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# Permutation and Combination Problems in Details for JEE Last updated date: 04th Dec 2023
Total views: 22.8k
Views today: 1.22k     ## Permutation and Combinations: Introduction

The word 'arrangement' can be seen in the definition of permutation. The term "permutation" refers to the simultaneous arrangement of some or all of the same or different things. In the case of permutations, the order in which the items are chosen is also considered. Combination refers to choosing one or more of the supplied items, which can be similar or unlike. Words like selection, collection, and committee can also be used instead of combinations. These two words, permutation and conjunction, are sometimes used interchangeably at an introductory level, which can be confusing. They are best described in this article, which goes something like this.

### Permutation and Combination Formulas

The situation is assessed to determine whether permutations or combinations should be used. The permutation and combination formulas are used as a result.

1. Combination Formula

• The number of distinct combinations of 'n' separate things that can be taken 'r' at the same time is given by:

${ }^{n} \mathrm{Cr}=\dfrac{n !}{r !(n-r) !}$

Where, ,$0 \leq r \leq n$

• Some Other Formulas

• If ' $n$ ' is even " $C_{r}$ is greatest for $r=\frac{n}{2}$

• ${ }^{n} C_{r}+{ }^{n} C_{r-1}={ }^{n+1} C_{r}$

• If ${ }^{n} C_{r}={ }^{n} C_{k-r}$, then $r=k$ or $n-r=k$

• ${ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}$

• $\dfrac{{ }^{n} C_{n-r}}{{ }^{n} C_{r-1}}=\dfrac{n-r+1}{r}$

2. Permutation Formula

• 'r' things taken from 'n' things can be arranged in a variety of ways.

${ }^{n} P_{r}=\dfrac{n !}{(n-p) !}$

Where, $0 \leq r \leq n$

• ‘n’ is the factorial of a natural number

$n !=1 \times 2 \times 3 \times 4 \times \ldots \ldots \times n$

• nr is the number of permutations of n different objects taken r at a time with repetition permitted.

• For r objects picked from n things, the relationship between permutation and combination is:

${ }^{n} P_{r}={ }^{n} C_{r} \times r !$

## Permutation and Combination Solved Examples

Example1: Determine the number of words made up of 5 letters from the letters in the word "INDEPENDENCE."

Solution:

We've offered 12 alphabets in this example:

 Alphabets Times E 4 D 2 P, I, C 1

Case1: When EEEDC (2 different and 3 alike)

Number of selection$= { }^{2} C_{1} \times{ }^{5} C_{2}=20$

Arrangement =  $\dfrac{5 !}{3 !}$

Thus, the number of permutations $=20 \times \dfrac{5 !}{3 !}=400$ .

Case 2: When EEEEP (4 alike and 1 different)

We only have one option for four people who are the same, namely E, and five alternatives for choosing one person who is different.

Number of selection= $1 \times{ }^{5} C_{1}=5$

Arrangement = $\dfrac{5 !}{4 !}$

Thus, the number of permutations $=5 \times \dfrac{5 !}{4 !}=25$ .

Case 3: When EENDI (2 alike and 3 different)

Number of selection $= { }^{3} C_{1} \times{ }^{5} C_{3}=30$

Arrangement $=\dfrac{5 !}{2 !}$

Thus, the number of permutations =$30 \times \dfrac{5 !}{2 !}=1800$ .

Case 4: When EEEDD (3 alike and 2 alike of different)

No. of selection $={ }^{2} C_{1} \times{ }^{5} C_{1}=4$

Arrangement=$\dfrac{5 !}{3 ! 2 !}$

Thus, the number of permutations =$4 \times \dfrac{5 !}{3 ! 2 !}=40$.

Case 5: When EDIPC (All five different)

Number of selection $={ }^{6} C_{5}=6$

Arrangement =$\dfrac{5 !}{1}$

Thus, the number of permutations = $6 \times \dfrac{5 !}{1}=720$ .

Case 6: When EENND (2 alike, other 2 alike and 1 different)

Number of selection = ${ }^{3} C_{2} \times 4=12$

Arrangement $=\dfrac{5 !}{2 ! 2 !}$

Thus, the number of permutations $=12 \times \frac{5 !}{2 ! 2 !}=360$.

As a consequence, sum all of the outcomes from the various situations together is:

= (400+ 25 + 1800 + 40 + 720 + 360)

= 3345

Example 2: Out of 7 consonants and 4 vowels, what number of expressions of 3 consonants and 2 vowels can be framed?

Solution:

Number of approaches to choosing 2 vowels from $4 = { }^{4} C_{2}$

Number of approaches to choosing 3 consonants from $7= { }^{7} C_{3}$

Number of approaches to choosing 3 consonants from 7 and 2 vowels from 4=

\begin{align} &{ }^{4} C_{2} \times{ }^{7} C_{3} \\ &=\dfrac{4 \times 3}{2 \times 1} \times \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \\ &=210 \end{align}

It implies that we can have 210 gatherings, where each gathering contains 5 complete letters with 3 consonants and 2 vowels.

Number of approaches to orchestrating 5 letters among themselves

\begin{align} &=5 ! \\ &=5 \times 4 \times 3 \times 2 \times 1 \\ &=120 \end{align}

Consequently, required number of ways $=120 \times 210=25200$

Out of 7 consonants and 4 vowels, 25200 words of 3 consonants and 2 vowels can be formed.

Example 3: In what number of alternative ways can the letters be arranged in the order that the vowels are always together within the letters of the word 'Optical'?

Solution:

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels ought to continuously meet up. Subsequently, these three vowels can be gathered and considered as a solitary letter.

Every one of the 3 vowels (OIA) are unique.

Number of ways of organizing these vowels among themselves

\begin{align} &=3 ! \\ &=3 \times 2 \times 1=6 \end{align}

Thus, we can accept all our letters as 5. This multitude of letters are unique.

Number of ways of organizing these letters = 5!

$=5 \times 4 \times 3 \times 2 \times 1=120$

Thus, required number of ways $= 120 \times 6=720$

The Required No. Of Ways is 720.

## Conclusion

The key distinction in mathematical concepts between permutation and combination is point order, placement and position, which are required in permutation but not in combinations. Combination refers to a wide range of possibilities for menu items, food, clothing, subjects and so on. Permutations are different ways of arranging objects, people, numbers, letters, colors and other things.

Within the defined requirements, a combination is an unordered set or pair of values. A single combination can result in a vast number of other combinations. A single permutation, on the other hand, only produces one result. Following the above discussion, it should be clear that permutation and combination are two distinct terms used in mathematics, statistics, research and daily life.

## FAQs on Permutation and Combination Problems in Details for JEE

1. What is the difference between permutation and combination?

Permutations are referred to as arrangements, while combinations are referred to as choices. Permutations and combinations are ways of counting the number of alternative outcomes in various scenarios. The concepts of counting are made up of permutations and combinations, which are used in a variety of circumstances. There are sum rules and product rules to use the calculation conveniently, according to the basic concept of calculation. In a permutation, the specifics matter because the order or sequence is crucial.

2. What is the weightage of  Permutation and Combinations in JEE?

As mentioned earlier, it is an important tool to solve problems. Each year, JEE tends to give one or two questions in the exam, mainly with permutation and combination  or concepts using range and all that. It is possible that you may not find any direct questions in JEE, but it is a very fundamental and equally important concept that will help you solve other problems in the exam. You may get a permutation and combination question at least once, be it a direct question or part of the solution. One can get permutation and combination problems with solutions pdf by simply searching on the internet.