# Tips And Tricks And Shortcuts For Permutation And Combination

## Tips and Tricks & Shortcuts for Permutation & Combination

**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**^{n}P_{r.}**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**^{n}C_{r}**.**

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

### Permutation and Combinations Tips and Tricks and Shortcuts

- 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

Order | Repetition | Formula |
---|---|---|

Permutation | Yes | n^{r
} |

Permutation | No | np_{r} |

Combination | Yes | r + n – 1_{Cr} |

Combination | No | n_{Cr} |

**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)

= (^{7}C_{3} x ^{4}C_{2}) |

= | 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 | = (^{6}C_{1} x ^{4}C_{3}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{4}) | |||||||||||||||||||

= (^{6}C_{1} x ^{4}C_{1}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{2}) | ||||||||||||||||||||

| ||||||||||||||||||||

= (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 | = (^{3}C_{1} x ^{6}C_{2}) + (^{3}C_{2} x ^{6}C_{1}) + (^{3}C_{3}) | |||||||||||||

| ||||||||||||||

= (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.

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

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

**Options:**

- 720
- 360
- 200
- 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:**

- 4
- 6
- 120
- 36

**Solution **We know that,

P_{r} = n!

P_{3} = 3!

P_{3} = 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:**

- 480
- 540
- 720
- 1540

**Solution **^{r + n – 1}C_{r =} ^{19 + 4 – 1}C_{19} =^{22}C_{19}

We know that, ^{n}C_{r} = \frac{n!}{(n-r)! r! }

^{22}C_{19} =\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:**

- 720
- 120
- 24
- 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**

**Read Also – Click for More on Permutation on Combination.**

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

Type 3:

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

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