Question

# The number of words can be formed with the letters of the word PATALIPUTRA in which the relative positions of the vowels and consonants remain same isA. 3600B. 360C. 2400D. 240

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Hint: Find the positions of vowels and consonants in the word PATALIPUTRA where vowels can be A, E, I, O, U and the rest of the English alphabets are consonants. Here we have to arrange the letters according to given conditions in the problem. For arrangement we use the permutation formula.

As we can see that in the word PATALIPUTRA there are 5 vowels and 6 consonants.
And the positions of vowels are 2nd, 4th, 6th, 8th and 11th.
And the positions of consonants are 1st, 3rd, 5th, 7th, 9th and 10th.
As we can see that vowels are A, A, I, U and A.
So, out of 5 vowels 3 are A (same).
Now, consonants are P, T, L, P, T and R out of 6, 2 are P (same) and 2 are T (same).
So, now as it is given in the question that the relative positions of vowels and consonants should be the same.
So, we can arrange vowels and consonants in their own positions only.
And as we know that if there are n numbers and out of them m are of the same type and r are same of another type.
Then number of ways of arranging n numbers will be $\dfrac{{{\text{n!}}}}{{{\text{m! r!}}}}$
So, here 5 vowels in the word PATALIPUTRA can be arranged themselves in $\dfrac{{{\text{5!}}}}{{{\text{3!}}}}$ as a letter A is repeated thrice.
The remaining 6 consonants can permute among themselves in $\dfrac{{{\text{6!}}}}{{{\text{2! 2!}}}}$ as two letters . P and T are repeated twice.
Hence, the total number of permutations are $\dfrac{{{\text{5!}}}}{{{\text{3!}}}}$ * $\dfrac{{{\text{6!}}}}{{{\text{2! 2!}}}}$ = 3600
So, the correct option will be A.

Note: Whenever we come up with this type of problem then first, we have to find the permutation for arranging vowels and then we will find the permutation of arranging the consonants and then we will multiply them both to get the required total number of ways.