Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?

A. 1/30
B. 1/15
C. 1/5
D. 2/5
E. 7/20



Kudos for a correct solution.
_________________

1 2 3 4 5
_ _ _ _ _

Total no of empty space is 5 so total no of ways A B C D and E can sit is 5 !

Solution 1:
Now condition is C&E will sit together
If they sit at 1st and 2nd place . # of ways will be 1*3! but C&E can be either at 1 or at 2 so # of ways will be 1*3!*2!
Similarly they can sit together at position 2 and 3 or 3 and 4 or 4 and 5

So total # ways will be 1*3!*2! + 1*3!*2! + 1*3!*2! + 1*3!*2! = 4 *1*3! *2 !

So probability is (4! * 2! ) / 5! = 2/5

Solution 2:
Consider C*E as 1 person . So we have 4 ppl and 4 places . # of ways they can sit is 4!
C&E can sit as C&E or E&C so total # of ways is 4! * 2!

So probability will be 2/5

MAGOOSH OFFICIAL SOLUTION:

First, we will count all the possible arrangements of the five children on the five seats, all the possible orders. This is 5! = 120. That’s the denominator.

Now, the more challenging part: we have to figure out how many arrangements there are involving C & E sitting together. This is a tricky problem to frame, so I’ll demonstrate the steps to follow. First, let’s look at the seats these two could be next to each other. There are four possible pairs of seats in which they could be next to each other
i. X X _ _ _
ii. _ X X _ _
iii. _ _ X X _
iv. _ _ _ X X

In each of those four cases, we could have either CE or EC, either order, so that’s 4 x 2 = 8 ways we could have just C & E sitting next to each other with the remaining three seats empty.

For the final step, we need to consider the other three children, A & B & D. In each of the eight cases, there are three blank seats waiting for those three, and those three could be put in any order in those blank seats. Three elements in any order —- that’s 3! = 6. Thus, the total number of arrangements in which C & E would be next to each other would be 8 x 6 = 48. This is our numerator.

The probability would be this number, 48, over the total number of arrangements of the children, 120.

Probability = 48/120 = 2/5.

Answer = D
_________________

We need to determine:

P(Childeric and Ethelred sit next to each other)

This probability is given by (Childeric and Ethelred sit next to each other)/(total number of ways to arrange the 5 children).

The total number of ways to arrange the 5 children is 5! = 120 ways.

The number of ways with Childeric next to Ethelred can be shown as:

[C-E] - A - B - D

We are treating C and E as a single entity because they must sit next to each other. Thus, we have 4 positions that can be arranged in 4! = 24 ways and we also see that [C-E] can be arranged in 2! = 2 ways, so the the total number of arrangements is 24 x 2 = 48.

So, the probability that Childeric and Ethelred sit next to each other is 48/120 = 2/5.

Answer: D
_________________