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stolyar
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There are a green ball, a blue ball, and 5 wthite ones. One 29 Oct 2003, 05:41
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There are a green ball, a blue ball, and 5 wthite ones. One arranges them in a row. In how many ways could one do so, provided that the blue ball lies to the right of the green one, not necessarily next to each other?

A 5040
B 120
C 42
D 21
E 7



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There are a green ball, a blue ball, and 5 wthite ones. One 29 Oct 2003, 09:22
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I think the answer is 21 ways.My reasoning is as follows.
If the Green ball is in the first position,then the blue ball can be put in six places to accomodate the rule;if the green ball is in second posn,then there are 5 ways(as the blue ball will be on the right of the green ball)...and so on..so 6+5+4+3+2+1 = 21.
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There are a green ball, a blue ball, and 5 wthite ones. One 29 Oct 2003, 23:14
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D is correct. Another way to go--the foregoing problem is symmetrical: the number of ways to lay the blue ball to the RIGHT of the green one equals to the number of ways to lay the blue ball to the LEFT of the green one.

The total number of unique combinations: 7!/5!=6*7=42
42/2=21
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There are a green ball, a blue ball, and 5 wthite ones. One 30 Oct 2003, 09:51
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I thought the answer should be 2520 but that answer choice is not given.

Here is my reasoning. What sudzpwc has indicated is just the total number of ways of arranging ONLY blue and green balls. The question is asking for number of ways to arrange ALL the balls (including white balls).

If we consider the arrangement of all balls, the total number of ways to arrange should be = 21 * 5! = 2520.

Stolyar any clarification?

Thanks
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There are a green ball, a blue ball, and 5 wthite ones. One 30 Oct 2003, 10:04
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stolyar
D is correct. Another way to go--the foregoing problem is symmetrical: the number of ways to lay the blue ball to the RIGHT of the green one equals to the number of ways to lay the blue ball to the LEFT of the green one.

The total number of unique combinations: 7!/5!=6*7=42
42/2=21


Hi Stolyar,

Thanks for the explanation.

I think the formula you have used for unique combination can not be applied here. Because it is NOT given in the question that the white balls are IDENTICAL. We can apply the unique combination formula ONLY if the white balls are identical.

If we take the white balls as NON IDENTICAL, then the solution I mentioned above would be applicable. Am I correct?
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There are a green ball, a blue ball, and 5 wthite ones. One 30 Oct 2003, 23:28
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if we treat white balls as nonidentical, then 7!/2=2520.
I assume them to be identical.



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