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Question: If\[{\rm{\;}}{{\rm{P}}_{\rm{n}}}\]denotes the number of n-letter words which can be formed using all...

If  Pn{\rm{\;}}{{\rm{P}}_{\rm{n}}}denotes the number of n-letter words which can be formed using all letters of the word “GOOGLE” under the condition that no two consecutive letters are identical. Then the value of PnPn−1\dfrac{{{{\rm{P}}_{\rm{n}}}}}{{{{\rm{P}}_{{\rm{n - 1}}}}}}is (where unlimited supply of each letter of the above word is given)-
A.4
B.5
C.3
D.A number greater than 25

Explanation

Solution

Here we have to use the concept of the permutation to form an n letter word with five letters in the word “GOOGLE”. So here firstly we have to see how many distinct letters are there to make a word as no two consecutive letters are identical. Then we have to apply the formula of permutation to find out the number of n letter words that can be formed. Then similarly finding the number of n−1{\rm{n - 1}} letter words that can be formed. Then by dividing them, we will get the required ratio.

Formula used:
Number of permutations of n things, taken r at a time, denoted by nPr=n!(n−r)!{}^{\rm{n}}{{\rm{P}}_{\rm{r}}} = \dfrac{{{\rm{n}}!}}{{{\rm{(n - r)}}!}}

Complete step-by-step answer:
The given word is “GOOGLE” whole letters are used to form words. So, there are only 4 distinct letters available for the formation of the words i.e. G, O, L, E.
The words that are formed is subjected to a condition of no two consecutive letters are identical. So, for the first letter of the word we have a choice of 4 letters but for all the remaining letters of the word we will have a choice of 3 letters.
So number of ways in which first letter of the word is selected=4P1=4!(4−1)!=4!(3)!=4{\rm{ = }}{}^4{{\rm{P}}_1} = \dfrac{{4!}}{{{\rm{(4}} - {\rm{1)}}!}} = \dfrac{{4!}}{{{\rm{(}}3{\rm{)}}!}} = 4
Now number of ways in which all the remaining letters of the word is selected =3P1=3!(3−1)!=3!(2)!=3{\rm{ = }}{}^3{{\rm{P}}_1} = \dfrac{{3!}}{{{\rm{(3}} - {\rm{1)}}!}} = \dfrac{{3!}}{{{\rm{(2)}}!}} = 3
Here 3 is multiplied n−1{\rm{n - 1}} times as for the first letter is selected in 4 ways and then all the remaining n−1{\rm{n - 1}} letters is selected in 3 ways. Therefore, we get
Therefore, number of ways n letter words that can be formed =4×3×3×.... = 4 \times 3 \times 3 \times ....
We can write number of ways n letter words that can be formed as =4×3(n−1) = 4 \times {3^{({\rm{n}} - 1)}}
⇒Pn=4×3(n−1)\Rightarrow {{\rm{P}}_{\rm{n}}} = 4 \times {3^{({\rm{n}} - 1)}}
Similarly, number of ways n−1{\rm{n - 1}} letter words that can be formed=4×3(n−2) = 4 \times {3^{({\rm{n}} - 2)}}
⇒Pn−1=4×3(n−2)\Rightarrow {{\rm{P}}_{{\rm{n - 1}}}} = 4 \times {3^{({\rm{n}} - 2)}}
Now we have to find their ration, we get
⇒PnPn−1=4×3(n−1)4×3(n−2)=4×3(n−2)×34×3(n−2)=3\Rightarrow \dfrac{{{{\rm{P}}_{\rm{n}}}}}{{{{\rm{P}}_{{\rm{n - 1}}}}}}{\rm{ = }}\dfrac{{4 \times {3^{({\rm{n}} - 1)}}}}{{4 \times {3^{({\rm{n}} - 2)}}}}{\rm{ = }}\dfrac{{4 \times {3^{({\rm{n}} - 2)}} \times 3}}{{4 \times {3^{({\rm{n}} - 2)}}}}{\rm{ = 3}}
⇒PnPn−1=3\Rightarrow \dfrac{{{{\rm{P}}_{\rm{n}}}}}{{{{\rm{P}}_{{\rm{n - 1}}}}}}{\rm{ = 3}}
Hence, option C is the correct option.

Note: Here we have to keep in mind to use the concept of permutation rather than combination. As permutations may be defined as the different ways in which a collection of items can be arranged. For example: The different ways in which the numbers 1, 2 and 3 can be grouped together, taken all at a time, are123,132,213231,312,321123,{\rm{ }}132,{\rm{ }}213{\rm{ }}231,{\rm{ }}312,{\rm{ }}321.
Combinations may be defined as the various ways in which objects from a set may be selected. For example: The different selections possible from the numbers 1, 2, 3 taking 2 at a time, are 12,23 and 31{\rm{12, 23\, and\, 31}}