What are Binomial Coefficients?

The positive integers that are coefficients in the binomial theorem are known as binomial coefficients. A combination is also known as a combinatorial number or a binomial coefficient. Combinations and permutations are used to compute the total number of possible outcomes from a given set. The only difference between the combination and permutation is that the order does not matter in combinations and matters in permutations. In this article, we will learn and solve examples related to binomial coefficients.

The general notation along with the formula of the binomial coefficient is given below:

 

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

We can read it as "n over k", or sometimes "n choose k". It is denoted as \binom{n}{k}.

It reflects the number of ways we can choose unordered k outcomes from "n" number of possibilities. Therefore, they are also known as the combination or combinatorial numbers. In the next section, we will see what are the properties of binomial coefficients.

 

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Properties of Binomial Coefficients

These coefficients are derived from Blaise Pascal's well-known triangle known as Pascal's triangle. They help in doing a combinatorial analysis. In this section, we will discuss some of the properties of binomial coefficients.

Symmetry Property

\binom{n}{x} = \binom {n} {n-x}

For example:

\binom{4}{2} = \binom {4} {4-2}

 

Special Cases

\binom{n}{n} = 1

For instance,

\binom{3}{3} = 1

 

Pascal's Rule

\binom{n + 1}{k} = \binom{n}{k} + \binom {n} {k - 1}

For instance,

\binom{12 + 1}{4} = \binom{12}{4} + \binom {12} {4 - 1}

 

Negated Upper Index of Binomial Coefficient

\binom{n}{k} = (-1) ^k \binom {(k - n - 1)} {k}

For instance:

\binom{8}{3} = (-1) ^3 \binom {(3 - 8 - 1)} {3}

Now, we know what are binomial coefficients and their properties, we will proceed to solve some of the examples related to binomial coefficients.

Example 1

Solve \binom{5}{3}

Solution

Use the following rule to solve this example:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

Substitute the values of n and k in the above formula like this:

\binom{5}{3} = \frac {5!} {3! (5-3!)}

\binom{5}{3} = \frac {5 \cdot 4 \cdot 3 \cdot 2} {3 \cdot 2 \cdot 2}

\binom{5}{3} = \frac {120} {12}

\binom{5}{3} = 10

 

Example 2

Solve \binom{8}{2}

Solution

Use the following rule to solve this example:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

Substitute the values of n and k in the above formula like this:

\binom{8}{2} = \frac {8!} {2! (8-2!)}

\binom{8}{2} = \frac {8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} { 2 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}

\binom{8}{2} = \frac {40320} {1440}

\binom{8}{2} = 28

 

Example 3

Solve 9 \choose 4

Solution

We know that 9 \choose 4 can be written as:

\binom{9}{4}

We will use the following formula to solve this example:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

To calculate the answer, substitute the values of n and k in the above formula:

\binom{9}{4} = \frac {9!} {4! (9-4!)}

\binom{9}{4} = \frac {9!} {4! (5!)}

\binom{9}{4} = \frac {9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} {4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 \cdot 3 \cdot 2}

= \frac{ 362880} {2880} = 126

 

Example 4

Suppose you have got 3 free tickets for a theme park and you have 7 friends and family members who want to come with you.

In how many ways can you make the groups of friends and family members to take with you to the theme park?

Solution

First, try to understand the problem. 7 people want to go with you to the theme park. But, you have just got 3 free tickets. It means you can only take 3 people with you. Now, in this question, we are asked to tell the number of groups that can be formed from this available data.

We can write this question like this in combinatorics form:

7 \choose 3

We will use the following formula to get the number of groups that can be formed from a group of 7 people:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

Substitute the values in the above formula to get all possible combinations of people who can accompany you to the park:

\binom{7}{3} = \frac {7!} {3! (7-3!)}

\binom{7}{3} = \frac {7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} {3 \cdot 2 \cdot 4 \cdot 3 \cdot 2}

\binom{7}{3} = \frac {5040} {144} = 35

Hence, you can form 35 different groups from the list of 7 people given the constraint that only 3 people can accompany you to the park.

 

Example 5

Suppose you have to paint a wall. From 10 colors, you can pick only 5 colors. How many combinations of colors are possible?

Solution

First, we will try to comprehend the problem. We have 10 colors and you have to choose a number of ways you can paint the wall by picking up 5 colors only. In other words, we can say that we can make different combinations by choosing 5 colors from 10 colors.

We can write this question like this in combinatorics form:

10 \choose 5

We will use the following formula to get the number of groups that can be formed from a group of 7 people:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

Substitute the values in the above formula to get the number of arrangements:

\binom{10}{5} = \frac {10!} {5! (10-5!)}

\binom{10}{5} = \frac {10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} {5 \cdot 4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 \cdot 3 \cdot 2}

\binom{10}{5} = \frac {3628800} {14400} = 252

Hence, you can paint the wall in 252 different ways by picking up 5 colors at a single time from a group of 10 colors.

 

Example 6

Sam has to choose a combination of three flavors of ice cream from 10 flavors. How many different combinations of ice cream can he choose from 10 flavors?

Solution

Let us understand this problem first by breaking down its elements.

There are 10 ice cream flavors and Sam has to choose 3 flavors. The problem asks us to tell the number of different combinations that can be made from 10 flavors of ice-cream given the fact that a person can have only 3 flavors.

We can write this question like this in combinatorics form:

10 \choose 3

We will use the following formula to get the number of ice-cream combinations that can be formed from 10 flavors:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

Substitute the values in the above formula:

\binom{10}{3} = \frac {10!} {3! (10-3!)}

\binom{10}{3} = \frac {10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} {3 \cdot 2 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}

\binom{10}{3} = \frac {3628800} {30240} = 120

Hence, Sam can make 120 different combinations of ice-cream flavors from 10 flavors, given that he can only choose 3 at a time.

 

Example 7

John has to make a group of 4 students in the class to represent the school in a science fair. There are total 15 students in the class. If a group should have four students, how many different combinations of groups John can make from a class of 15 students by picking 4 students at random?

Solution

Let us understand this problem first by breaking down its elements.

There are 15 students in the class. The group should be constituted of 4 students and John needs to select these four students from a class of 15 students.

We can write this question like this in combinatorics form:

15 \choose 4

We will use the following formula to get the number of ice-cream combinations that can be formed from 10 flavors:

\binom{n}{k} = \frac {n!} {k! (n-k!)}, where n \geq k \geq 0

Substitute the values in the above formula:

\binom{15}{4} = \frac {15!} {4! (15-4!)}

\binom{15}{4} = \frac {15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 } {4 \cdot 3 \cdot 2 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}

\binom{15}{4}  = 1365

The binomial coefficient equals to 1365, that shows that John can make the group in 1365 ways from a class of 15 students, given that the group has only four members.

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Rafia Shabbir