# Permutation And Combination Problems

## Permutation And Combination Problems

When the order doesn't matter, it is a **Combination**. When the order does matter it is a **Permutation**. In other words A Permutation is an ordered Combination.

**Combination**When the order doesn't matter, it is a Combination. A Permutation is an ordered Combination.**Permutation**When the order does matter it is a Permutation. To help you to remember, think "Permutation ... Position"

The Different arrangements which can be made by taking some or all of the given things or objects at a time is called Permutation.

Each of the different selections or groups which is made by taking some or all of a number of things or objects at a time is called combination .

### Types of Permutation

There are basically two types of permutation:

**Repetition is Allowed:**such as the lock above. It could be "333".**No Repetition:**for example the first three people in a running race. You can't be first and second.

### Permutations with Repetition

These are the easiest to calculate. When a thing has n different types ... we have n choices each time! For example: choosing 3 of those things, the permutations are: n * n * n (n multiplied 3 times)

More generally: choosing r of something that has n different types, the permutations are: n × n × ... (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilities for the second choice, and so on, multiplying each time.)

Which is easier to write down using an exponent of r: n × n × ... (r times) = n ^{r}

**Example: **Iin the lock above, there are 10 numbers to choose from (0,1,2,3,4,5,6,7,8,9) and we choose 3 of them: 10 × 10 × ... (3 times) = 103 = 1,000 permutations

## So, the formula is simply:

n^{r}where n is the number of things to choose from, and we choose r of them, repetition is allowed, and order matters.

**Question 1: **In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

**(A)**920**(B)**825**(C)**720**(D)**610

**Answer: **(C) 720

## Solution

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA). Hence we can assume total letters as 5 and all these letters are different. Number of ways to arrange these letters = 5! = 5 * * 3 * 2 * 1 = 120 All the 3 vowels (OIA) are different Number of ways to arrange these vowels among themselves = 3! = 3* 2 * 1 = 6 Hence, required number of ways = 120 * 6 = 720

**Question 2: **In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

**(A)**810**(B)**1440**(C)**2880**(D)**50400

**Answer: **(D) 50400

## Solution

In the word 'CORPORATION', we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO). This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters = 7!/2! = 2520 Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5!/3! = 20 ways Required number of ways = (2520 x 20) = 50400.

**Question 3: **In how many ways can the letters of the word 'LEADER' be arranged?

**(A)**72**(B)**144**(C)**360**(D)**720

**Answer: **(C) 360

## Solution

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. Required number of ways = 6!/(1!)(2!)(1!)(1!)(1!) = 360

**Question 1: **In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

**(A)**32**(B)**48**(C)**36**(D)**60

**Answer: **(C) 36

## Solution

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants. Let us mark these positions as under: (1) (2) (3) (4) (5) (6) Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5 Number of ways of arranging the vowels =^{3}P_{3}= 3! = 6 Also, the 3 consonants can be arranged at the remaining 3 positions. Number of ways of these arrangements =^{3}P_{3}= 3! = 6 Total number of ways = (6 x 6) = 36.