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Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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04 Mar 2015, 04:28
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58% (01:13) correct 42% (01:46) wrong based on 158 sessions
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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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04 Mar 2015, 04:52
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



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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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04 Mar 2015, 04:56
Bunuel wrote: 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. Ah pretty tricky question. So the total arrangement will be 5! =120 We can make Childeric and Ethelred sit next to each other in [1st and 2nd] or [2nd and 3rd] or [3rd and 4th] or [4th and 5th] 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.
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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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04 Mar 2015, 19:01
Bunuel wrote: 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. 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.
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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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08 Mar 2015, 14:22
Bunuel wrote: 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. 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
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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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30 Aug 2016, 00:03
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



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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre
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08 Nov 2017, 17:42
Bunuel wrote: 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 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: [CE]  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 [CE] 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
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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelre &nbs
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