6 horses {A,B,C,D,E,F} Participate in a race. If there are no draws in the rate then in how many ways can the race end such that horse A always finishes ahead of C and B both the horses?
A) 720
B) 360
C) 240
D) 120
E) 48
Source: www.GMATinsight.com
Answer: Option C
6 horses {A,B,C,D,E,F} Participate in a race. If there are n
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Hello GMATinsight and Regor.
I would like to give you a more detailed solution.
We have 6 horses and we can have the following cases:
- Horse A finishes 1st: here, we have to organize 5 horses in 5 places, the number of ways to do that is A*5*4*3*2*1=5!=120.
- Horse A finishes 2nd: First, we have to pick the horse for the first place but we can not select horses B or C. So, we have 3 options for the first place (D, E, F) and the number of ways to select a horse for the first place is 3. Then we have to organize 4 horses in the last 4 places. The total number of ways in this case is 3*A*4*3*2*1=72.
- Horse A finishes 3rd: as in the case before, we have to pick 2 horses for the first 2 places but we can not select horses B or C. So, we have 3 options again and we can organize this 2 horses in 3*2=6 different ways. Now we have to organize the last 3 horses in the last 3 places. The total number of ways in this case is 3*2*A*3*2*1=36.
- Horse A finishes 4th: Now, we have to pick 3 horses for the first 3 places but we can not select horses B or C, the number of ways to organize this 3 horses is 3*2*1=6. Finally, we have to organize 2 horses (B and C) in the last 2 places. The total number of ways in this case is 3*2*1*A*2*1=12.
You should noticed that A can not finish in the 5th place or in the 6th place.
In conclusion, the total number of ways this race can end is 120+72+36+12=240.
So, the correct answer is C.
I hope this help you. I'm available if you'd like any follow up.
I would like to give you a more detailed solution.
We have 6 horses and we can have the following cases:
- Horse A finishes 1st: here, we have to organize 5 horses in 5 places, the number of ways to do that is A*5*4*3*2*1=5!=120.
- Horse A finishes 2nd: First, we have to pick the horse for the first place but we can not select horses B or C. So, we have 3 options for the first place (D, E, F) and the number of ways to select a horse for the first place is 3. Then we have to organize 4 horses in the last 4 places. The total number of ways in this case is 3*A*4*3*2*1=72.
- Horse A finishes 3rd: as in the case before, we have to pick 2 horses for the first 2 places but we can not select horses B or C. So, we have 3 options again and we can organize this 2 horses in 3*2=6 different ways. Now we have to organize the last 3 horses in the last 3 places. The total number of ways in this case is 3*2*A*3*2*1=36.
- Horse A finishes 4th: Now, we have to pick 3 horses for the first 3 places but we can not select horses B or C, the number of ways to organize this 3 horses is 3*2*1=6. Finally, we have to organize 2 horses (B and C) in the last 2 places. The total number of ways in this case is 3*2*1*A*2*1=12.
You should noticed that A can not finish in the 5th place or in the 6th place.
In conclusion, the total number of ways this race can end is 120+72+36+12=240.
So, the correct answer is C.
I hope this help you. I'm available if you'd like any follow up.
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There are 6! = 720 ways to arrange the six horses if there are no restrictions on how they should be placed.GMATinsight wrote: ↑Thu Sep 28, 2017 5:27 am6 horses {A,B,C,D,E,F} Participate in a race. If there are no draws in the rate then in how many ways can the race end such that horse A always finishes ahead of C and B both the horses?
A) 720
B) 360
C) 240
D) 120
E) 48
Source: www.GMATinsight.com
Answer: Option C
Now if we focus only on the three horses A, B and C, we can arrange them in the following ways:
ABC, ACB, BAC, BCA, CAB, CBA
We see that 2/6, or ⅓, of the time, horse A is placed ahead of both B and C; therefore, 1/3 of all 720 ways, or 240 ways, will have A placing ahead of both B and C regardless of the placement of the other 3 horses.
Answer: C
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