Question

The letters of the word SOCIETY are placed at random in a row. The probability that the three vowels come together is?
 $$\begin{gathered} (a){\displaystyle \frac{1}{6}} \\ (b){\displaystyle \frac{1}{7}} \\ (c){\displaystyle \frac{2}{7}} \\ (d){\displaystyle \frac{5}{6}} \end{gathered}$$

Answer 1

Hint: In this question use the fact that there are in total 5 vowels in the English alphabet which are given as (A, E, I, O, U), so this implies we have three vowels (O, I, and E) in the word SOCIETY. Consider these three vowels as one and find the total number of letters remaining in the word. Apply suitable concepts of probability, this will help approaching the problem.

Complete step-by-step answer:
Given word:
SOCIETY
As we see there are seven letters in the word SOCIETY.
Now as we see that there are no repeated letters in the given word SOCIETY.
So the total number of seven letter words which are possible form the given word are (7!).
Now as we know that there are five vowels in the English alphabet which are given as (A, E, I, O, U).
So in the given word SOCIETY there are three vowels (O, I, and E).
Now consider these three vowels as one, so the total letters remaining in the word are (7 – 3 + 1) = 5
So the total number of words possible from these five letters are (5!).
And the number of arrangements of the vowels are (3!), as there are 3 vowels in the given word.
So the total number of possible words containing all vowels together are $\left(5!\times 3!\right)$ .
So the favorable number of outcomes are $\left(5!\times 3!\right)$ and the total number of outcomes are (7!).
Now these words can arrange in rows, so that in one row there is only one word.
Now as we know that the probability is the ratio of the favorable number of outcomes to the total number of outcomes, so the probability when the letters of the word SOCIETY are placed at random in a row, the probability that the three vowels come together is,
 $\Rightarrow P={\displaystyle \frac{\text{favorable number of outcomes}}{\text{total number of outcomes}}}$
Now substitute the values we have,
 $\Rightarrow P={\displaystyle \frac{\left(5!\times 3!\right)}{\text{7!}}}={\displaystyle \frac{\left(5!\times 3.2.1\right)}{7.6.5!}}={\displaystyle \frac{1}{7}}$
So this is the required probability so that the three vowels come together.

Note:The trick concept in the above problem was that when 3 vowels are taken into one then we have remaining 5 words so there 5 words can be rearranged amongst themselves in 5! Ways, but the 3 vowels which are although taken as one can also be permuted amongst themselves and thus 3!. Since we have to take the entire case of vowels being together thus it has to be 3! Times 5! Number of favorable cases.