In Mathematics, the permutation can be explained as the arrangement of objects in a particular order. It is an ordered Combination where order of arrangement is important. It is the all possible ways of doing something where order is taken into consideration.

Let us consider an example. For instance, there are 6 teams A, B, C, D, E and F participating in a contest. In how many ways can we award a 1st $(Gold)$, $2nd\ (Silver)$ and $3rd\ (Bronze)$ place prize among six teams?

- Gold medal: 6 choices: A B C D E F. Example, A wins the contest for the Gold.
- Silver medal: 5 choices: B C D E F. Example, B wins the contest for the silver.
- Bronze medal: 4 choices: C D E F. Example, C wins the contest for the bronze.

First, we had 6 choices, then 5 and then 4. Total no. of options: $6 \times 5 \times 4$ = 120.

We should order three teams out of 6 teams. For doing this, we began with all 6 options and took them away one at a time $(5,\ then\ 4)$ until all medals are all given away.

We know that 6! = $6 \times 5 \times 4 \times 3 \times 2 \times 1$

And, 3! = 3 $\times 2 \times 1$

So, while doing $\frac{6!}{3!}$ we get:

To derive a general formula for permutation, let us see how we can represent $\frac{6!}{3!}$ in total number of teams and number of medals.

$\frac{6!}{3!}$ can also be written as $\frac{6!}{(6-3)!}$.

The total number of teams contesting is 6, the total number of teams awarded are 3.

If n represents total number of teams, r represents number of teams awarded, and then the number of ways the teams can be ordered can be given by $\frac{n!}{(n-r)!}$. Here the same team cannot take two awards. So the order is not repeated.

When we consider examples like licence plate, password or codes, the objects $(terms\ or\ numbers)$ can repeat in the order. Such arrangements are considered as permutations with repetitions.

When we choose r of n things, then the permutations are $n \times n \times n...(r\ times)$ = n

The notation for permutation is P(n, r) or

1. Identify if the arrangement is with repetition or without repetition.

2. If it is with repetition, then the permutation of selecting is calculated by:

$n \times n \times n...(r\ times)$ = n^{r}

where, n is the total number of objects and r is the number of objects in arrangement.

3. If it is an arrangement without repetition, then the permutation of selecting is calculated by:

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

where, n is the total number of objects and r is the number of objects in arrangement.

Solution: Since the arrangement has no repetitions, we find the permutation without repetitions.

Here, n = Total number of letters given = 5

r = Number of letters chosen = 3

Then the number of possible ways is given by:

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

= $\frac{5!}{(5-3)!}$ = $\frac{5!}{2!}$ = $\frac{(5 \times 4 \times 3 \times 2 \times 1)}{(2 \times 1)}$ = 60

There are 60 ways possible 3 - letter arrangement from r, i, g, h, t if each letter is used only once.= $\frac{5!}{(5-3)!}$ = $\frac{5!}{2!}$ = $\frac{(5 \times 4 \times 3 \times 2 \times 1)}{(2 \times 1)}$ = 60

Solution: A password can be set with repetition of digits. So the arrangement is with repetition.

Here we have total number of digits $( 0,\ 1,\ 2...9)$, n = 10

and the number of digits chosen, r = 3.

Thus the possible ways is given by:

$P(n,\ r) = n \times n \times n...(r\ times)$ = n^{r}

= 10^{3 }= 1000 ways.

Thus there are 1000 ways to set a 3 digit password.
= 10

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