How many different words ending and beginning with a consonant can be formed with the letters of the word ‘EQUATION’?
$\left( A \right)$ 720
$\left( B \right)$ 4320
$\left( C \right)$ 1440
$\left( D \right)$ None of these

Hint – In this particular question first find out the different consonants and vowels in the given word, then choose one of the consonant letter from the given consonants using combination rule (i.e. to select r objects out of n objects we use ${}^n{C_r}$) for the extreme left and from the remaining consonants choose again one consonants for the extreme right again by combination rule then arrange the middle words, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given word is
EQUATION
As we see there are 8 letters in the given word.
Now in 8 letters the vowels are (E, U, A, I and O) and consonants are (Q, T and N).
So there are 5 vowels and 3 consonants are present in the given word.
Now we have to form a word ending and beginning with a consonant.
So first choose end letter from the given 3 consonant letters, so the number of ways to choose end letter = ${}^3{C_1}$
Now remaining consonant letters are (3 – 1) = 2, so the number of ways to choose beginning letter = ${}^2{C_1}$
Now end and begin letters are choses so the remaining letters are (8 – 2) = 6.
So these letters are present in the middle of the words as there are no repeated letters so the number of ways to arrange these letters are (6!)
So the total number of different words ending and beginning with a consonant is the multiplication of the above calculated values = ${}^3{C_1} \times {}^2{C_1} \times 6!$
Now simplify this according to the property ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so we have,
$ \Rightarrow \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \times \left( {6.5.4.3.2.1} \right)$
$ \Rightarrow \dfrac{{3.2!}}{{2!}} \times \dfrac{{2.1}}{1} \times \left( {6.5.4.3.2.1} \right)$
$ \Rightarrow 3 \times 2 \times \left( {6.5.4.3.2.1} \right)$
$ \Rightarrow 6 \times \left( {720} \right) = 4320$
So this is the required answer.
Hence option (B) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that the combination rule of choosing r objects out of n objects and the formula of combination which is stated above so first chose the number of ways of end letter and begin letter to be filled by the given consonants and then arrange the middle letters of the words from the remaining letters as above and then multiply all these values we will get the required answer.