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. $\frac{9}{20}$
• B. $\frac{9}{40}$
• C. $\frac{7}{20}$
• D. $\frac{11}{40}$

Answer 1

Answer: (D)

$\frac{11}{40}$

We have $3$ Vowels and $3$ Consonants. Words with at least five Letter have to be formed

For Five Lettered Word:

Two cases possible, (2 Vowels, 3 Consonants), and (3 Vowels, 2 Consonants).

Restriction: No consonants falls between any two vowels in the word

Case 1: 2 Vowels, 3 Consonants, with Restriction

Total number of words: ${}^3{C}_2\times 2!\times 4!=144$

Case 2: 3 Vowels, 2 Consonants, with Restriction

Total number of words: ${}^3{C}_2\times 3!\times 3!=108$

Total number of 5 letter words with restriction = $144+108=252$

Total number of 5 letter words without restriction = ${}^6{C}_5\times 5!=720$

6 Lettered words:

Total number of 6 letter words with restriction = $3!\times 4!=144$

Total number of 6 letter words without restriction = $6!=720$

The probability that no constant falls between any two vowels in the word:

$p=\frac{\text{Total number of}5\text{and}6\text{lettered words with restriction}}{\text{Total number of}5\text{and}6\text{lettered words without restriction}}$

$p=\frac{252+144}{720\times 2}=\frac{11}{40}$