# Binomial Coefficient:

#### What is the Binomial Coefficient:

From the binomial expansion $$\binom{n}{0}$$, $$\binom{n}{1}$$, $$\binom{n}{3}$$......$$\binom{n}{n}$$ are the binomial coefficients and the sum of binomial coefficients can be written in the form of formula as given below to calculate and find the value of the binomial coefficient.

#### Binomial Coefficient Formula:

The given formula of binomial coefficient can be used to find or calculate the value binomial coefficient easily.$$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ Where $$n = 0, 1, 2, 3, 4,.......n$$

$$k = 0, 1, 2, 3, 4,.......n$$

and $$(n \ge k)$$

Note: In the binomial coefficient, the value of "k" changes from "0" to "n", hence the number of terms in the binomial expansion $$(a + b)^n$$ will always be $$(n + 1)$$.

#### Binomial Coefficient Examples:

Example(1): Evaluate the binomial coefficient $$\binom{15}{5} ?$$

Solution: Given $$n = 15$$ and $$k = 5$$.

By putting these values in the binomial coefficient formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{15}{5} = \frac{15!}{(15 - 5)! \ 5!}$$ $$\binom{15}{5} = \frac{15!}{10! \ 5!}$$ The $$15!$$ is expanded till $$10!$$ so that it could be canceled with the same value, given in the numerator. $$\binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10!}{10! \ 5!}$$ $$\binom{15}{5} = \frac{15 \times 14 \times 13 \times 12 \times 11}{5!}$$ $$\binom{15}{5} = 3003$$

Example(2): Evaluate the binomial coefficient $$\binom{10}{0} ?$$

Solution: Given $$n = 10$$ and $$k = 0$$.

By putting these values in the binomial coefficient formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{10}{0} = \frac{10!}{(10 - 0)! \ 0!}$$ $$\binom{10}{0} = \frac{10!}{10! \ 0!}$$ $$\binom{10}{0} = 1$$

Example(3): Evaluate the binomial coefficient $$\binom{5}{5} ?$$

Solution: Given $$n = 5$$ and $$k = 5$$.

By putting these values in the binomial coefficient formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{5}{5} = \frac{5!}{(5 - 5)! \ 5!}$$ $$\binom{5}{5} = \frac{5!}{0! \ 5!}$$ $$\binom{5}{5} = 1$$

#### What is "n choose k":

The symbol $$\binom{n}{k}$$ is read as "n choose k", which means "k" number of objects can be chosen from the total number of objects "n". The total number of ways to choose the objects is $$\binom{n}{k}$$, here the order of choosing the ways of objects are not important.

For example, for $$\binom{5}{3}$$, there are 10 ways to choose 3 objects from the total number of 5 objects. (If we solve $$\binom{5}{3}$$ then we will get 10) Let the total number of 5 objects are {a, b, c, d, e} and we can choose 3 objects from that in 10 different ways {a, b, c}, {a, c, d}, {a, d, e}, {a, c, e}, {a, b, d} {a, b, e}, {b, c, d}, {b, d, e}, {c, d, e}, {b, c, e}.

#### n choose k formula:

To solve any "n choose k" we can use binomial coefficient formula because they are the same and the formula is $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ The examples to calculate any "n choose k" are given below.

Example(1): Evaluate the value of "2 choose 2"?

Solution: Given $$n = 2$$, and $$k = 2$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{2}{2} = \frac{2!}{(2 - 2)! \ 2!}$$ $$\binom{2}{2} = \frac{2!}{0! \ 2!}$$ $$\binom{2}{2} = \frac{2!}{1 \ 2!}$$ $$\binom{2}{2} = \frac{1}{1}$$ $$\binom{2}{2} = 1$$

Example(2): Evaluate the value of "1 choose 1"?

Solution: Given $$n = 1$$, and $$k = 1$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{1}{1} = \frac{1!}{(1 - 1)! \ 1!}$$ $$\binom{1}{1} = \frac{1!}{0! \ 1!}$$ $$\binom{1}{1} = \frac{1}{1}$$ $$\binom{1}{1} = 1$$

Example(3): Evaluate the value of "n choose n"?

Solution: Given $$n = n$$, and $$k = n$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{n}{n} = \frac{n!}{(n - n)! \ n!}$$ $$\binom{n}{n} = \frac{n!}{0! \ n!}$$ $$\binom{n}{n} = \frac{1}{1}$$ $$\binom{n}{n} = 1$$

Example(4): Evaluate the value of "2n choose n"?

Solution: Given $$n = 2n$$, and $$k = n$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{2n}{n} = \frac{(2n)!}{(2n - n)! \ n!}$$ $$\binom{2n}{n} = \frac{(2n)!}{n! \ n!}$$ $$\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$$ This is true for all $$n \ge 0$$.

In the same way, you can find the values of other similar "n choose k" questions.

#### More Examples:

Example (1): Evaluate the value of "4 choose 2"?

Solution: Given $$n = 4$$, and $$k = 2$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{4}{2} = \frac{4!}{(4 - 2)! \ 2!}$$ $$\binom{4}{2} = \frac{4!}{2! \ 2!}$$ $$\binom{4}{2} = \frac{4 \times 3 \times 2!}{2! \ 2!}$$ $$\binom{4}{2} = \frac{4 \times 3}{2 \times 1}$$ $$\binom{4}{2} = 6$$

Example (2): Evaluate the value of "2 choose 1"?

Solution: Given $$n = 2$$, and $$k = 1$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{2}{1} = \frac{2!}{(2 - 1)! \ 1!}$$ $$\binom{2}{1} = \frac{2!}{1! \ 1!}$$ $$\binom{2}{1} = \frac{2 \times 1}{1 \ 1}$$ $$\binom{2}{1} = \frac{2}{1}$$ $$\binom{2}{1} = 2$$

Example (3): Evaluate the value of "7 choose 2"?

Solution: Given $$n = 7$$, and $$k = 2$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{7}{2} = \frac{7!}{(7 - 2)! \ 2!}$$ $$\binom{7}{2} = \frac{7!}{5! \ 2!}$$ $$\binom{7}{2} = \frac{7 \times 6 \times 5!}{5! \ 2!}$$ $$\binom{7}{2} = \frac{7 \times 6}{2!}$$ $$\binom{7}{2} = \frac{42}{2}$$ $$\binom{7}{2} = 21$$

Example (4): Evaluate the value of "6 choose 0"?

Solution: Given $$n = 6$$, and $$k = 0$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{6}{0} = \frac{6!}{(6 - 0)! \ 0!}$$ $$\binom{6}{0} = \frac{6!}{6! \ 0!}$$ $$\binom{6}{0} = \frac{6!}{6! \ 1}$$ $$\binom{6}{0} = 1$$

Example (5): Evaluate the value of "8 choose 5"?

Solution: Given $$n = 8$$, and $$k = 5$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{8}{5} = \frac{8!}{(8 - 5)! \ 5!}$$ $$\binom{8}{5} = \frac{8!}{3! \ 5!}$$ $$\binom{8}{5} = \frac{8 \times 7 \times 6 \times 5!}{3! \ 5!}$$ $$\binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$$ $$\binom{8}{5} = 56$$

Example(6): Evaluate the value of "1 choose 0"?

Solution: Given $$n = 1$$, and $$k = 0$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{1}{0} = \frac{1!}{(1 - 0)! \ 0!}$$ $$\binom{1}{0} = \frac{1!}{1! \ 0!}$$ $$\binom{1}{0} = \frac{1}{1}$$ $$\binom{1}{0} = 1$$

Example (7): Evaluate the value of "100 choose 4"?

Solution: Given $$n = 8$$, and $$k = 5$$.

By putting the values of n and k in the formula.$$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{100}{4} = \frac{100!}{(100 - 4)! \ 4!}$$ $$\binom{100}{4} = \frac{100!}{96! \ 4!}$$ $$\binom{100}{4} = \frac{100 \times 99 \times 98 \times 97 \times 96!}{96! \ 4!}$$ $$\binom{100}{4} = \frac{100 \times 99 \times 98 \times 97}{4 \times 3 \times 2 \times 1}$$ $$\binom{100}{4} = 3,921,225$$

Example (8): Evaluate the value of "100 choose 100"?

Solution: Given $$n = 8$$, and $$k = 5$$.

By putting the values of n and k in the formula. $$\binom{n}{k} = \frac{n!}{(n - k)! \ k!}$$ $$\binom{100}{100} = \frac{100!}{(100 - 100)! \ 100!}$$ $$\binom{100}{100} = \frac{100!}{0! \ 100!}$$ $$\binom{100}{100} = 1$$

Note: From the above examples, it is clear that whenever the values of "n" and "k" are the same, the answer will always be "1".