Combinatorics is the field of mathematics involving combinations and permutations.
Permutations are the various order-sensitive subsets of a larger set. Customarily, the
larger set is said to have n elements, whereas the subset contains k elements (A = {a1, a2,
a3, … , an-1, an} and a permutation is = {b1, b2, b3, …, bk-1, bk}  A.) The number of
permutations would be denoted nPk with these variables.
Combinations, on the other hand, are order-insensitive subsets. All permutations
of a set that have different orders but contain all the same elements count for only one
permutation. Thus, there are fewer combinations than permutations of a given set. The
total numbers of combinations can be denoted by nCk , but the nomenclature “n choose k”
 n
or   is equally acceptable.
k
The primary operator in combinatorics is the factorial. The factorial of a number is by
definition equal to all the integers between one and the number inclusive, multiplied
together. The factorial of n is written n!.
The factorials relevance in combinatorics is due to its importance in finding the
number of possible orders of elements in a set. In other words, nPn = n!. IT only works if
all elements are distinct. The proof of this statement is rather simple. Let’s sat we have a
set of n=6 elements. It’s a word: ABCDEF. What if we want to find the number of
possible “words” we can form from all of these letters used once? We can consider it as
thought there are six blank spaces to fill with these letters. In the first place, we can place
any of the six. In the second, only five letters remain to be placed. The 4, and 3, and then
2, until finally there is only one letter left from the set to be put in the last blank space.
Thus there are 6  5  4  3  2  1 total permutations with n elements. This means 6P6= 6!.
If you have any number n of distinct elements comprising a set, nPn will equal n!.
The case is different if a set of n elements contains some non-distinct elements.
We can’t use the simple formula for n distinct elements , because you cannot tell the
difference between, for instance, the two N’s in the word BANNED. Thus, for every
distinct permutation, there are 2! non-distinct permutations with the two N’s reversed.
There must be
6!
= 360 distinct permutations of the word BANNED. Now, if we have
2!
the word BANANA, which has more than one set of indistinct permutations for every
distinct permutation. There are
(# letters )
6!
=
= 120 distinct permutations of the
(# A' s)( N ' s)
3!2!
word BANANA.
With this information, we can now figure out the number of permutations of a set
of n distinct elements, for any k. To find the formula, we must look at how many
elements we don’t care about (they can be considered indistinct), how many we do care
about (distinct), and how many we can choose out of. We are choosing out of a set of n
elements. Although the elements left outside of a permutation (n-k elements) may be
distinct, we do not care about their order. There are k elements in each permutation, and
these are the ones whose orders are relevant. Thus, nPk =
n!
k!(n  k )!
If we are trying to find the number of combinations, n choose k, then the only
distinction is that the order of the k elements to be chosen does not matter, so the formula
 n
n!
is nCk =   =
.
 k  k!(n  k )!