Question

The total number of words that can be made by writing the letters of the word PARAMETER so that no vowel is between two consonants is :
A. 1440
B. 1800
C. 2160
D. None of these

Answer 1

Hint: To find the total number of words that can be made without taking vowel between two consonants, we need to find the number of spots which can justify the condition given in the question. Then we will find the number of ways the vowel and consonants separately. Then we will find the total number of words by multiplying all the possible ways.

Complete step-by-step solution -
The word given in the question is ‘PARAMETER’.
It consists of 5 consonants i.e. P, R, M, T, R, and 4 vowels i.e. A, A, E, E.
The condition given the question is that no vowel is between two consonants which means that the word must consist of the consonants being in a single block and the vowels flowing around them.
So, the spots the consonants can start are :
$\begin{aligned} & CCCCCVVVV \\\ & VCCCCCVVV \\\ & VVCCCCCVV \\\ & VVVCCCCCV \\\ & VVVVCCCCC \\\ \end{aligned}$
Therefore, the spots that the consonants can start are in 5 ways ……… (1)
Now, we will take the vowels together that in A, A, E, E then we can arrange them 4! Ways.
As the ‘A’ and ‘E’ are repeating twice, so we will divide it by $2!\times 2!$.
So, we get –
$\begin{aligned} & \Rightarrow \dfrac{4!}{2!\times 2!}=\dfrac{4\times 3\times 2\times 1}{2\times 2} \\\ & =\dfrac{24}{4} \\\ \end{aligned}$
By cancelling the common factors from numerator and denominator, we get –
$=6$ ways …………………………. (2)
Now, we will take all the consonants that in P, R, M, T, R then we can arrange them in 5! Ways. As ‘R’ is repeating twice, so we will divide it by 2!.
So, we get –
$\begin{aligned} & \Rightarrow \dfrac{5!}{2!}=\dfrac{5\times 4\times 3\times 2\times 1}{2\times 1} \\\ & =\dfrac{120}{2} \\\ \end{aligned}$
By cancelling the common factors from numerator and denominator, we get –
$=60$ ways …………………………… (3)
By taking equation (1), (2), and (3), we can find the total number of ways.
So, the total numbers of words are $5\times 6\times 60=1800$
Therefore, the total numbers of words are 1800
Hence, option B. is the correct answer.

Note: We can also solve this problem in another way.
There are 5 letters which can be arranged in 5! Ways as A and E are repeated, so we divide it by $2!\times 2!$.
Now, 5 consonants can be arranged in 5! Ways, as R is repeating twice we should divide it by 2!.
$\therefore$ The number of ways are
$\begin{aligned} & \Rightarrow \dfrac{5!\times 5!}{2!\times 2!\times 2!}=\dfrac{5\times 4\times 3\times 2\times 1\times 5\times 4\times 3\times 2\times 1}{2\times 2\times 2} \\\ & =\dfrac{14400}{8} \\\ & =1800 \\\ \end{aligned}$