Permutations Examples & Word Problems

# Permutations Examples

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

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:

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:

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:

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:

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:

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.