Combination in Math: Definition, Formula and Example

So far our task was always to “arrange” objects i.e. to place them in a specific order among themselves. Sometimes we would be interested in only “selecting” a few objects out of the given objects. In this case, we just need to “select” and we do not need to “arrange” them in an order. The selection of objects is called combinations. In real life, we use combinations for making lottery numbers and selecting nominees for student council.

In this article, we will cover the Introduction to Combinations. This topic falls under the broader category of Permutations and combinations, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE.

What is Combination?

The meaning of combination is selection. Suppose we want to select two objects from four distinct objects a, b, c, and d. This can be stated as a number of combinations of four different objects taken two at a time.

Here we have six different combinations ab, ac, ad, bc, bd, cd. In other words, we can say that there are six ways in which we can select two objects from four distinct objects.

Combination Formula

We can generalize this concept for r object to be selected from given $n$ objects as

$\begin{array}{rl}& {}^n{C}_r\times r!={}^n{P}_r\\ & {}^n{C}_r=\frac{{}^n{P}_r}{r!}\\ & {}^n{C}_r=\frac{n!}{(n-r)!r!}\end{array}$

Where $0\le r\le n$, and $r$ is a whole number.
Let's derive the value of ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$, and its relation with permutation notation.

Let's say we want to arrange 2 objects out of 5 objects: A, B, C, D, and E then using the concept of permutation we can do this in ${}^5{\mathrm{P}}_2$ ways.

We can calculate the same thing by another method: by selecting 2 things out of 5 , which can be done as ${}^5{\mathrm{C}}_2$ and then arranging the 2 selected things which can be done in 2 ! ways. So we have

$\begin{array}{rl}& {}^5{\mathrm{C}}_2\times 2!={}^5{\mathrm{P}}_2\\ & {}^5{\mathrm{C}}_2=\frac{{}^5{\mathrm{P}}_2}{2!}\\ & {}^5{\mathrm{C}}_2=\frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}\end{array}$

Relation Between Combination And Permutations

The combination is a type of permutation where the order of the selection is not considered. Hence, the count of permutations is always more than the number of the combination.

The notation of selecting r objects from n given object is ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$.

_{Corresponding to each combination of ${}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}$, we have r ! permutations because r objects in every combination can be rearranged in r! ways.}

_{So, the relation between combinations and permutations is given by}

_{${}^n{C}_r\times r!={}^n{P}_r$}

Example: In ICC World Cup 2019 total of 10 teams participated and each team has to play one game in the league stage with all other teams before qualifying for the semifinals, how many total games will be played in the league stage?

Solution: For playing a game we need to select two teams. So this is a simple problem of selecting two teams, so this can be done in

$\begin{array}{rl}& {}^{10}{\mathrm{C}}_2=\frac{10!}{(10-2)!2!}\\ & \frac{10\times 9\times 8!}{8!\times 2!}=\frac{10\times 9}{2}=45\end{array}$

Hence in total 45 games will be played in the league stage.

Recommended Video Based on Introductions of Combinations

Solved Examples Based on Introductions of Combinations

Example 1: Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples, and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is
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Solution: Three cases are possible

$\begin{array}{rl}& \text{ R }1\mathrm{W}3\mathrm{O}+2\mathrm{R}1\mathrm{W}2\mathrm{O}+1\mathrm{R}2\mathrm{W}2\mathrm{O}\\ & {}^7{\mathrm{C}}_1\cdot {}^5{\mathrm{C}}_1\cdot {}^8{\mathrm{C}}_3+{}^7{\mathrm{C}}_2\cdot {}^5{\mathrm{C}}_1\cdot {}^8{\mathrm{C}}_2+{}^7{\mathrm{C}}_1\cdot {}^5{\mathrm{C}}_2\cdot {}^8{\mathrm{C}}_2\\ & =6860\end{array}$

Hence, the answer is 6860 .

Example 2: Let $S=\{1,2,3,5,7,10,11\}$. The number of non-empty subsets of $S$ that have the sum of all elements as a multiple of 3 , is
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Solution
No. of element $1=\{3\}$
No. of element $2=\{(3\mathrm{K}+1),(3\mathrm{k}+2)\}$
(3) $(3)=9$

No. of element $3=\{3\mathrm{k},3\mathrm{k}+1,3\mathrm{K}+2\}=$ (1) (3) (3) $=9$ $=\{(3\mathrm{k}+1),(3\mathrm{k}+1),(3\mathrm{k}+1)\}=1$

$=\{(3\mathrm{K}+2),(3\mathrm{k}+2),(3\mathrm{k}+2)\}=\frac{1}{11}$

No. of element $4=\{3\mathrm{k},3\mathrm{k}+1,3\mathrm{k}+1,3\mathrm{k}+1\}\to 1$
$=\{3\mathrm{k},3\mathrm{k}+2,3\mathrm{k}+2,3\mathrm{k}+2\}\to 1$

$=(3\mathrm{k}+1,3\mathrm{k}+2,3\mathrm{k}+2,3\mathrm{k}+1\}\to {}^3{\mathrm{C}}_2\times {}^3{\mathrm{C}}_2=9$

No. of element $5=9$, no. of element $6=1$, no. of element $7=1$
Total $=43$
Hence, the answer is 43 .

Example 3: Let the digits $\mathrm{a},\mathrm{b}$, and c be in A.P. Nine-digit numbers are to be formed using each of these three digits thrice such that three consecutive digits are in A.P. at least once. How many such numbers can be formed?
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Solution
$abc$ or cba
$abc$
$cba$

$\frac{{}^7{C}_1\times 2\times 6!}{2!2!2!}=1260$

Hence, the answer is 1260.

Example 4: A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is $p$. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colour is q . If $\mathrm{p}:\mathrm{q}=\mathrm{m}:\mathrm{n}$, where m and n are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to $$
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Solution

$\begin{array}{rl}& \begin{array}{rl}& p=1\cdot \frac{1}{6}\\ & \mathrm{q}=({}^6{\mathrm{C}}_1\cdot \frac{1}{6}\cdot \frac{1}{6}\cdot \frac{1}{6}\cdot \frac{5}{6})\frac{4!}{3!}=\frac{5}{216}\times 4=\frac{5}{54}\\ & \frac{p}{q}=\frac{1/6}{5/54}=\frac{9}{5}\\ & \mathrm{m}=9\\ & \mathrm{n}=5\\ & m+n=9+5=14\end{array}\\ & \text{ Hence, the answer is }14\end{array}$

Example 5: A class contains b boys and $g$ girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168 , then $b+3g$ is equal to $$
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Solution

$\begin{array}{rl}& \text{ Number of ways }={}^{\mathrm{b}}{\mathrm{C}}_3\cdot {}^{\mathrm{g}}{\mathrm{C}}_2=168\\ & \Rightarrow \frac{\mathrm{b}(\mathrm{b}-1)(\mathrm{b}-2)}{6}\cdot \frac{\mathrm{g}(\mathrm{g}-1)}{2}=168\\ & \Rightarrow \mathrm{b}(\mathrm{b}-1)(\mathrm{b}-2)\mathrm{g}(\mathrm{g}-1)=8\cdot 7\cdot 6\cdot 3\cdot 2\\ & \Rightarrow \mathrm{b}=8,\mathrm{g}=3\\ & \Rightarrow \mathrm{b}+3\mathrm{g}=8+9=17\end{array}$

Hence, the answer is 17.