Question

# If the four-letter words (need not to be meaningful) are to be formed using the letter from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is: A. $110$ B. $59$ C. $\dfrac{{11!}}{{{{(2!)}^3}}}$ D. $56$

It’s given to us that first and fourth latter has to be R and E respectively. In the word “MEDITERRANEAN”, we have$13$places out of which$2$is fixed. So, we are remaining with$11$places and the remaining letters are$2{\text{ times }}N$,$2{\text{ times }}E$,$2{\text{ times }}A$and$1$time$M,D,I,T,R$. With all these letters we need to fill two places. Remember that we have already fixed two places. We can break down this problem into two cases.
Case One: Both the places, which we need to fill, are the same. That is EE, AA and NN. The total count for this case is$3$.
Case Two: Both the places, which we need to fill, are not the same. Then we need to form two letters from 8 letters without repetition. Number of such word will be$\left( {\begin{array}{*{20}{c}} 8 \\ 2 \end{array}} \right) = \dfrac{{8!}}{{2!\left( {8 - 2} \right)!}} = \dfrac{{8 \times 7 \times 6!}}{{2 \times 6!}} = \dfrac{{56}}{2} = 28$. Since, EA and AE both make different sense but this formula of combination have not counted this way. So, we need to double this number. that is$28 \times 2 = 56$.
Hence, the total number of such four-letter words with R as first place and E as fourth place, will be$56 + 3 = 59$
Note: The hack in this question was to understand that, we need to multiply $28$ by $2$. To develop this understanding, we need to understand the core concept of combination. It’s a tool to count, By how many ways we can choose $2$ letters out of $8$ letters. That’s it. It has nothing to do, whether EA and AE are the same or not. It’ll just count them as $1$. Which is not correct in our case, that's why we multiplied $28$ by $2$.