Question

One 5 letter word is to be formed taking all letters S, A, P, T and E. What is the probability that the word formed will contain all vowels together.

Answer 1

Hint: The first step to solve these types of questions is to find the number of elements present in the sample space and after that we will find the number of elements or words in which the vowels are together, with the help of permutation or simple principle of multiplication. The formula for calculating the probability of an event in the respective experiment should be used here, the formula is,
$P(A)=\dfrac{n(A)}{n(S)}$

Complete step-by-step answer:
Now, P(A) is the probability of occurrence of Event A,
n(A) is the number of all outcomes in favour of event A and in this case it is number of word having all the vowels together, in this case we are going to find the possible words in which all the vowels are together and for this we will assume temporarily that there are four letters and one letter comprises of all the vowels (in this case A and E) and our answer will be ${}^{4}{{P}_{4}}=\dfrac{4!}{\left( 4-4 \right)!}=\dfrac{4!}{0!}=\dfrac{4\times 3\times 2\times 1}{1}=24$ and multiply it by two because there are two vowels and they can be arranged in two ways.
n(S) is the number of all possible outcomes in the experiment or the number of elements in sample space which we can find by using permutations and will come out to be ${}^{5}{{P}_{5}}=\dfrac{5!}{\left( 5-5 \right)!}=\dfrac{5!}{0!}=\dfrac{5\times 4\times 3\times 2\times 1}{1}=120$ by using the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
At the end put the values in the probability formula and find the probability of forming a word with all vowels together.

We have to select 1 out of 5 alphabets for each position in the word and the number of choices will vary after each selection because repetition is not allowed.
So, we have 5 letters S, A, P, T and E and we have to construct 5 letter words by using them.
In this question we are going to use the principle of multiplication as there are 5 jobs to be done one after the other and when every position of the word is filled with some letter then only we can say that the whole job is completed and also after selecting one letter we have many other choices left.
The left-most position is first in the word and the right-most is the last position.
If we are fixing a letter to a position then we will write that letter below the position space and above it we will write 1 because there is only one choice and that is the fixed letter.

\\_ & \\_ & \\_ & \\_ \\\ \end{matrix}\begin{matrix} {} \\\ {} \\\ \end{matrix}\\_$$ As in the beginning we have 5 choices of letters, so we will write the number 5 in the first position. $$\underset{\scriptscriptstyle-}{5}\begin{matrix} {} \\\ {} \\\ \end{matrix}\begin{matrix} \\_ & \\_ & \\_ & \\_ \\\ \end{matrix}$$ Now as repetition is not allowed we have only 4 choices left, so in the second position we will write 4. $$\underset{\scriptscriptstyle-}{5}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{4}\begin{matrix} {} \\\ {} \\\ \end{matrix}\begin{matrix} \\_ & \\_ & \\_ \\\ \end{matrix}$$ Now there are three letters left, so the third position will have three choices. $$\underset{\scriptscriptstyle-}{5}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{4}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{3}\begin{matrix} {} \\\ {} \\\ \end{matrix}\begin{matrix} \\_ & \\_ \\\ \end{matrix}$$ For the second position there are two letters left and so are the two choices. $$\underset{\scriptscriptstyle-}{5}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{4}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{3}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{2}\begin{matrix} {} \\\ {} \\\ \end{matrix}\\_$$ Now there is only one choice of letter left, so the last position will be having only one last choice and we will put there 1. $$\underset{\scriptscriptstyle-}{5}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{4}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{3}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{2}\begin{matrix} {} \\\ {} \\\ \end{matrix}\underset{\scriptscriptstyle-}{1}$$ Now these numbers are multiplied as told in the discussions done above. So we will get, $$\underset{\scriptscriptstyle-}{5}\times \underset{\scriptscriptstyle-}{4}\times \underset{\scriptscriptstyle-}{3}\times \underset{\scriptscriptstyle-}{2}\times \underset{\scriptscriptstyle-}{1}=5!$$ $$5!=120$$ Hence there are a total 120 possible arrangements of words made from 5 letters. Now we need to find the number of words in which all the vowels are together. As there are two vowels in the given 5 letters S, A, P, T and E. the vowels are A and E. In order to solve this portion we have to consider the two vowels as one letter and we will assume that there are only 4 letters and proceed with the question. But there will be two cases because there are two possibilities of arranging these two letters even when they are together and assumed to be a single letter. In Case-1 we assume that letters A and E are together but A is in first place and E is in second place of the letter formed by merging these two letters, and it will look like  In Case-2 we assume that letters E and A are together but now E is in first place and A is in second place of the letter formed by merging these two letters, and it will look like  As, now there are 4 letters in each case so the possible arrangements of letters to form words will be the same. In that way we only need to calculate the number of arrangements in any of the one cases mentioned above and then multiplying it by 2 to get the final answer which includes arrangements of both cases. So to find the number of possible arrangements we are going to follow the method we used above to find the arrangements of 5 letters to form words. Hence directly jumping to the conclusion, we get, $$\underset{\scriptscriptstyle-}{4}\times \underset{\scriptscriptstyle-}{3}\times \underset{\scriptscriptstyle-}{2}\times \underset{\scriptscriptstyle-}{1}=4!$$ $$4!=24$$ So, now we have got 24 arrangements in Case-1 and 24 arrangements in Case-2, which means we have got a total of 48 arrangements by both the cases, $$24\times 2=48$$. Now we have to find the probability that the word formed will have all vowels together, so we have got the value of total outcomes which is $$n\left( S \right)=120$$ and also the value of favourable outcomes which is $$n\left( A \right)=48$$. Therefore the probability of occurrence of the event is, $$P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$$ $$P\left( A \right)=\dfrac{48}{120}$$ $$P\left( A \right)=\dfrac{2}{5}$$ Therefore the probability that the word formed, by using the given 5 letters, will have all vowels together is $$\dfrac{2}{5}$$. Note: Instead of using the block method we can also use the permutations by using the formula $${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$$ which means permutations of n distinct objects taken r at a time. Remember that the merging of vowels to one letter is temporary and to remove this assumption after using it we will multiply the answer obtained by solving one with the factorial of the number of letters merged to form a single temporary letter. Like in this case there were two letters and we multiplied the answer by $$2$$ or $$2!$$ but if there were three letters to be merged then we would multiply our answer by $$3!$$ or $$6$$ and so on. Special consideration is that when dealing with words and numbers in permutations and combinations always look out for the same letters or digits and the number of times they are repeated because to get the correct answer then you will have to divide the answer by the factorial of the number of times that letter or digit is occurring. Example if we need to find all possible words made using the letters of word ROOTS then the answer will be $$\dfrac{5!}{2!}$$ because the letter O is repeating itself only once and there are two O, so some cases will repeat because of them which we do not want in our answer.