Question

How many seven letter words can be formed using the word “ARTICLE” ,so that the order of vowels remains the same.

Answer 1

As the question asked that vowels remain same order as in the word “ARTICLE” so we can arrange all the consonants latter but order of vowel can not change so order of vowels is “A,I,E “ always.First we find total arrangements and divide it by arrangements of vowels and we get our answer and we know that total arrangements of $n$ letter word is $n!$ always.

Complete step-by-step answer:
Given word “ARTICLE” is a seven latter word that contains three vowels and four consonant
Now Three vowels = A, I, E
Four consonant = R, C, L, T
So the word “ARTICLE” contains seven latter so arrangements of all seven latter is called total arrangements.
Total arrangements = $7!$
And arrangements of vowels = $3!$ ( we have only three vowels and order of vowels can not change)
Now to find our required answer we assume it as $x$
So Total arrangements = arrangements of vowels $\times$ $x$
So $7! = 4! \times x$
And from here $x = \dfrac{{7!}}{{4!}}$
Now we expand factorials
$x = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}}$
From this we get
$x = 7 \times 6 \times 5 \times 4$
Or $x = 840$
So total $840$ words would be formed by word “ARTICLE” having order of vowels is same

So, the correct answer is “Option B”.

Note: If question say that order of vowels as well as consonant is same then find how many seven letter word can be formed by using word “ARTICLE”
So for this we assume our required answer as $x$
Total arrangements = arrangements of vowels $\times$ arrangements of consonants $\times$ $x$
So $7! = 3! \times 4! \times x$
Or it becomes
$x = \dfrac{{7!}}{{3! \times 4!}}$
Now expand factorials
$x = \dfrac{{7 \times 6 \times 5 \times 4!}}{{3 \times 2 \times 1 \times 4!}}$
This gives
$x = 35$