Question

The first 10 letters of English alphabet are written in a row at random. What is the probability that there are exactly three letters in between A and B?

Answer 1

The first step in solving the given question is to calculate the total number of ways to select the 10 letters of the English alphabets by the formula $n!$, n! is the product of all integers from 1 to n. After this, we will fix the positions of A and B and then we will calculate the probability of 3 letters present in between A and B. To calculate this probability, we will use the formula $^{n}{{C}_{r}}$ whose value is calculated as $\dfrac{n!}{r!\left( n-r \right)!}$. Then we will calculate the probability of the required arrangement as $\dfrac{p}{n}$, here p is the number of possible ways, and n is the total number of ways.

Complete step-by-step solution:
As we already said the first step is to calculate the total number of arrangements of the given 10 letters of English alphabet and they are written randomly which means that repetition is allowed. The total number of ways to select 10 random alphabets is $10!$.
Now, we will calculate the number of possible ways. Let us say A and B are given two positions randomly, the number of ways to do so will be $2!$. The 3 letters which are between A and B can be selected from the remaining 8 by the formula so, the total number of ways to select 3 letters will be $^{8}{{C}_{3}}3!$.
There are remaining 5 letters now and let us assume that the arrangement we calculated is one single letter. So, we can say that there are 7 terms now in total. So, the total number of ways to arrange them will be $6!\times 2!\times 3!{{\times }^{8}}{{C}_{3}}$
Hence the probability is $\dfrac{6!\times 2!\times 3!{{\times }^{8}}{{C}_{3}}}{10!}$, we need to simplify this to get a fraction. We will do this by writing algebraic meanings of each term.
$\dfrac{6!\times 2!\times 3!{{\times }^{8}}{{C}_{3}}}{10!}=\dfrac{\left( 6\times 5\times 4\times 3\times 2\times 1 \right)\left( 2\times 1 \right)\left( 3\times 2\times 1 \right)\dfrac{8!}{3!\left( 8-3 \right)!}}{\left( 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right)}$
Further simplifying this, we get
$\dfrac{\left( 6\times 5\times 4\times 3\times 2\times 1 \right)\left( 2\times 1 \right)\left( 3\times 2\times 1 \right)\dfrac{8!}{3!\times 5!}}{\left( 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right)}$
Simplifying all the factorial terms, we get
$\dfrac{\left( 6\times 5\times 4\times 3\times 2\times 1 \right)\left( 2\times 1 \right)\left( 3\times 2\times 1 \right)\dfrac{\left( 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right)}{\left( 3\times 2\times 1 \right)\left( 5\times 4\times 3\times 2\times 1 \right)}}{\left( 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 \right)}$
As we can see that the above expression has many common factors in numerator and denominator, cancelling out all them we get the probability as $\dfrac{2}{15}$.

Note: In all probability questions, we calculate the total number of ways, and the number of favourable ways and calculate their ratio. These types of questions are easy to solve, but calculation mistakes should be avoided.