Question

The number of words of four letters containing equal number of vowels and consonants, repetition being allowed, is

• A. ${105}^2$
• B. $210\times 243$
• C. $105\times 243$
• D. none of these

Answer 1

Answer: (B)

$210\times 243$

Given, number of vowels =number of consonants
So, there will be 2 cases (i) no repetition(number of vowels =number of consonants=2)
(ii) both the vowel and consonant will be repeated twice(number of vowels =number of consonants=1)
The method that will be followed involves selection of 2 places out of the 4 places which can be done in 6 ways and then applying the above 2 cases
case (i): there will be 21 ways for secting the first consonant and 20 ways for the second and There will be 5 ways of selecting the first vowel and 4 ways of selecting the second.
that makes number of words in case (i) = $6(21\times 20\times 5\times 4)$
case (ii): we have to select only 1 out of the 21 consonants and 5 vowels and both of them will be repeated twice
$⟹$number ofwords in case (ii) =$6(21\times 5)$
\therefore number of words$=6(21\times 20\times 5\times 4)$+$6(21\times 5)$
$=6(21\times 5)(20\times 4+1)$
$=210\times 243$