Counting and Probability Problem URGENT PLEASE HELP!!

a) We have three standard 6-sided dice, colored red, yellow, and green. In how many ways can we roll them to get a sum of 9? The dice are colored so that a red 2, yellow 3, and green 4 is a different roll from red 3, yellow 4, and green 2.

Notice

Roll Total         Frequency

 3                          1

 4                          3

 5                          6

 6                        10

 7                        15

 8                        21

 9                        25

10                       27        and the  rolls of  11-18   will be in the same descending frequency order from 27 back to 1

So....there are 25 ways to roll a "9"

1 2 6     2 1 6     2 6 1    3 5 1     5 1 3

1 3 5     2 2 5     3 1 5    4 1 4     5 2 2

1 4 4     2 3 4     3 2 4    4 2 3     5 3 1

1 5 3     2 4 3     3 3 3    4 3 2     6 1 2

1 6 2     2 5 2     3 4 2    4 4 1     6 2 1

And notice that, no matter that the dice have different colors, that every possibility has been accounted for.

For instance.....for a roll of 2-3-4....we have 6 different possibilities  as denoted in the table :

Red 2  Yellow 3 Grren  4

Red 2  Yellow 4 Green 3

Red 3  Yellow 2 Green 4

Red 3  Yellow 4 Green 3

Red 4  Yellow 2 Green 3

Red 4  Yellow 3 Green 2

So....there sre 25  distinguishable ways to roll a "9' with three different colored dice.

(b) We have 10 standard 6-sided dice, all different colors. In how many ways can we roll them to get a sum of 20?

Like the last question , this just boils down to the number of ways to roll a "20"  with 10 dice.....

The probability =    21307 /   15116544   =   85228 / 6^10  =     85228 /  60466176

https://www.wolframalpha.com/input/?i=probability+of+rolling+a+20+with+10+dice

So.....the number  of ways of rolling a "20"  with 10  dice = 85228

a) It is essentially the same question as many in the same vein: Since they want the NUMBER of ways of obtaining a 9 with the roll of 3 dice, then the answer is the same as other examples of this sort: It is the coefficient of n^9 in the expansion of this series, which is 25!:

[n+n^2+n^3+n^4+n^5+n^6]^3.

b) Similarly here. It is the coefficient of n^20 in the expansion of above series to 10th power:

[n+n^2+n^3+n^4+n^5+n^6]^10. n^20 =85,228 different ways.

Yes!. You have to learn how to calculate the coefficient of any term in the series. In the case of "b", you have to calculate n^20, which is the 41st term from n^60, counting down.

KAIDON is definetly taking AOPS classes like Mellie.

Just thought you should know this is a repost. Check here:

http://web2.0calc.com/questions/a-we-have-three-standard-6-sided-dice-colored-red-yellow-and-green-in-how-many-ways-can-we-roll-them-to-get-a-sum-of-9-the-dice-are-c