There are 5 keys in a key ring. If two more keys are to be added in

Question

There are 5 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent?

A. 1/7
B. 1/6
C. 2/7
D. 1/3
E. 1/2

vipulduhoon77 · Answer

Total arrangements for 7 non identical rings = (7-1)!

total arrangement with 2 specific rings adjacent = 2 * (6-1)!

2 because the rings can take two places among themsleves. and 6-1! is the ways in which 6 items can be arranged in a circle.

hence the answer is 1/3/

edouardp · Answer

Is this correct?

I did 7C2= 21, then counted manually that there are 7 ways for the keys to be adjacent on a circular key chain : 
X _ _ _ _ _ X
XX_ _ _ _ _
_XX_ _ _ _ 
etc...

Hence 7/21 = 1/3

ZIX · Answer

Here's an easier approach:

This is a circular arrangement question. The first key can be at any one of the 7 places. Therefore, probability is  \frac{7}{7} = 1 . Now, for the second key to be adjacent, it can be placed either immediate left OR right side of the first key.

There's a 1/6 chance for it be placed on the immediate left. The chance is the same for the right side. Therefore, the probability is:

\frac{1}{6} + \frac{1}{6} = \frac{1}{3}

Kinshook · Answer

There are 5 keys in a key ring. If two more keys are to be added in the ring at random, what is the probability that two keys will be adjacent?

Total ways to 7 keys in a key ring = 6!
Number of ways in which 2 keys wil be adjacent = 2*5!

The probability that two keys will be adjacent = 2*5!/6! = 1/3

IMO D

Gmat232323 · Answer

After adding the two rings, there are 7 positions so total number of arrangements = 7!

We need to fill 2 positions such that 2 rings are always together.

_ _ _ _ _ _ _

Case 1: We put one of the rings in the first position = 2 ways
The next position can be filled only in 1 way as the other key has to be adjacent to the first one.
The remaining 5 positions can be arranged in 5! ways. 
Total arrangements = 2*1*5!

There will be 7 such cases as the 2 keys can be placed in 7 different places.

xx_ _ _ _ _, _xx_ _ _ _, _ _xx_ _ _, _ _ _xx_ _,......,_ _ _ _ _xx

Total arrangements for all cases = 5!*2*7

Probability = Favourable outcomes / Total outcomes =  (5!*2*7)/7! = 1/3

Answer is D.

Bunuel   Could you please let me know if my method is correct?

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