Question

In how many ways can you arrange $26$ letters?

Answer 1

We solve the problem using simple logic. We take $26$ boxes which are laid side by side one after the other. One box can have only one letter inside it. The first box can be filled by any one letter out of the total $26$ letters. So, the number of ways to fill the first box will be $26$ . The second box can be filled by any one letter out of the total $25$ letters. So, the total number of ways to fill all the boxes will be $26\times 25\times 24\times 23\times 22\times ...\times 1$ .

Complete step by step solution:
Let us imagine that there are $26$ boxes which are laid side by side one after the other. One box can have only one letter inside it. So, the number of ways we can arrange $26$ letters will be equivalent to the number of ways of filling the boxes by different letters. The first box can be filled by any one letter out of the total $26$ letters. So, the number of ways to fill the first box will be $26$ .
Now, one letter has been used. So, the second box can be filled by any one letter out of the total $25$ letters. So, the number of ways to fill the second box will be $25$ . For, each of the $26$ ways of filling the first box, the second box can be filled in $25$ ways. The total ways of filling the first two boxes will then be $26\times 25$ . Similarly, the total ways of filling the first three boxes will then be $26\times 25\times 24$ , the total ways of filling the first four boxes will then be $26\times 25\times 24\times 23$ and so on till the last box, where only one letter can be put.
This means that the total ways of filling the $26$ boxes will then be $26\times 25\times 24\times 23\times 22\times ...\times 1$ , which is nothing but $26!$ .
Thus, we can conclude that we can arrange $26$ letters in $26!$ ways.

Note: The permutation of r objects out of n objects is denoted in two ways, which are ${}^{n}{{P}_{r}}$ and $P\left( n,r \right)$ . The corresponding formula for the number of permutations possible is given by,
$P\left( n,r \right)=\dfrac{n!}{\left( n-r \right)!}$
Putting, $n,r=26$ , we get,
${}^{26}{{P}_{26}}=\dfrac{26!}{\left( 26-26 \right)!}=26!$