# 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.

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) 19
• ARITRA

Type 3:
There are Four types of sweets and there are 19 places, I think the answer would have been 19^4. 27
• vignesh

type 3 belongs to combination with repetion, so they proceeded with nCr 8
• Yashwi

no it;s right 3