This question aims to find the possible number of ways $10$ positions can be assigned to the players out of a team of $13$.

A mathematical method that is used to work out the number of potential groupings in a set when the order of grouping is required. An ordinary mathematical problem involves selecting only a few items from a set of items in a specific order. Most commonly, the permutations are perplexed with another method called combinations.Â In combinations, however, the order of the selected items does not affect the selection.

Permutation and combinations each necessitate a set of numbers. Moreover, the sequence of the numbers isÂ important in permutations. SequencingÂ has no importance in combinations. For instance, in permutation, the order is important, as it is in a combination while opening a lock. There are also multiple kinds of permutations. There are numerous ways to write a set of numbers. Permutations with recurrence, on the other hand, can be found. Specifically, the number of total permutations when the numbers cannot be utilized or can be used more than once.

## Expert Answer

In the given problem:

$n=13$Â andÂ $r=10$

The order of choosing the players is important because a dissimilar order leads to dissimilar positions for dissimilar players and so the permutation will be used in this case. So the number of ways players can be chosen are:

${}^{13}P_{10}$

Since, ${}^{n}P_{r}=\dfrac{n!}{(n-r)!}$

Substitute the values of $n$ and $r$ in the above formula as:

${}^{13}P_{10}=\dfrac{13!}{(13-10)!}$

$=\dfrac{13!}{3!}$

$=\dfrac{13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3!}{3!}$

$=13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4$

$=1037836800$

So, there are $1037836800$ ways to assign the $10$ positions to the players.

## Example 1

Find the maximum number of different permutations of the digits $1,2,3,4$ and $5$ that can be used if no digit is used more than once in making a number plate starting with $2$ digits.

### Solution

Number of total digits $(n)=5$

Digits required in making a number plate $(r)=2$

We are required to find ${}^{5}P_{2}$.

Now,Â ${}^{5}P_{2}=\dfrac{5!}{(5-2)!}$

$=\dfrac{5!}{3!}$

$=\dfrac{5\cdot 4\cdot 3!}{3!}$

$=5\cdot 4$

$=20$

## Example 2

Work out the permutations of the letters in the word COMPUTER.

### Solution

Total in the word COMPUTER is $(n)=6$

Since each letter is distinct, so the number of permutations will be:

${}^{8}P_{8}=\dfrac{8!}{(8-8)!}$

$=\dfrac{5!}{0!}$

Since, $0!=1$ so:

${}^{8}P_{8}=8!$

$=8\cdot 7\cdot 6\cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$

$=40320$