A word of 6 letters is formed from a set of 16 different letters of English alphabets (with replacement). Find the probability that exactly two letters are repeated.
The answer is $\frac{18080}{16^6}$.
I would prefer different types of approaches to this problem.
Initial approach
I first thought of selecting 4 letters out of 16 and selected 2 out of them for repeating them and arranged. But that is not the answer.
Best Answer
There are ${16\choose 2}=120$ ways to select the two letters appearing repeatedly. Each selected pair is an ordered pair according to the order of letters in the English alphabet.
The selected letters occur in multiplicities $(x,y)$ where $x\geq2$, $y\geq2$. The possible multiplicities are
(i) $\quad(4,2)$, $(3,3)$, $(2,4)$, occupying all six places,
(ii) $\quad(3,2)$, $(2,3)$, occupying five places, and
(iii) $\quad (2,2)$, occupying four places.
In the cases (i)$(4,2)$ and (i)$(2,4)$ we can select the places of the first letter in ${6\choose4}={6\choose 2}=15$ ways, and in case (i)$(3,3)$ we can select the places of the first letter in ${6\choose 3}=20$ ways. This gives a total of $50$ possibilities for case (i).
In case (ii)$(3,2)$ we can select the places of the first letter in ${6\choose3}=20$ ways, then we can select the place to be left temporarily empty in $3$ ways, and finally we can fill this empty place with any of the $14$ remaining letters, making a total of $840$ possibilities. This has to be doubled to take care of the case (ii)$(2,3)$ as well, so that we get $1680$ possibilities in all for case (ii).
In case (iii) we can place the first chosen letter in ${6\choose2}=15$ ways, then the second chosen letter in ${4\choose2}=6$ ways. Now there are two places left empty. For the lefthand empty spot we have $14$ choices among the remaining letters, and for the righthand empty spot we have $13$ choices. In this way case (iii) amounts to $15\cdot 6\cdot 14\cdot 13=16380$ possibilities.
In all we obtain $$120(50+1680+16380)=2173200$$ admissible words. This has to be divided by $16^6$ in order to arrive at the requested probability. The resulting value is $p=0.1295$.
The result we arrived at is different from the answer in your book. Maybe the exact meaning of "repeated" has to be clarified.