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Studying for a final:

1) Suppose we have four indistinguishable red balls, 6 indistinguishable blue balls, and 2 indistinguishable green balls. How many different color patterns can be obtained by arranging these balls in a straight line?

I did ${12\choose4} {12\choose6} {12\choose2} = 30,187,080$ but that's definitely not correct.

2) A fair, ordinary six-sided die is colored red on one face, blue on two faces, and green on the remaining three faces. Find an explicit expression (but do not simplify it) for the probability that, in the 12 rolls fo this die, red will come up 4 times, blue will come up 6 times, and green will come up 2 times? Hint: What is the probability of observing the sequence RBBRGBBRGBRB?

I'm not really sure what to do and the hint only made me even more confused.

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2 Answers 2

up vote 1 down vote accepted

(1) You started out okay, but then you went astray. There are $\binom{12}4$ ways to choose the positions of the red balls, but after they've been placed, only $12-4=8$ positions remain open. There are $\binom86$ ways to choose $6$ of these $8$ open positions for the blue balls, and once you've done that, you don't have to make any further choices: the $2$ green balls have to go into the two spots that remain available. The final answer is therefore $\binom{12}4\binom86=13,860$ arrangements.

(2) Each possible sequence of $12$ rolls of the die corresponds to one of the $13,860$ possible arrangements of a string of $4$ red, $6$ blue, and $2$ green balls. Thus, you know from (1) that there are $13,860$ ways to roll the die $12$ times and get $4$ red, $6$ blue, and $2$ green faces. Now the probability on any roll of getting the red face is $\frac16$, the probability of getting a blue face is $\frac26=\frac13$, and the probability of getting a green face is $\frac36=\frac12$. The probability of getting any particular string of $4$ red, $6$ blue, and $2$ green faces is therefore $\left(\frac16\right)^4\left(\frac13\right)^6\left(\frac12\right)^2$, and your final answer is $$13860\left(\frac16\right)^4\left(\frac13\right)^6\left(\frac12\right)^2=\frac{13860}{6^43^62^2}\;.$$

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The first answer is wrong, and soon you will see why. We can choose the places for the red balls in $\binom{12}{4}$ ways. For each of these ways, there are $8$ empty spots left, not $12$. We can fill these spots with blues in $\binom{8}{6}$ ways, for a total of $$\binom{12}{4}\binom{8}{6}.$$ (If you prefer, there are then $2$ empty spots left, which we can fill with green in $\binom{2}{2}$ ways. Of course, that does not affect anything, since $\binom{2}{2}=1$.)

For the second question, find the probability of the sequence they gave you. It is just a multiplication. Now how many sequences of that type are there, with the right numbers of red, blue, green? You just counted them in the first part of the problem!

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So in reality it's (12 choose 4) (8 choose 6) (2 choose 2)? Obviously (2 choose 2) is 1. Makes sense. By the same token, that should be equal to (12 choose 6) (6 choose 4) (2 choose 2). Both are 13,860. Great, thanks :) –  switz May 3 '12 at 18:14
    
Yes, as a matter of computing strategy I would use $\binom{12}{2}\binom{10}{4}$. Less work. –  André Nicolas May 3 '12 at 18:18
    
Note that it is equal to the multinomial coefficient ${12 \choose 6,4,2} = \frac{12!}{6!4!2!}$ (see Wikipedia, section Interpretations) which is a generalization of the binomial coefficient. –  TMM May 3 '12 at 18:18

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