Question
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The number of words of four letters containing an 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
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Hint: First of all calculate the ways in which 2 vowels and 2 consonants will be filled in 4 letter words. Then calculate the number of ways in which each place can be filled by consonants and vowels. Finally, multiply all these values to get the total number of words.

Complete step-by-step answer:
Here, we have to find the number of words of four letters containing an equal number of vowels and consonants, and repetition is allowed. First of all, we have to find the number of consonants and vowels in 4 letter words such that both of them are equal.
_ _ _ _
Since we have only 4 places to fill with an equal number of vowels and consonants, therefore, we have only 1 way after doing that and that is we have to put 2 vowels at any 2 places out of 4 and put 2 consonants in remaining 2 places in the 4 letter word.
We can fill two vowels in 2 places out of 4 in the following ways:
V _ _ V
_ V V _
V V _ _ etc.
Here V is for vowels. So basically, we need to choose 2 out of 4 places for vowels. So, we get,
The total number of ways of filling the vowels in 2 out of 4 places \[=4{{C}_{2}}\]
\[\begin{align}
  & =\dfrac{4!}{2!\left( 4-2 \right)!} \\
 & =\dfrac{4!}{2!2!} \\
 & =\dfrac{4\times 3}{2} \\
 & =6\text{ ways} \\
\end{align}\]
Now, we know that we have a total of 5 vowels in the English language that is a, e, i, o, u. So, the first place can be filled by any vowel in 5 ways. Similarly, as the repetition is allowed, second place can also be filled by any vowels in 5 ways.
Hence, we get the total ways in which 2 places can be filled by vowels = 5 x 5 = 25 ways.
Now, we also know that there are a total of 21 consonants in the English language. So, the first place can be filled by any consonants in 21 ways.
Similarly, as the repetition is allowed, second place can also be filled by any consonants in 21 ways.
So, we get the total ways in which two remaining places can be filled by consonants = 21 x 21 = 441 ways.
So, we get the total number of words of four letters containing an equal number of vowels and consonants and repetition being allowed \[=4{{C}_{2}}\times 25\times {{21}^{2}}\]
\[=6\times 25\times 441\]
\[=150\times 441\text{ words}\]
Hence, option (d) is the right answer.

Note: Here, students often make this mistake of multiplying \[4{{C}_{2}}\] two times, one for selecting 2 places for vowels and other for selecting 2 places for consonants, but this is wrong. When we choose 2 places to fill the vowels/consonants, we will automatically get 2 remaining places to fill consonants/vowels and we need not select them. So \[4{{C}_{2}}\] will come only once. This point must be kept in mind.