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The organizers of a weeklong fair have hired exactly five security
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30 Oct 2014, 08:12
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Re: The organizers of a weeklong fair have hired exactly five security
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30 Oct 2014, 16:22
Bunuel wrote: Tough and Tricky questions: Combinations. The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night? (A) 9 (B) 7 (C) 5 (D) 3 (E) 2 For any given day, only the guards patrolling on the previous day won't be available. So, 2 guards who patrolled on Friday won't be available. We are thus left with 3 guards. To choose 2 out of 3, we will have 3 different pairs. Ans  D
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The organizers of a weeklong fair have hired exactly five security
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30 Mar 2017, 18:42
Bunuel wrote: Tough and Tricky questions: Combinations. The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night? (A) 9 (B) 7 (C) 5 (D) 3 (E) 2 OFFICIAL EXPLANATION This question is not as complicated as it may initially seem. The trick is to recognize a recurring pattern in the assignment of the guards. First, we have five guards (let's call them a, b, c, d, and e) and we have to break them down into pairs. So how many pairs are possible in a group of five distinct entities? We could use the combinations formula: \(\frac{n!}{{k!(nk)!}}\), where n is the number of items you are selecting from (the pool) and k is the number of items you are selecting (the subgroup). Here we would get \(\frac{5!}{{2!(52)!}} = \frac{5*4*3*2*1}{{2*(3*2*1)}}=\frac{20}{2}=10\) . So there are 10 different pairs in a group of 5 individuals. However, in this particular case, it is actually more helpful to write them out (since there are only 5 guards and 10 pairs, it is not so onerous): ab, ac, ad, ae, bc, bd, be, cd, ce, de. Now, on the first night (Monday), any one of the ten pairs may be assigned, since no one has worked yet. Let's say that pair ab is assigned to work the first night. That means no pair containing either a or b may be assigned on Tuesday night. That rules out 7 of the 10 pairs, leaving only cd, ce, and de available for assignment. If, say, cd were assigned on Tuesday, then on Wednesday no pair containing either c or d could be assigned. This leaves only 3 pairs available for Wednesday: ab, ae, and be. At this point the savvy test taker will realize that on any given night after the first, including Saturday, only 3 pairs will be available for assignment. Those test takers who are really on the ball may have realized right away that the assignment of any two guards on any night necessarily rules out 7 of the 10 pairs for the next night, leaving only 3 pairs available on all nights after Monday. The correct answer is Choice D; 3 different pairs will be available to patrol the grounds on Saturday night.
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Re: The organizers of a weeklong fair have hired exactly five security
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30 Mar 2017, 23:08
Bunuel wrote: Tough and Tricky questions: Combinations. The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night? (A) 9 (B) 7 (C) 5 (D) 3 (E) 2 Maybe I was just lucky but my solution is following: It doesn't matter when the fair begins. We know that today is Saturday and 2 guards patroled last night so we have just 3 guards avaliable. So we have to pick 2 out of 3: 3!/(2!*1!)=3. Am I wrong?



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Re: The organizers of a weeklong fair have hired exactly five security
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31 Mar 2017, 01:51
malevg wrote: Bunuel wrote: Tough and Tricky questions: Combinations. The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night? (A) 9 (B) 7 (C) 5 (D) 3 (E) 2 Maybe I was just lucky but my solution is following: It doesn't matter when the fair begins. We know that today is Saturday and 2 guards patroled last night so we have just 3 guards avaliable. So we have to pick 2 out of 3: 3!/(2!*1!)=3. Am I wrong? No, you are not wrong. The only constraint is that "no guards are assigned consecutive nights". So the 2 guards of Friday are not available. The other 3 are. Just choose 2 of them.
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Re: The organizers of a weeklong fair have hired exactly five security
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04 Apr 2017, 15:11
Bunuel wrote: The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night?
(A) 9 (B) 7 (C) 5 (D) 3 (E) 2
Note that in this question, the guards that served Monday through Thursday are irrelevant; we only need to take into account that the guards that served on Friday cannot serve on Saturday. Since two guards served on Friday, we are free to choose any two guards from the remaining three to serve on Saturday. This choice can be made in 3C2 = 3 ways. Answer: D
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Re: The organizers of a weeklong fair have hired exactly five security
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09 Apr 2017, 12:57
ziyuen wrote: Bunuel wrote: Tough and Tricky questions: Combinations. The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night? (A) 9 (B) 7 (C) 5 (D) 3 (E) 2 OFFICIAL EXPLANATION This question is not as complicated as it may initially seem. The trick is to recognize a recurring pattern in the assignment of the guards. First, we have five guards (let's call them a, b, c, d, and e) and we have to break them down into pairs. So how many pairs are possible in a group of five distinct entities? We could use the combinations formula: \(\frac{n!}{{k!(nk)!}}\), where n is the number of items you are selecting from (the pool) and k is the number of items you are selecting (the subgroup). Here we would get \(\frac{5!}{{2!(52)!}} = \frac{5*4*3*2*1}{{2*(3*2*1)}}=\frac{20}{2}=10\) . So there are 10 different pairs in a group of 5 individuals. However, in this particular case, it is actually more helpful to write them out (since there are only 5 guards and 10 pairs, it is not so onerous): ab, ac, ad, ae, bc, bd, be, cd, ce, de. Now, on the first night (Monday), any one of the ten pairs may be assigned, since no one has worked yet. Let's say that pair ab is assigned to work the first night. That means no pair containing either a or b may be assigned on Tuesday night. That rules out 7 of the 10 pairs, leaving only cd, ce, and de available for assignment. If, say, cd were assigned on Tuesday, then on Wednesday no pair containing either c or d could be assigned. This leaves only 3 pairs available for Wednesday: ab, ae, and be. At this point the savvy test taker will realize that on any given night after the first, including Saturday, only 3 pairs will be available for assignment. Those test takers who are really on the ball may have realized right away that the assignment of any two guards on any night necessarily rules out 7 of the 10 pairs for the next night, leaving only 3 pairs available on all nights after Monday. The correct answer is Choice D; 3 different pairs will be available to patrol the grounds on Saturday night. Isn't the most straightforward approach to realize that if one pair works on a particular night, say Wednesday, then the eligible pair for the following night, Thursday, can only be made from the three guards that did not work on the Wednesday. Thus, the pair can be made from three persons. This results in 3C2 = 3/1 = 3 pairs?



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Re: The organizers of a weeklong fair have hired exactly five security
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12 Jun 2017, 23:37
Bunuel wrote: Tough and Tricky questions: Combinations. The organizers of a weeklong fair have hired exactly five security guards to patrol the fairgrounds at night for the duration of the event. Exactly two guards are assigned to patrol the grounds every night, with no guard assigned consecutive nights. If the fair begins on a Monday, how many different pairs of guards will be available to patrol the fairgrounds on the following Saturday night? (A) 9 (B) 7 (C) 5 (D) 3 (E) 2 BunuelThis has got me confused..in my opinion..the answer should be 10. Because any pair can be made to patrol on Saturday night..(A,B),(A,C),(D,E),(C,D)...and so on.. Then I saw the answer...and although it is correct that because "Some" two persons will be used up on Friday night and cannot be used to make a pair on next day..we will have "Some" three persons to choose two from. My question to you is..as you can see..when the answer is 10 or 3, the situations are the same..What would be the language of the question to which my answer(10) would be correct. Please help
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Re: The organizers of a weeklong fair have hired exactly five security
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Re: The organizers of a weeklong fair have hired exactly five security
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