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I am having a bit of a problem with the following question: There are 16 balls, 5 red, 8 blue, and 3 green. The result is the ordered list of the first 4 balls only (although all balls have been drawn). I am able to determine the number of possible results, and the number of results if all the first four marbles are the same color. The issue is how to determine the number of combinations of when both the first and fourth balls are one color (say red) and the number of combinations where the first two balls drawn are different colors.

If possible, I would like to see the "how" you determine this, since I have been running through these problems multiple times and am becoming frustrated.

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Let me get this straight. You want to know how many ways I can select 4 balls, in order, from those 16 balls, such that the first and fourth are the same color, and such that the first two balls are different colors. For example, two such elements would be RBGR and RGGR. So the different numbers mean nothing. Is that correct? –  mixedmath Jul 25 '12 at 1:40
    
Are marbles the same as balls? More important, are the 5 red objects distinguishable from each other? similarly, the 8 blue? the 3 green? –  Gerry Myerson Jul 25 '12 at 1:41

1 Answer 1

First, it doesn't matter whether we consider the first and fourth balls or the first and second balls, so the question is basically asking you to divide the combinations up into the ones that have the same or different colours for any fixed pair of balls. The sum of the two numbers must be the total number of combinations, which you already know, so we only have to determine one of the two, whichever we find easier. I find it easier to think about the three cases of two balls of the same colour than about the three cases of two balls of different colours, so I'll go for that.

If the two fixed balls are red, you have $3$ red, $8$ blue and $3$ green left. Since you knew how to calculate the number of combinations for $5$ red, $8$ blue and $3$ green, you can probably calculate this number, too. Likewise, if the two fixed balls are blue, you have $3$ red, $6$ blue and $3$ green left, and if they're both green you have $5$ red, $8$ blue and $1$ green left. Then you just have to add those three numbers.

There was a question in the comments whether balls of the same colour are considered to be distinguishable: I assumed in this answer that they aren't, but I suspect that you're being asked to calculate a probability anyway; if so, the factor $5!8!3!$ for the number of permutations of balls of the same colour appears in both the numerator and the denominator if you treat them as distinguishable, so it cancels and the probability is the same, as one might expect.

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