# The first 12 letters of the English alphabets are written down at random. The probability that there are 4 letters between A and B is: A.$\dfrac{7}{{33}}$B.$\dfrac{{12}}{{33}}$C.$\dfrac{{14}}{{33}}$D.$\dfrac{7}{{66}}$

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Hint: First we will check the total number of ways to select the 12 letters of the English alphabets by the formula n! Then, we will fix the positions of A and B and then we will calculate the probability of 4 letters lying between A and B by the formula ${}^n{C_r}$where n is the total ways and r is the selected ways in which we are required to calculate the probability. Then we will calculate the probability of the required arrangement by the formula: possible ways / total number of ways.

The 4 letters which are between A and B can be selected from the remaining 10 by the formula ${}^n{C_r}$ so, the total number of ways to select 4 letters will be ${}^{10}{C_4} \times 4!$
Now, we can see that these events occur simultaneously after each other. So, the total number of ways will be the product of all the above events i.e., the ways will be $2! \times 4! \times {}^{10}{C_4}$.
There will be remaining 6 letters now and let us assume that the arrangement we just calculated as one individual letter. So, we can say that there are 7 letters now in total. So, the total number of ways to arrange them will be $7! \times 2! \times 4! \times {}^{10}{C_4}$.
Therefore, the required probability will be equal to $\dfrac{{7! \times 2! \times 4! \times {}^{10}{C_4}}}{{12!}}$by the formula of required probability = total possible ways / total number of ways.
Hence, we get $probability = \dfrac{{7!2!4!10!}}{{12!6!4!}} = \dfrac{7}{{66}}$ .