Question

A word of $6$ letters is formed from a set of $16$ different letters of the English alphabet (with replacement). Find out the probability that exactly two letters are repeated.

Answer 1

First we will calculate the number of ways to pick $5$ letters from a set of $16$ different letters and then we will observe whether repetition is allowed or not. If repetition is allowed, we will calculate the number of ways to pick $5$ letters from a set of $16$ different letters with repetition. Now from the number of ways of picking $5$ letters from a set of $16$ different letters with repetition, we will calculate the number arrangements that are possible to form a $6$ letter word. This will result in a number of possible events for the given condition. Now we will need to calculate the total number of events in a sample space. For this we will calculate the number of ways to pick $6$ letters from a set of $16$ different letters with repeating $2$ letters. Now this will be our total number of events in the sample space. To calculate the probability of the given condition we will find the ratio of the number of possible events for the given condition and the total number of events in the sample space.

Complete step-by-step answer:
Given that, a word of $6$ letters is formed from a set of $16$ different letters of the English alphabet (with replacement)
First, we will choose $5$ letters from $16$ different letters in ${{5}^{16}}$ different ways.
Given that replacement/repetition is allowed, so the number of ways to pick $5$ letters from $16$ different letters with repetition is $5\times {{5}^{16}}$.
Now all the letters are arranged in $6!$ to form a word. Hence, the number of possible cases for the given condition is given by
$n\left( A \right)=5\times {{5}^{16}}\times 6!$.
To find the probability we need the total number of events in the sample space.
We can pick $6$ letters from $16$ letters in ${{16}^{6}}$.
We need to divide the obtained value by $2!$ since they have mentioned that two letters are repeated.
$\therefore P\left( A \right)=\dfrac{{{5}^{16}}\times 5\times 6!}{2!\times {{16}^{6}}}$

Note: We can also solve this problem in another method. The first four letters, non-repeated, are chosen in certain arrangements: $16\times 15\times 14\times 13$
There are $12$ possibilities for the repeated letter, and there are ${}^{6}{{C}_{2}}$ ways to place that letter in two positions.
So, the probability is $16\times 15\times 14\times 13\times {}^{6}{{C}_{2}}$.