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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

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

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

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

Hi

The correct answer to that question is 5!*6C4 or 15*5!

Regret the error.

KSB
4gmat.com

[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]