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In how many different words can the letters of the word “OPTICAL” be arranged so that the vowels always come together?
A). 120
B). 720
C). 4320
D). 2160
E) None of these

Last updated date: 21st Jul 2024
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Hint: Here the given question is of permutation and combination word problem where we need to solve for the number of words obtained by using the condition in the question, for which we need to study the word problem theory in the combination.
Formulae Used: The possible word which can be formed by using the word of “N” letters, we have:
\[ \Rightarrow N!\]
Here “!” represents a factorial sign.

Complete step-by-step solution:
Here in the given question we are having one condition which says that the vowels in the word “OPTICAL” should come together and the other letters of the word need to be arranged in any ways, and the total possible words are need to be obtained here.
To solve this question here first we see the vowels in the word:
\[ \Rightarrow Vowels = OIA\]
Now the rest letters are:
\[ \Rightarrow letters = PTCL\]
Now we have the condition that the vowels should come together, hence the word can be written as:
\[ \Rightarrow word = (OIA)PTCL\]
Here in the above word we have 5 letters because vowels will come together hence counted as one, and rest we have four letters, now possible words for five letters is given as:
\[ \Rightarrow 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\]
Here the number of vowels are three and can interchanged their position too, hence we need to consider that too, so for three letters we have:
\[ \Rightarrow 3! = 3 \times 2 \times 1 = 6\]
So the total words comes too be:
\[ \Rightarrow 120 \times 6 = 720\]
Therefore, option (B) is correct.

Note: Here the given question is to solve by combination rule, we can write the words also to see the total words according to the condition, but here the possibilities are seven hundred and twenty hence it might be not possible to write each and every word without forgetting any possibility or without repeating any word hence we need to solve such question by combination theory of words.