General and Middle Terms in Binomial Expansion

Edited By Komal Miglani | Updated on Oct 15, 2024 02:19 PM IST

A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. There are many situations in which we have to find permutations under some restrictions. We have the following two main types of restrictions. First, particular objects are separated and second, Particular objects are always together. In real life, we use permutation for solving questions related to probability and statistics.

JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

JEE Main 2025: Maths Formulas | Study Materials

JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics

This Story also Contains

Geometrical Permutations: definitions
Necklace, Garland formation: Explanation
Permutation under restriction
Summary
Solved Examples Based on Geometrical Permutations

General and Middle Terms in Binomial Expansion

In this article, we will cover the Geometrical Permutations. 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.

Geometrical Permutations: definitions

A geometric permutation is a permutation formed by arranging the objects or set of objects around the geometry like a circle.

Let's say there is a round table with 6 chairs all identical and 6 persons have to sit. For the first person there is only one choice to make as all chairs are identical, so wherever he may sit doesn't matter. Now when $1^{\text {st }}$ person has sat, $2^{\text {nd }}$ person with respect to $1^{\text {st }}$ has five choices to sit, directly opposite or left or $2^{\text {nd }}$ from left or right or $2^{\text {nd }}$ from right to $1^{\text {st }}$ person, in the same way, $3^{\text {rd }}$ person will have 4 choices, $4^{\text {th }}$ person will have 3 choices, $5^{\text {th }}$ person will have 2 choices, and last person 1 choice, so in that way total $5 \times 4 \times 3 \times 2 \times 1=(6-1)$ ! permutations are possible.

We can generalize this result as an object that can be arranged along with a circular table in ( $n-1)$ ! Ways

Example: In how many ways 7 people can be arranged along a circular table having 7 identical chairs?

Solution: using the above concept, it can be done in (7-1)! = 6 !.

Necklace, Garland formation: Explanation

If clockwise and anticlockwise permutations are the same in a circular permutation, (as in the case of garlands or necklace formation), the number of permutations becomes $(1 / 2)(n-1)$ !

Since as in the case of necklace and garland if we flip the bead or garland then anticlockwise arrangements become clockwise but they are identical because the objects are identical, hence two arrangements are reduced to one causing the total number of permutations to be halved.

Necklace, Garland Type Questions

Example 1: Find the ways in which 10 different beads can be arranged to form a necklace.

Solution: Using circular permutations the total number of permutations = $(10-1)!$

Now since clockwise and anticlockwise arrangements give the same permutation the total number of permutations becomes (1/2)(9!)

If clockwise and anticlockwise permutations are the same in a circular permutation, (as in the case of garlands or necklace formation), the number of permutations becomes $(1 / 2)(n-1)$ !

Example 2 : Find the ways in which 10 different beads can be arranged to form a necklace.

Solution: Using circular permutations the total number of permutations = $(10-1)!$

Now since clockwise and anticlockwise arrangements give the same permutation the total number of permutations becomes $(1 / 2)(9!)$

Permutation under restriction

There are many situations in which we have to find permutations under some restrictions. We have the following two main types of restrictions.

(i) Particular objects are separated

In this case, we have to arrange $n$ objects in which no two of r particular objects are together. Here first we find the number of permutations of ( $n$ r) objects. Arrangements of these objects create ( $n-r+1$ ) places or gaps (between the objects, before and after the objects). Now we can arrange particular objects in these gaps so that objects are separated.

(ii) Particular objects are always together

In this case, we have to arrange n objects of which r particular objects are always together. Here we consider particular objects as one group and first find the number of permutations of ( $n-r+1$ ) objects which contains the group of $r$ particular objects and multiply it with the number of arrangements of r particular objects in the group.

Summary

The permutations represent the number of distinct arrangements of objects where the order matters. The calculation of permutations using factorial notation provides a precise method to quantify and analyze sequential arrangements. Understanding of permutation is necessary for solving complex problems.

Solved Examples Based on Geometrical Permutations

Example 1: The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together, is:

[JEE MAINS 2023]
Solution

$$
\begin{aligned}
& 6!\times{ }^7 C_5 \times 5! \\
& \Rightarrow 720 \times 21 \times 120 \\
& \Rightarrow 2 \times 360 \times 7 \times 3 \times 120 \\
& \Rightarrow 126 \times(5!)^2
\end{aligned}
$$

Hence, the answer is $126(5!)^2$

Example 2: Consider a rectangle ABCD having 5,7,6,9 points in the interior of the line segment AB, CD, BC, and DA respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta - \alpha)$ is equal to: [JEE MAINS 2021]

Solution

Example 2: Consider a rectangle $A B C D$ having $5,7,6,9$ points in the interior of the line segment $\mathrm{AB}, \mathrm{CD}, \mathrm{BC}$, and DA respectively. Let $\alpha$ be the number of triangles having these points from different sides as vertices and $\beta$ be the number of quadrilaterals having these points from different sides as vertices. Then $(\beta-\alpha)$ is equal to:
[JEE MAINS 2021]

Solution

\begin{aligned}
&\begin{aligned}
\alpha & =\text { Number of triangles } \\
\alpha & =5 \cdot 6 \cdot 7+5 \cdot 7 \cdot 9+5 \cdot 6 \cdot 9+6 \cdot 7 \cdot 9 \\
& =210+315+270+378 \\
& =1173 \\
\beta & =\text { Number of Quadrilateral } \\
\beta & =5 \cdot 6 \cdot 7 \cdot 9=1890 \\
\beta & -\alpha=1890-1173=717
\end{aligned}\\
&\text { Hence, the answer is } 717
\end{aligned}

Example 3 : If in a regular polygon, the number of diagonals is 54 , then the number of sides of this polygon is :
[JEE MAINS 2015]

Solution: The number of diagonals of n sided convex polygon is ${ }^n C_2-n$ (Where $n>3$ )

Now,
no. of diagonals $={ }^n \mathrm{C}_2-n=54$
$\Rightarrow \frac{n(n-1)}{2}-n=54$
$\Rightarrow n^2-n-2 n-108=0$
$\Rightarrow n^2-3 n-108=0$
$\Rightarrow n^2-12 n+9 n-108=0$
$\Rightarrow(n-12)(n+9)=0$
$\Rightarrow n=12$
Hence, the answer is 12

Example 4: The number of ways in which 6 people can sit around a round table for dinner is

Solution: As we have learned in circular permutations, we fix the position of one object and then arrange the other objects in all possible ways.

Now, we will fix any one person at any place, for that number of ways will be 1.

After that other fives can be arranged w.r.t him in (6-1)! $=5!$ i.e. 120 ways.
Hence, the answer is 120
Example 5 : The Number of different garlands that can be made using 5 flowers is

Solution: To arrange 5 flowers in a circle, the number of ways is $4!=24$
But as this is a garland with clockwise and anticlockwise arrangements being the same, so total number of garlands will be half of 24 , which is 12 .

Hence, the answer is 12