In the word "EQUATION" there are 3 consonants [Q,T,N] so there are $3!$ ways to arrange and 5 vowels [E,U,A,I,O] so $5!$ ways to arrange. In whole for both first consonant and then vowels or vice-versa, there are $2!$ ways . so in total $2! \cdot 5! \cdot 3!=1440$ ways.
We want the number of letter arrangements that start and end with a consonant. Let's first see that with letter arrangements, we're working with permutations [we care about the order of things]. The general formula is:
#P_[n,k]=[n!]/[[n-k]!]; n="population", k="picks"#
First let's work with the end letters. We only want consonants [there are 3 of them], which gives:
#P_[3,2]=[3!]/[1!]=6#
For the letters in between the end letters, there are 6 of them [5 vowels and the one consonant we didn't use] and can be placed anywhere. That's#6! = 720#.
How many arrangements are there of the word EQUATION?
Therefore, 1440 words with or without meaning, can be formed using all the letters of the word 'EQUATION', at a time so that the vowels and consonants occur together.
How many consonants are in a word?
There are 24 consonant sounds in most English accents, conveyed by 21 letters of the regular English alphabet [sometimes in combination, e.g., ch and th].
How many 5 letter words can be formed from the word EQUATION?
We know that nPr=n! [n−r]! Therefore , 15120 five letter words can be forms with the letters of the word EQUATIONS without repetition.
How many words of 4 letters beginning with A or E can be formed with the letters of the word equator '?
Hence, The total number of four-letter words that can be formed is 270. Therefore, option D. is the correct answer.