Question

A word of at least 5 letters is made at random from 3 vowels and 3 consonants, all the letters being different. The probability that no consonant falls between any two vowels in the word, is
(a) $ \dfrac{9}{20} $
(b) $ \dfrac{9}{40} $
(c) $ \dfrac{7}{20} $
(d) $ \dfrac{11}{40} $

Answer 1

Hint: We start solving the problem by finding the total number of 5 letter words and 6 letter words formed using the given 3 vowels and 3 consonants using the fact that the total number of ways of arranging n objects in r places is $ {}^{n}{{P}_{r}} $ . We then find the total number of 5 letters and 6 letter words formed by satisfying the criteria that no consonant falls between any two vowels in the word by assuming that a word with all vowels together as a single letter. We then make use of the fact probability = $ \dfrac{\text{Number of favorable words formed}}{\text{Total number of words formed}} $ to get the required answer.

Complete step by step answer:
According to the problem, we are given that a word of at least 5 letters is made at random from 3 vowels and 3 consonants, all the letters being different. We need to find the probability that no consonant falls between any two vowels in the word.
Let us find the total number of 5 letters and 6 letter words formed using the given letters.
We know that the total number of ways of arranging n objects in r places is $ {}^{n}{{P}_{r}} $ .
So, the total number of 5 letter words formed using the given letters is $ {}^{6}{{P}_{5}}=6\times 5\times 4\times 3\times 2=720 $ ---(1).
We know that the total number of ways of arranging n objects in n places is $ n! $.
So, the total number of 6 letter words formed using the given letters is $ 6!=6\times 5\times 4\times 3\times 2\times 1=720 $ ---(2).
From equations (1) and (2), we have found the total number of 5 letters and 6 letter words formed is $ 720+720=1440 $ ---(3).
Now, let us find the total number of 5 letter words that no consonant falls between any two vowels in the word; this will be possible only if all the vowels are together.
Let us assume that 2 vowels are present in the 5 letter words.
Since the two vowels are always together, we assume them as a single letter. So, we will have 4 letters to arrange in four places and two vowels arranging between them.
So, the required way is $ {}^{3}{{P}_{2}}\times 4!=144 $ ---(4).
Here $ {}^{3}{{P}_{2}} $ is the number of ways to choose and arrange two vowels, $ 4! $ is to arrange all the four places after assuming the vowel as a word.
Let us assume that 3 vowels are present in the 5 letter words.
Since the three vowels are always together, we assume them as a single letter. So, we will have 3 letters to arrange in three places and two vowels arranging between them.
So, the required way is $ {}^{3}{{c}_{2}}\times 3!\times 3!=108 $ ---(5).
Here $ {}^{3}{{P}_{2}} $ is the number of ways to choose two consonants, $ 3! $ is to arrange the vowels internally and $ 3! $ is to arrange all the three places after assuming the vowel as a word.
Now, let us find the total number of 6 letter words that no consonant falls between any two vowels in the word; this will be possible only if all the vowels are together.
Since the three vowels are always together, we assume them as a single letter. So, we will have 4 letters to arrange in four places and two vowels arranging between them.
So, the required way is $ 3!\times 4!=144 $ ---(6).
Here $ 3! $ is to arrange the vowels internally and $ 4! $ is to arrange all the four places after assuming the vowel as a word.
Now, the total number of ways of favorable letters is $ 144+144+108=396 $ .
We know that the probability = $ \dfrac{\text{Number of favourable words formed}}{\text{Total number of words formed}} $ .
 $ \Rightarrow $ Probability = $ \dfrac{11}{40} $ .
 $ \therefore $ The correct option for the given problem is (d).

Note:
 We should not confuse and make calculation mistakes while solving this problem. We should not confuse between $ {}^{n}{{C}_{r}} $ and $ {}^{n}{{P}_{r}} $ while solving this problem. We can also solve this problem by subtracting the probability of unfavorable cases from 1 which also gives the same answer. Here we should consider both 5 letter and 6 letter words as it is clearly mentioned at least in the problem. Similarly, we can expect the problems to find the probability that there is no vowel present in all the three-letter words formed.