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jfa1013
Associate
Joined: 18 Oct 2005
Posts: 3
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 Posted: Tue Oct 18, 2005 4:38 am Post subject: New Quiz |
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Hi All--
I'm new to this site, and I have a question on #4 from the fortnightly quiz.
4 In how many ways can 4 identical red balls and 5 different green balls be arranged in a line such that no two red balls are together?
I understand why the 5! should be multiplied by some number, but I can't figure out why that number is 6. Any help would be greatly appreciated.
Jared |
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mysticzephyr
Associate
Joined: 30 Oct 2005
Posts: 1
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| Posted: Sun Oct 30, 2005 2:29 am Post subject: Free Quiz problem 4 |
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I disagree with the answer to quiz question 4. According to my calcs the answer should be 15 times 5!
5! comes from the ways the 5 different green balls can be ordered. Within each of the 5! permutations, the 4 identical red balls can be placed around the green balls as follows:
GRGRGRGRG
GRGRGRGGR
GRGRGGRGR
GRGGRGRGR
GGRGRGRGR
RGRGRGRGG
RGRGRGGRG
RGRGGRGRG
RGGRGRGRG
RGRGRGGGR
RGRGGGRGR
RGGGRGRGR
RGRGGRGGR
RGGRGRGGR
RGGRGGRGR |
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jfa1013
Associate
Joined: 18 Oct 2005
Posts: 3
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| Posted: Fri Nov 04, 2005 1:35 am Post subject: I figured out the answer to the question I posted... |
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Assume that you place the distinct green balls first. Such that:
G G G G G
From this, there are six different slots in which the red balls can be placed--in this way:
1 G 2 G 3 G 4 G 5 G 6
The green balls can be arranged in 5! ways since each ball is distinct. Thus, the answer is (5!)(6).
I hope that this response makes sense.
Jared |
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jfa1013
Associate
Joined: 18 Oct 2005
Posts: 3
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| Posted: Fri Nov 04, 2005 1:35 am Post subject: I figured out the answer to the question I posted... |
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On second thought, after looking at your answer, I agree with you. There should be 5! possibilites for the green balls leaving 6 vacant spots for the red balls. Since each red ball is the same, the order does not matter so there should be 6C4 or 15 possibilities for the red balls.
I agree. I think it should be (5!)(6C4).
Jared |
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gmat
Site Admin
Joined: 25 Dec 2004
Posts: 4
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| Posted: Wed Nov 09, 2005 10:24 am Post subject: |
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Hi
The correct answer to that question is 5!*6C4 or 15*5!
Regret the error.
KSB
4gmat.com |
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shaji
Senior Consultant
Joined: 25 Apr 2005
Posts: 34
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| Posted: Wed Nov 09, 2005 2:46 pm Post subject: PRECISELY CORRECT!!!. |
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[Hi;
6*5! looked closed to (6C4*)5! so missed being noticed. Nevertheless, noticed now. Better late than never.
Now spot the fallacy!!!.
1) No red balls are together=15*5!
2) Two red balls together=85*5!
3)Three red balls together=30*5!
4)4 red balls together=6*5!
The sum of all the above=136*5! which exceeds 126*5!
Note 126*5! is the total number of ways in which the balls can be arranged.
Regards;
Shaji.
quote="jfa1013"]Hi All--
I'm new to this site, and I have a question on #4 from the fortnightly quiz.
4 In how many ways can 4 identical red balls and 5 different green balls be arranged in a line such that no two red balls are together?
I understand why the 5! should be multiplied by some number, but I can't figure out why that number is 6. Any help would be greatly appreciated.
Jared[/quote] |
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