways.
Using standard notation of combinations, this is represented as $\xacomb{n}{r}$ and is commonly read as $\unicode{0x201C} n\ choose\ r \unicode{0x201D}$.