# Tips And Tricks And Shortcuts For Permutation And Combination

## Tips and Tricks, and Shortcuts for Permutation and Combination

Tips and Tricks for Permutation and Combination has been discussed on this page to help student practice shortcuts while solving questions:

• Permutation: The different arrangements of a given number of things by taking some or all at a time, are called permutations. This is denoted by nPr.
• Combination: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination. This is denoted by nCr.

Here, are rapid and easy tips and tricks and shortcuts on Permutation and Combination questions swiftly, easily, and efficiently in competitive exams and recruitment exams.

### Tips and Tricks and Shortcuts for Permutation and Combination

• Use permutations if a problem calls for the number of arrangements of objects and different orders are to be counted.
• Use combinations if a problem calls for the number of ways of selecting objects and the order of selection is not to be counted.
• Summary of formula to use

OrderRepetitionFormula
PermutationYesnr
PermutationNonpr
CombinationYesr + n – 1Cr
CombinationNonCr

## Types of Tips and Tricks and Shortcuts for Permutation and Combination

In how many ways can a word be arranged.

Tricks and Tips on type 1 Question

This is Permutation Question.

Let us take this ahead as an example –

In how many ways can the letters of the word ‘LEADER’ be arranged?

Count number of Occurances

• L – 1
• E – 2
• A – 1
• D – 1
• R – 1

Total Unique Occurrences – 6(as E repeated 2 times)

Direct Formula = (Unique Occurrences)!/(Each Individual Unique Occurrences)

so = 6!/(1!)(2!)(1!)(1!)(1!) = 360

In how many ways x objects out of a and y objects out of b can be arranged.

Tips and Tricks type 2 problems

Let us take this as well with an example –

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

 = (7C3 x 4C2)
 = 7 x 6 x 5 x 4 x 3 3 x 2 x 1 2 x 1

= 210.Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.

Required number of ways = (210 x 120) = 25200.

There are x objects and y objects, a from x has be selected and b from y. How many ways can it be done when N Number of objects from x should always be selected

Tricks and Tips Type 3 Problems

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required number
of ways
= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)

 = (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5 2 x 1 2 x 1 3 x 2 x 1 2 x 1
= (24 + 90 + 80 + 15)
= 209.

Coloured Ball Questions

Tricks and Tips Type 4 Problems

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

 = 3 x 6 x 5 + 3 x 2 x 6 + 1 2 x 1 2 x 1
= (45 + 18 + 1)
= 64.

Circular Combinations Problems

If 6 people are going to sitting at a round table, but Sam will not sit next to Suzie, how many different ways can the group of 6 sit?

Couple of of ways of doing this:

First:
a. Total circular permutations = (6-1)! = 5! = 120.
b. Ways in which Sam and Suzie sit together = 2! * 4! = 2*24 = 48
Required ways = Total – Together = 120 – 48 = 72.

Second:
a. We have total of 6 places. Fix Suzie. Now Sam can’t sit at either seat beside her. So number of places where Sam can sit = 5-2 = 3.
For the other 4 people we can arrange them in 4! ways in 4 seats.
So total ways = 3 * 4! = 72.

## Questions on Types of Tips and Tricks and Shortcuts for Permutation and Combination

### Type 1: Different ways to arrange (with repetition)

Question 1 In how many ways can the letters of the word ‘LEADER’ be arranged?

Options:

1. 720
2. 360
3. 200
4. 120

Solution      Letter ‘E’ appears twice and all other letters 1L, 1A, 1D and 1R appears once in the word.

Required number of ways = $\frac{6!}{2!}$ =$\frac{ 6 × 5 × 4 × 3 × 2 × 1}{2 × 1} = \frac{720}{2}$ = 360

Correct option: 2

### Type 2: Different ways to arrange (without repetition)

Question 1 How many different ways are there to arrange your first three classes if they are Math, English, and Hindi?

Options:

1. 4
2. 6
3. 120
4. 36

Solution      We know that,

Pr = n!

P3 = 3!

P3 = 6

Correct option: 2

### Type 3: Different ways to select (with repetition)

Question 1 In a shop there are 4 types of sweets. In how many ways can Shekhar buy 19 sweets?

Options:

1. 480
2. 540
3. 720
4. 1540

Solution      r + n – 1Cr = 19 + 4 – 1C19 =22C19

We know that, nCr = $\frac{n!}{(n-r)! r! }$

22C19 =$\frac{22!}{(22-19)! 19! }$ = 1540

### Type 4: Different ways to select (without repetition)

Question 1 How many different 4 digit numbers can be formed using the digits 2,3,4,5,6,7,8 no digit being repeated in any number

Options:

1. 720
2. 120
3. 24
4. 840

Solution:    The thousand place can be filled in 7 ways, the hundredth place can be filled in 6 ways, the tens place can be filled in 5 ways, and the ones place can be filled in 5 ways.

Total ways = 7*6*5*4 = 840

Correct option: 4

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### 4 comments on “Tips And Tricks And Shortcuts For Permutation And Combination”

• Shravana

Note clear with type 3 ques.I also think it is 19^4(19X19X19X19ways)

• ARITRA

Type 3:
There are Four types of sweets and there are 19 places, I think the answer would have been 19^4.

• vignesh

type 3 belongs to combination with repetion, so they proceeded with nCr