Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre

Question

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 .

aviram · Accepted Answer

For such questions, use counting techniques to solve the problem. The total arrangements (total outcomes since we are looking for a probability) is 5!, which is our denominator. For the numerator (desired scenarios), club C and E together. So we have A, B, D, CE. These can be arranged in 4! ways and CE can also be arranged as EC, multiply 4! by 2. Therefore, we get 4!*2/5*4! = 2/5, which is answer D.

Gnpth · Answer

Ah pretty tricky question.

So the total arrangement will be 5! =120

We can make Childeric and Ethelred sit next to each other in   or   or   or

that's 4 and we can make Ethelred sit first and then Childeric. so total will be
4*2= 8.

Now we can make Anaxagoras, Beatrice and Desdemona sit in other 3 open places in 3! ways= 6

So total will be 8*6= 48

The probability will be  \frac{48}{120} =  \frac{12}{30} =  \frac{4}{10}  =  \frac{2}{5}

Answer is D.

abhinav008 · Answer

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

Bunuel · Answer

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

Argo · Answer

Total number of arrangements for 5 children to sit in 5 chairs = 5!

Now, 
consider kids C&E as single entity.
Since we consider 2 children as a single entity, we are left with 4 chairs, as the adjacent chairs in any case will be occupied by the two kids, whom we considered as a single entity. Hence we are left with 4 chairs and 4 children = 4!

But since it is unordered we can place C&E or E&C within these two chairs. Hence 2*4!

2*4!/5! = 2/5

JeffTargetTestPrep · Answer

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:

- 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   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

Jubile87 · Answer

All of the answers are very helpful! My process to a little over 2 mins, so be advised.

Let
A = Anaxagoras
B = Beatrice
C = Childeric
D = Desdemona
E = Ethelred.

Our formula is "desired outcomes"/"total outcomes"

Desired outcomes:
Here are the ways that CE can be next to one another.
ABD CE  -- CE at end
 CE ABD -- CE at start
A CE BD -- CE left mid
AB CE D -- CE right mid

There are 4 ways, so we will multiply 4 by 2 because they could also be EC instead of CE.

Also, the other people could be in different orders, as well. There are 3 people, so 3! will suffice.

Total possibilities: 5!

(3! * 8)/5! = 2/5 
ANS: D

Basshead · Answer

Total outcomes = 5! = 120

Consider CE as stuck together (A, B, CE, D).

Number of ways we can arrange (A, B, CE, D): 4! = 24.

Number of ways we can arrange CE = 2 (CE and EC)

24 * 2 = 48

48/120 = 2/5

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards