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A. $110$

B. $59$

C. $\dfrac{{11!}}{{{{(2!)}^3}}}$

D. $56$

Answer

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Hint: Use combinations, break down the problem in different cases. When the remaining two numbers are the same and when they are not.

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 $2$.

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 $2$.