Question

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.

Answer 1

will give medal

Answer 2

@kirbykirby you must be working real hard on this one.

Answer 3

First,, think of all the possible permutations you can get from rolling 3 dice:
111
112
113 ...
121
122...
...
999

There should be \(\large 6 \times 6 \times 6=6^3\)
Now to think of the permutations that give "9" as a sum, just consider the COMBINATIONS first:
126
135
144
225
234
... etc. And each of those combinations has 3! permutations since each die is unique. So try to find the unique combinations that sum to 9, and then find the permutations for that.

Answer 4

and then what ?

Answer 5

then the probability is \[ \large \frac{\#~of~permutations~summing~to~9}{6^3} \]

Answer 6

i am only in 7 th grade and havent learned this yet. My math teacher gave this as an extra 15 extra points on my report card. I really need help with this so please dumb it as much as possible because i dont really understand this.

Answer 7

do you know the difference between permutations and combinations?

Answer 8

havent even heard of those before now

Answer 9

oye. Have you done probability problems before?

Answer 10

no this is for extra credit somy grade can be at least a B in math

Answer 11

this is very important to me and i just dont understand

Answer 12

oye.. there is a lot to explain then if you haven't done any basic probabilities or combination/permutation problems .. this problem is quite involved if you haven't seen these basics yet :o

Answer 13

Maybe this will give you a primer to start: http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Answer 14

can you just explain the end until you actually answer the problem please? Really are being helpful

Answer 15

@kirbykirby ?

Answer 16

Well does it make sense that there are \(\large 6^3\) total possible permutations ("arrangements") of the die

Answer 17

yes

Answer 18

do you understand how you can enumerate all possible permutations to get the sum 9? Like you can have (1, 2, 6), (1, 6, 2), (2, 1, 6), (2, 6, 1), (6, 1, 2), (6, 2, 1)
You can just say that this is one combination of (1, 2, 6) which has 3! = 6 perumtations

Answer 19

no

Answer 20

thanks but i have to go to school now. Can you please just leave the answer with a good explanation in the comments and i will get back to you. Thanks alot. Really appreciate it. Bye

Answer 21

@kirbykirby ? are you there?

Answer 22

Yes I am typing a big paragraph...

Answer 23

well a combination doesn't consider the order of the numbers to be important.
(1, 2, 6), (1, 6, 2), (2, 1, 6), (2, 6, 1), (6, 1, 2), (6, 2, 1) is just ONE combination since it doesn't matter what order the numbers are in... for this reason I will just use (1, 2, 6) to denote a combination (for all of these)

A permutation though considers the order to be important. That's why in the above, all the 6 different arrangements are considered different, and so you have SIX permutations. As a further note... a permutation like (1, 4, 4) and (1, 4, 4) will be different if in the first set, the two 4's are of different color to the two 4's in the second set. That's why you can think of the two 4's as being unique elements.

Now, the only thing you need to consider are the COMBINATIONS that give you a sum of "9" because then, you just now that for each of these combinations, there will be 6 permutations for each of them...
SO if you found say 10 combinations that gave the sum of 9 (I didn't actually count this.. I'm just using 10 as an example), then you will have 10 * 6 = 60 permutations. And thus the overall probability would be \(\Large \frac{10*6}{6^3}\)

Answer 24

thanks @kirbykirby