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saroj

· started a discussion

· 1 Months ago

pl explain more (Question: 181518.0 )

pl explain more

Question:
The letters of the word ‘ANSWER’ are arranged at random. Find the probability of having exactly two letters in between A and E.
Options:
A) \(\cfrac{1}{5}\)
B) \(\cfrac{1}{4}\)
C) \(\cfrac{1}{10}\)
D) \(\cfrac{1}{2}\)
E) None of these
Solution:
Ans: (a)

Given word is ‘ANSWER’

Total number of arrangements = n (S) = 6! = 720

The possible position of A, E are 1, 4; 2, 5 or 3, 6. For each case the number of possible words 2(4!).

∴ The total number of words is 6(4!) = 144.

∴ Required probability = \(\cfrac{144}{720}=\cfrac{1}{5}\)

Knowledge Expert

· commented

· 1 Months ago

Totally there are 6 letters.

A,E can occupy 3 positions like (1,4) (2,5) (3,6). In this A and E can interchange among themselves.

In remaining positions the 4 other letters can arrange themselves in 4! ways. So total ways where A and E have 2 letters between them = 6 * 4!

Required probability = (6 * 4!)/(6!) = 1/5
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