Combination

The title of this article is in itself explanatory as to what is expected to be learned from this article. You must have seen that to denote the combination in mathematics or statistics, etc., as . This symbol means “the number of ways to choose ‘r’ objects from a total of ‘n’ objects. Let’s take a few examples to understand how to read this symbol.

$$ \text{The number of ways to choose 5 unique objects from a set of 10 unique objects is} \newline \text{$^{10}C_5$ or $\binom{10}{5}$} $$

$$ \text{The number of ways to choose 3 unique objects from a set of 7 unique objects is} \newline \text{$^{7}C_3$ or $\binom{7}{3}$} $$

$$ \text{The number of ways to choose 9 unique objects from a set of 12 unique objects is} \newline \text{$^{12}C_9$ or $\binom{12}{9}$} $$

{Important note: For understanding combination, one needs to understand the concept of factorials well enough to proceed with this.}

Now, having understood the way to write combinations using the abovementioned symbols, let’s take a look at the following images:

The above image is known as Pascal’s triangle, where from the third row, the number in the middle is the sum of the numbers in the above row (to the left and the right of the middle number in the row below). For example, 3 in the third row (left) is the sum of 1 and 2 in the second row (left position).
Similarly, 20 in the middle position in the last row is the sum of 10 and 10 in the fifth row.

Now let’s take a look at the same triangle using the symbol of combination.

This image above is of the same Pascal triangle, but using the combination symbols. Compare this image with the first one and you will notice that the value of the combination symbol matches with the numbers in the triangle above. Which means:
$$ \text{1. The number of ways to choose 1 unique objects from a set of 6 unique objects is 6} \newline \text{$^{6}C_1$ or $\binom{6}{1}$ = 6} $$
$$ \text{2. The number of ways to choose 3 unique objects from a set of 6 unique objects is 20} \newline \text{$^{6}C_3$ or $\binom{6}{3}$ = 20} $$
$$ \text{3. The number of ways to choose 3 unique objects from a set of 5 unique objects is 10} \newline \text{$^{5}C_3$ or $\binom{5}{3}$ = 10} $$
$$ \text{4. The number of ways to choose 3 unique objects from a set of 4 unique objects is 4} \newline \text{$^{4}C_3$ or $\binom{4}{3}$ = 4} $$

Practice with these two triangles so that you will start getting more clarity about the concept using the tools given above. You can keep making Pascal’s triangle by repeating the process to any value, but this will eventually become brute force work, and if you have to find a value like , it will be very lengthy to write down the entire triangle for 50 rows and then find the 21st element in that row by summing up the 20th and 22nd terms of the 49th row.

To avoid this level of time-consuming work, we will understand and practice the formulas that are as follows:
$$ {\large{^nC_r}} = \frac{n!}{{r!}\times{(n-r)!}} $$ (One needs to be familiar with the concept of factorials to understand and practice this formula.)
Also, note that:
$$ {\large{^nC_r}} = {\large{^nC_{n-r}}} $$
The above identity can be verified from the images for Pascal’s triangle given above to develop the intuition for why this is true. (To further understand this, use a few cards from the deck (3 or 4) and arrange them in a straight line and pick up to verify the number of possible combinations.)

Let’s practice by taking a few examples to understand the formula:
$$ ^6C_3 = \frac{6!}{{3!}\times{(6-3)!}} = \frac{6!}{{3!}\times{3!}} $$
$$ ^7C_2 = \frac{7!}{{2!}\times{(7-2)!}} = \frac{7!}{{2!}\times{5!}}$$
$$ ^8C_5 = \frac{8!}{{5!}\times{(8-5)!}} = \frac{8!}{{5!}\times{3!}} $$
$$ ^{10}C_3 = \frac{10!}{{3!}\times{(10-3)!}} = \frac{10!}{{3!}\times{7!}} $$

Remember: n! can also be written as or and so on.

Similarly, you can verify that:
$$ ^{10}C_3 = {^{10}C_7} $$
$$ ^7C_3 = {^7C_4} $$
$$ ^5C_3 = {^5C_2} $$
$$ ^{1000}C_0 = {^{1000}C_{1000}} $$

This might feel like too much practice, but let me mention its applications so that you can understand why practicing this extensively is important. This part of mathematics is used in various problems in combinatorics and in binomial expansion (a special case of multinomial expansion) and the Leibniz rule (used in finding Laguerre Polynomials for finding wavefunctions) for finding the nth derivative of the product of two functions with respect to one variable like: $$ \frac{d^n}{{dx}^n} \cdot {(u \cdot v)} $$

Combinations provide a powerful way to count selections where the order of elements does not matter. Mastery of combinations not only strengthens problem-solving skills in probability and discrete mathematics but also builds intuition for real-world scenarios such as team selection, data analysis, and decision-making. All the best, and keep learning!