Question

The number of different ordered permutations of all the letters of the word ‘PERMUTATION’ such that two consecutive letters in the arrangement are neither both vowels nor both identical is?
(a) $63\times 6!\times 5!$
(b) $57\times 5!\times 5!$
(c) $33\times 6!\times 5!$
(d) $7\times 7!\times 5!$

Answer 1

Hint: First arrange all the 6 consonants present in the given word. Divide the number of ways with $2!$ because of the repetition of T twice. Now, arrange the 5 vowels present in between these consonants such that there should not be two vowels appearing side by side. Find the possible number of such arrangements. Now, focus on the repeated letter T and find how many times they appear together in the above arrangements by considering them as one unit. Subtract the number of such arrangements from the number of arrangements obtained in the first step to get the answer.

Complete step by step solution:
Here we have been provided with the word ‘PERMUTATION’ and we are asked to arrange the letters of this word such that no two vowels and the identical letter appears consecutively.
Now, we can see that there are 5 vowels (A, E, I, O, U) in the given word and 6 consonants (P, R, M, N, T, T) in which T is appearing twice. We need to arrange the letters so that no two vowels appear adjacent to each other and also both the T’s should not be adjacent. So first let us arrange these 6 consonants in the following manner: -
_ P _ R _ M _ N _ T _ T _
Here we can arrange the six consonants in $6!$ ways but since T is repeating twice so effectively the number of arrangements will be $\dfrac{6!}{2!}$. Now, we need to arrange the 5 vowels for which there are 7 vacant places (denoted by _), so we need to select 5 places out of 7 places first.
$\Rightarrow $ Number of ways to select 5 places out of 7 places = ${}^{7}{{C}_{5}}$
$\Rightarrow $ Number of ways to arrange these 5 vowels among themselves = $5!$
Therefore, total number of arrangements possible = $\dfrac{6!}{2!}\times {}^{7}{{C}_{5}}\times 5!.........\left( i \right)$
As you can see that in the above number of cases some cases will also be there in which the two T’s will appear twice. Like in the above sample of arrangement of consonants if we will not fill the space between the two T’s then they will appear consecutively and make our answer wrong. So we need to find and subtract such possible arrangements from relation (i).
Let us consider that these two T’s appear together, so we can consider them as one unit.
_ P _ R _ M _ N _ TT _
That means we have effectively 5 consonants in which no letter is repeating, so these 5 letters can be arranged in $5!$ ways. Now, we have 6 places left to be filled by 5 vowels so we need to select 5 places from the 6 vacant places.
$\Rightarrow $ Number of ways to select 5 places out of 6 places = ${}^{6}{{C}_{5}}$
$\Rightarrow $ Number of ways to arrange these 5 vowels among themselves = $5!$
Therefore, total number of cases in which the two T’s will appear together = \[5!\times {}^{6}{{C}_{5}}\times 5!...........\left( ii \right)\]
Subtracting the value in (ii) from (i) we get,
$\Rightarrow $ Number of possible ways of arrangement of the letters according to the given question = $\dfrac{6!}{2!}\times {}^{7}{{C}_{5}}\times 5!-5!\times {}^{6}{{C}_{5}}\times 5!$.
On simplifying using the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ we get,
$\Rightarrow $ Possible ways = $63\times 5!\times 5!-6\times 5!\times 5!$
$\Rightarrow $ Possible ways = $57\times 5!\times 5!$
Hence option (b) is the correct answer.

Note: It is a bit complicated problem so you need to understand the question carefully. We have arranged the consonants first so that we might get the clear idea about the places in which vowels can be filled. Remember the combination formula of selection of r objects from a total of n objects given as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.