Permutation is the arrangement of a given set of numbers or things in a certain order. There can be two types of permutation based on if repetition of elements or numbers are allowed or not. The formula for permutation of choosing and arranging non-repeating $r$ elements from a set of $n$ elements can be given as,

$_{n}^{r}\textrm{P}=\frac{n!}{(n-r)!}$

If five digits 1, 2, 3, 4, 5 are being given and a three digit code has to be made from it if the repetition of digits is allowed then how many such codes can be formed.

As repetition is allowed, we have five options for each digit of the code. Hence, the required number of ways code can be formed is, $5\times 5\times 5$ = $125$.

If three alphabets are to be chosen from A, B, C, D and E such that repetition is not allowed then in how many ways it can be done?

The number of ways three alphabets can be chosen from five will be,

$_{5}^{3}\textrm{P}=\frac{5!}{(5-3)!}$ = $\frac{5\times 4\times 3\times 2\times 1}{2\times 1}$ = 60.

Hence, there are 60 possible ways.

$_{n}^{r}\textrm{P}=\frac{n!}{(n-r)!}$

**Example 1:**If five digits 1, 2, 3, 4, 5 are being given and a three digit code has to be made from it if the repetition of digits is allowed then how many such codes can be formed.

**Solution:**As repetition is allowed, we have five options for each digit of the code. Hence, the required number of ways code can be formed is, $5\times 5\times 5$ = $125$.

**Example 2:**If three alphabets are to be chosen from A, B, C, D and E such that repetition is not allowed then in how many ways it can be done?

**Solution:**The number of ways three alphabets can be chosen from five will be,

$_{5}^{3}\textrm{P}=\frac{5!}{(5-3)!}$ = $\frac{5\times 4\times 3\times 2\times 1}{2\times 1}$ = 60.

Hence, there are 60 possible ways.

## Permutations Word Problems

**Problem 1:**

In how many ways can the letters of the word APPLE can be rearranged?

Solution:

Solution:

Total number of alphabets in APPLE = 5.

Number of repeated alphabets = 2

Number of ways APPLE can be rearranged = $\frac{5!}{2!}$ = 60.

The word APPLE can be rearranged in 60 ways.

**Problem 2:**

10 students have appeared in a test in which the top three will get a prize. How many possible ways are there to get the prize winners?

Solution:

Solution:

We need to choose and arrange 3 persons out of 10. Hence, the number of possible ways will be

$_{10}^{3}\textrm{P}$ = $\frac{10!}{(10-3)!}$ = $10\times 9\times 8$ = $720$.

**Problem 3:**

Ellie want to change her password which is ELLIE9 but with same letters and number. In how many ways she can do that?

Solution:

Solution:

Total number of letters = 6.

Repeated letters = 2 Ls and 2 Es.

Number of times ELLIE9 can be rearranged = $\frac{6!}{2!2!}$ = $6\times 5\times 3\times 2\times 1$ = $180$.

But the password need to be changed. So, the number of ways new password can be made = $180 - 1 = 179$.

**Problem 4:**

In how many ways the word HOLIDAY can be rearranged such that the letter I will always come to the left of letter L?

Solution:

Solution:

Number of letters in HOLIDAY = 7 and there is no repetition of letters. Hence, the number of ways all letters can be arranged is 7!.

The number of ways the letters are arranged such that I will come left of L will be, $\frac{7!}{2}$ as in half of the arrangements L will be right of I and in other half it will be on left of I.

**Problem 5:**

There are 6 people who will sit in a row but out of them Ronnie will always be left of Annie and Rachel will always be right of Annie. In how many ways such arrangement can be done?

Solution:

Solution:

The total number of ways of 6 people being seated in a row will be 6!.

Now, with the given constraint the total number of ways will be $\frac{6!}{3!}$ = $6\times 5\times 4$ = $120$.

It implies that out of 6 people arrangement of arrangement of 3 people is predefined.