Question

The letters of the word ARTICLE are arranged in all possible ways at random. The probability of arranging them so that the vowels may occupy even places is
a. $ \dfrac{1}{2} $
b. $ \dfrac{1}{6} $
c. $ \dfrac{1}{35} $
d. $ \dfrac{1}{4} $

Answer 1

Hint: Here in this question we consider the concept of permutation and probability where we are arranging the word ARTICLE in all possible ways where the vowels of the word ARTICLE lies in the even place of the word. And then we have to find the probability

Complete step-by-step answer:
Now consider the ARTICLE, where it contains 7 letters
Roughing we write the table for the word ARTICLE

A

R

T

I

C

L

E

The even places are

e

e

e

In the above figure where the boxes having e represents the even places
Total number of words containing in the word ARTICLE is 7
The number of possible ways to arrange randomly is $ 7! $
Where factorial of n is given by $ n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1 $
The number of even places we are having is 3
The word ARTICLE contains 3 vowels namely A, E and I
The number of possible ways to arrange the vowels is $ 3! $
Then the rest of letters in the word ARTICLE are arranged, the possible ways is $ 4! $
Therefore, the total possibilities of ways where the vowels are placed in the even places is $ 3! \times 4! $
Apply the formula of factorial of n we have
 $ 3! \times 4! = (3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) $
 $ \Rightarrow 3! \times 4! = 144 $
Now we have to find the probability of arranging the vowels occupy the even place is the ratio of the number of possible ways the vowels occupy the even place to the number of ways the word ARTICLE arranged , that is
The probability $ = \dfrac{{3! \times 4!}}{{7!}} $
Apply the factorial of n formula we have
 $ = \dfrac{{(3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1)}}{{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}} $
Cancelling the similar terms on both numerator and denominator we have
 $ = \dfrac{{(3 \times 2 \times 1)}}{{(7 \times 6 \times 5)}} $
On further simplification we have
Probability $ = \dfrac{1}{{35}} $
So, the correct answer is “Option c”.

Note: Candidate must know we have to use the permutation concept or combination concept to solve the given problem because it is the first and main thing to solve the problem. Here we arrange the things in the possible ways so we are using the concept permutation. The formula $ n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1 $ is applied for this concept.