Applications of Combinations

Edited By Komal Miglani | Updated on Sep 12, 2024 09:34 AM IST

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. In real life, we use combinations for making lottery numbers and selecting nominees for student council.

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This Story also Contains

Restricted Combination
Number of Parallelograms
Number of rectangles and squares
Solved Examples Based on Applications of Combinations
Example 1: The number of bijective functions , such that , is [JEE MAINS 2022]
Summary

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.

The notation of selecting r objects from n given object is $\mathrm{^nC_r}$ . Let’s derive the value of $\mathrm{^nC_r}$ , and its relation with permutation notation.

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

$\\\mathrm{^nC_r \times r!=^nP_r}\\\ \\\mathrm{^nC_r=\frac{^nP_r}{r!}} \\\mathrm{^nC_r = \frac{n!}{(n-r)!r!}}$

Where 0 ≤ r ≤ n, and r is a whole number.

Now we have the value of $\mathrm{^nC_r} .$

Applications of Combinations

Let us take an example of selecting things from two or more different groups:

Out of 5 men and 6 women in how many ways can a committee of 5 members be selected such that at least 2 members are women?

Solution:

The following cases are possible for at least 2 women,

2 women + 3 men = $\\\mathrm{^6C_2\times ^5C_3}$

3 women + 2 men = $\\\mathrm{^6C_3\times ^5C_2}$

4 women + 1 men = $\\\mathrm{^6C_4\times ^5C_1}$

5 women = $\\\mathrm{^6C_5}$

So, the total number of ways $\\\mathrm{=^6C_2\times ^5C_3+^6C_3\times ^5C_2 + ^6C_4\times ^5C_1 + ^6C_5 = 431}$

Restricted Combination

The number of selection of r objects from n different objects:

When k particular things are always included = $\mathrm{^{n-k}C_{r-k}}$

This can be comprehended as taking out those k things that have to be included which can be done in 1 way and then finding the ways in which r-k objects can be selected from remaining n - k things, and putting those k things (which are already taken out) in r-k selected objects.

k particular things are never included = $\mathrm{^{n-k}C_{r}}$

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This can be comprehended as taking out k things that are not to be selected which can be done in 1 way and then finding the ways of selecting r things from n-k things.

The number of ways selecting r things out of n different things such that p particular objects are always included and q particular objects are always excluded = $\mathrm{^{n-p-q}C_{r-p}}$

This can be comprehended as taking out the q objects that should not be selected and putting it out and then taking out p objects that have to be selected and then finding ways of selecting r-p objects out of n- p-q objects and putting back p objects in r-p selected objects.

Example: In how many ways a cricket team can be selected out of 16 players such that 5 certain players must be included in the team?

Solution: Since 5 certain players have to be included so we need to select 11-5 = 6 players from 16 - 5 = 11 players.

So we can select the team in $\\\mathrm{^{16-5}C_{11-5}=^{11}C_6=\frac{11!}{5!6!}}$

If there are n points in the plane and out of which no three are collinear then,

Total No. of lines that can be formed using these n points = ⁿC₂
Total No. of triangles that can be formed using these n points = ⁿC₃
Total no. of Diagonals that can be formed in n-sided polygon = ⁿC₂ - n

If there are n points in the plane and out of which m points are collinear, then,

The total No. of different lines that can be formed by joining these n points is $\mathrm{^nC_2-^mC_2+1}$

The total No. of different triangles that can be formed by joining these n points is $\mathrm{^nC_3-^mC_3}$

The total No. of different quadrilaterals formed by joining these n points is $\mathrm{^nC_4-\left (^mC_3 \cdot^{n-m}C_1+^mC_4\right )}$

Number of Parallelograms

If m parallel lines in a plane are intersected by the family of other n parallel lines, then the total number of parallelograms formed is

$\mathrm{^mC_2\cdot ^nC_2=\frac{mn(m-1)(n-1)}{4}}$

Number of rectangles and squares

The number of rectangles of any size in a square of size n x n is $\sum_{r=1}^{n}r^3$ and number of squares of any size is $\sum_{r=1}^{n}r^2$ .
In a rectangle of size n x p (n < p) number of rectangles of any size is $\frac{np}{4}(n+1)(p+1)$ .

To determine the number of ways to reach in the shortest way from point A to B.

When considering the possible paths or shortest path one can observe that the total number of steps in the forward direction is 6-R(Right) and in the upward direction is 4-U(Upward)

Now, If we arrange these 6 Rs and 4 Us in any way, it comes out to be the shortest path.

Or one can say that first find all the possible steps and arrange them to get the total number of possible ways.

Using "u" and "r" we can write out a path:

r r r r r r u u u u

r r r u u u u r r r

and others......

Hence, the total number of ways is $\frac{10!}{4!6!}$ or, $^{10}C_4$

Solved Examples Based on Applications of Combinations

Example 1: The number of bijective functions $\mathrm{f}:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots, 100\}$ , such that $\mathrm{\mathrm{f}(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \geq f(99)}$ , is [JEE MAINS 2022]

Solution: For the bijective function

$\mathrm{f(3)>f(9)>\ldots>f(99)}$

First, select 17 images for these 17 inputs in $\mathrm{^{50}C}_{33}$ ways and divide them in 1 way.

Now, select images for the rest of the 33 inputs in 33! ways.

$\mathrm{\therefore \text { Total }={ }^{50} \mathrm{C}_{17} \cdot 33 !}\\$

$=\frac{50 !}{17 !}\\$

$=50 P_{33}$

Hence, the answer is ${ }^{50} P_{33}$

Example 2: Five numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are randomly selected from the numbers $1,2,3, \ldots ., 18$ and are arranged in increasing order $\left(x_{1}<x_{2}<x_{3}<x_{4}<x_{5}\right)$ . The probability that $x_{2}=7$ and $x_{4}=11$ is? [JEE MAINS 2022]

Solution: $\mathrm{\text { If } \quad x_{2}=7, \quad x_{1}<7 \Rightarrow x, \in\{1,2,3,4,5,6\}} \\$

$\mathrm{x_{3} \in\left\{8,9,10\right\}}$ , $\mathrm{x_{5} \in\left\{12,13,14,15,16,17,18\right\}}$

$\mathrm{\text { So Probability }= \frac{{ }^{6} C_{1} \times{ }^{3} C_{1} \times \frac{7}{7} C_{1}}{18 C_{5}}} \\$

$\mathrm{= \frac{6 \times 3 \times 7\times 8 \times 4 \times 8 \times 2 \times 1}{18 \times 17 \times+6 \times 15 \times 14}=\frac{1}{68}}$

Hence, the answer is 1/ 68

Example 3: The sum of all the elements of the set $\mathrm{\{\alpha \in\{1,2, \ldots . .100\}: H C F(\alpha, 24)=1\} }$ is: [JEE MAINS 2022]

Solution

$\mathrm{\begin{aligned} &\alpha \in\{1,2,3, \ldots, 100\} \quad ; \mathcal{H C F}(2,24)=1\\ &\text { Sum }=[1+2+3+\ldots .100]-[2+4+\ldots 100]-[3+6+\ldots-99]+[6+12+\ldots .96]\\ &=\frac{100 \times 101}{2}-\frac{2 \times 50 \times 51}{2}-3 \times \frac{33 \times 34}{2}+\frac{6 \times 16 \times 17}{2}\\ &=5050-2550-(99)(17)+(48)(17)\\ &=2500-1683+816\\ &=3316-1683\\ &=\quad 1633 \end{aligned}}$

Hence, the answer is 1633.

Example 4: There are ten boys $B_{1}, B_{2}, \ldots, B_{10}$ and five girls $G_{1}, G_{2}, \ldots, G_{5}$ in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both $B_{1}$ and $B_{2}$ together should not be the members of a group, is_____________. [JEE MAINS 2022]

Solution: $\mathrm{n(B)=10 ,\quad n(a)=5}$

The number of ways of forming a group of 3 girls and 3 boys

$\mathrm{={ }^{10} C_{3} \times{ }^{5} C_{3}} \\$

$\mathrm{=\frac{10 \times 9 \times 8}{3 \times 2} \times \frac{5 \times 4}{2}=1200}$

The number of ways where two particular boys $\mathrm{B_{1}\: and\: B_{2}}$ be members of a group together

$\mathrm{=^8{C_{1}} \times^{5 } {C_{3}}=8 \times 10=80}$

Number of ways boys $\mathrm{B_{1}\: and\: B_{2}}$ are not in the same group together

Required no.of selections = $\mathrm{1200-80=1120}$

Hence, the answer is $\mathrm{1120}$ .

Example 5: Let $\mathrm{P}_{1}, \mathrm{P}_{2} \ldots, \mathrm{P}_{15}$ be 15 points on a circle. The number of distinct triangles formed by points $\mathrm{P}_{i}, \mathrm{P}_{j}, \mathrm{P}_{k}$ such that $i+j+k \neq 15$ , is : [JEE MAINS 2021a]

Solution: Total no. of Triangles
$= \: ^{15}C_{3}= \frac{15\times 14\times 13}{3\times 2\times 1}= 455$

$Cases\: when\: i+j+k= 15\: are :$

$i= 1,j= 2,k= 12$

$i= 1,j= 3,k= 11$

.....

$i= 1,j= 6,k= 8$

$i= 2,j= 3,k= 10$

$i= 2,j= 4,k= 9$

$i= 2,j= 5,k= 8$

$i= 2,j= 6,k= 7$

$i= 3,j= 4,k= 8$

$i= 3,j= 5,k= 7$

$i= 4,j= 5,k= 6$

These are 12 cases

So Desired no. of triangles $= 455-12= 443$

Hence, the answer is 443

Summary

Understanding the principles of combinations equips individuals with powerful tools to solve complex problems involving selection, grouping, and probability calculations efficiently. These applications underscore the fundamental role of combinations in both theoretical and practical domains. Knowledge of combinations is required for data analysis, algorithm design, or decision-making processes that involve combinatorial reasoning.