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.

Complete step-by-step answer:
 First of all, we are given 12 letters of English alphabet and they are chosen randomly. Given that A and B are two letters from them, we need to find the probability that there are 4 letters between A and B.
Let us first calculate the total number of ways to select 12 random alphabets. The total number of ways will be n! i. e., 12!
Now, we are told that A and B are two letters from the 12 selected. So, there are 10 remaining alphabets now. Let us say A and B are given two positions such that there are 4 letters in between them. So, the number of ways to do so will be 2!
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 $probability = \dfrac{{7!2!4!10!}}{{12!6!4!}} = \dfrac{7}{{66}}$ .
Hence, the probability of A and B having 4 letters in between is 7/66.

Therefore, option(D) is correct.

Note: You should take care of using formulae and especially while deducing the number of ways for 4 letters to lie between A and B. You may get confused in the final steps where you need to use multiplication by thinking that addition should be used. These events have occurred one after and they complete the probability, so they are bound to be multiplied instead of adding.