JEE Main Syllabus 2025 PDF - Subject Wise Topics

# Introduction Of Combinations - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

## Quick Facts

• INTRODUCTION OF COMBINATIONS is considered one the most difficult concept.

• 100 Questions around this concept.

## Solve by difficulty

The value of $\sum_{r=1}^{15}r^{2}\, \left ( \frac{{}^{15}\textrm{C}_{r}}{{}^{15}\textrm{C}_{r-1}} \right )$ is equal to:

n is selected from the set {1,2,3......49} and the number 2n+3n+5n is formed. Total number of ways of selecting n so that the formed number is divisible by 4 is equal to

If $\mathrm{ n C r: n C r+1=1: 4}$ and $\mathrm{ n C r+1: n C r+2=4: 5}$, determine the values of n and r.

${ }^{n-1} C_r=\left(k^2-8\right)^n C_{r+1}$ if and only if:

Let $\alpha=\frac{(4 !) !}{(4 !)^{3 !}}$ and $\beta=\frac{(5 !) !}{(5 !)^{4 !}}$ Then :

The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is

## Concepts Covered - 1

INTRODUCTION OF COMBINATIONS

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” 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.

E.g., we need to select 4 students out of the 15 students who will represent the college at a quiz or we need to form an academic committee of 3 professors from 10 professors. In this case, who is selected “first”, who is selected “second” and so on does not matter. The words “first” and “second” implicitly implies an “ordering”. What matters in the case of selection is only the composition of the final “group”.

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

Let's say we want to arrange 2 objects out of 5 objects : A,B,C,D,E then using the concept of permutation we can do this in $\mathrm{^5P_2}$ ways.

We can calculate the same thing by another method: by selecting 2 things out of 5, which can be done as $\mathrm{^5C_2}$ and then arrange the 2 selected thing which can be done in 2! ways. So we have

$\\\mathrm{^5C_2 \times 2!=^5P_2}$

$\\\mathrm{^5C_2=\frac{^5P_2}{2!}} \\\mathrm{^5C_2 = \frac{5!}{(5-2)!2!}=\frac{5!}{3!2!}}$

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 $\mathrm{^nC_r} .$

Example: In ICC World Cup 2019 total 10 teams participated and each team has to play one game in the league stage with all other teams before qualifying for the semifinals, so 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

$\\\mathrm{^{10}C_2 =\frac{10!}{(10-2)!2!}}\\ \\\mathrm{\frac{10\times 9\times 8!}{8! \times 2!}=\frac{10\times 9}{2}=45}$

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

## Study it with Videos

INTRODUCTION OF COMBINATIONS

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