Question

Find the number of ways in which the letters of the word ‘ARGUMENT’ can be arranged so that only consonants are placed at both ends, options are:
a) 14400
b) 41000
c) 15000
d) 14800

Answer 1

Hint:In this question, have to arrange alphabets of the words in a particular manner which means we have to apply permutation and combination to solve the question. We simply apply the formula and obtain our result.

Complete solution step by step:
Firstly we write down the word which has to be arranged i.e.
ARGUMENT
Now we count the number of consonants and vowels of the word for a better understanding
Vowels – A, U, E = 3
Consonants – R, G, M, N, T = 5
So we make a box and fix its ends with consonants and see how it can be arranged

Consonant

Consonant

This means we have to select 2 consonants out of 5 i.e. we are supposed to take a combination of 2 from 5 which translates into –
$C(5,2)$ So we write the formula of combination and put values to solve
\[
{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\
{}^5{C_2} = \dfrac{{5!}}{{2!\left( {5 - 2} \right)!}} = \dfrac{{5!}}{{2!3!}} = \dfrac{{5 \times 4 \times 3
\times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}} = 10 \\
\]
Now solving for rest of the places we apply the permutation formula i.e.
${}^n{P_r} = \dfrac{{n!}}{{r!}}$
We have 2 end places for the selection of consonants so we have \[{}^2{P_1} = \dfrac{{2!}}{{1!}} = 2\]
This will give us the total number of arrangements for these places
$10 \times 2 = 20$
Now we deal with the rest of the places which are 6 boxes. So to fill these boxes we have 8 letters from which 2 are fixed at both the ends so we have 6 letters and due to the fixed consonants the repetition is not allowed means boxes will be filled with choices like this

Consonant

6

5

4

3

2

1

Consonant

\[ \Rightarrow \;6!\, = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\]
With this the total number of ways in which the words can be arranged is
$10 \times 2 \times 720 = 14400$
Hence, option ‘a’ is correct.

Note: Permutation is for indexing (for listing of data or things) whereas; combination is used for grouping things in a particular manner. In permutation, order matters while arranging the data while in combination order doesn't matter. We used both the tools in the question due to the given condition.