Question

In how many ways can the letter of the word DELHI arranged, so that the vowels occupy only the even places?
A. 2
B. 6
C. 12
D. 36

Answer 1

Hint: Here, use the rules of permutation as this is a question of arrangement. Keep in mind the positions of vowels i.e. vowels must be in second and fourth places. So first find the number of ways for consonants and then find the number of ways for vowels, and multiply both the results.

Complete step-by-step answer:
We are given the word DELHI, which has 5 letters.
Also there are two vowels E and I.
And it is given that all vowels should occupy only the even places.
Even places in the word consisting of 5 letters means second and fourth places.
Number of ways for two vowels to take second and fourth places (which are even places) can be given as ${}^2{P_2}$.
[Permutation formula: According to permutation formula, if we have total n object and we have to choose p objects then it can be done in ${}^n{P_r}$ ways i.e. $\dfrac{{n!}}{{(n - r)!}}$ ]
And also the number of ways to arrange the remaining 3 letters which are consonants (D, L, H) in three places (i.e. odd places) can be given as ${}^3{P_3}$
Now, ${}^2{P_2} = \dfrac{{2!}}{{(2 - 2)!}} = \dfrac{{2!}}{{0!}} = 2$ [Since, 0! = 1]
And ${}^3{P_3} = \dfrac{{3!}}{{(3 - 3)!}} = \dfrac{{3 \times 2}}{{0!}} = 6$ [Since, 0! = 1]
In permutation ‘and’ means multiplication and ‘or’ means addition. Here both the above arrangement for vowels and for consonants will happen together, therefore, the two results will be multiplied.
So, the total number of ways to arrange the word DELHI arranged, so that the vowels occupy only the even places is 2 × 6 = 12.
Hence, the correct option is (C).

Note: In these types of questions simply apply the rules of permutation, with given conditions. Do not try to write all the possible results as it may become very large in quantity. Hit and trial method will not work for more questions.