anonymous
  • anonymous
If we have 10 regular dice, how many ways can we roll them so the numbers add to 20?
Mathematics
jamiebookeater
  • jamiebookeater
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ganeshie8
  • ganeshie8
Hey!
anonymous
  • anonymous
hi!
ganeshie8
  • ganeshie8
@dan815

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anonymous
  • anonymous
60
anonymous
  • anonymous
How did you find that @aaronacg07 ?
anonymous
  • anonymous
for 2 or 3 dice it would be easy but for 10 it's a bit difficult!
anonymous
  • anonymous
Now I am hoping this question isn't asking for raw permutations of each potential roll, but is treating this like a situation where all 10 are thrown at once... thus any given set of numbers on the dice is only ONE possible way and not every possible permutation of those numbers Start off by thinking about the fewest 6's you need (or the most 1's)... a max/min roll of sorts to get 20... so 8 1's and 2 6's will give you a 20 The symmetry here gives me an idea... if I can find all the possible ways of getting 10 out of 5 dice starting with 4 1's and a 6 then that number will be the same for both sets of 5 die: 1 1 1 1 6 1 1 1 2 5 1 1 1 3 4 1 1 2 2 4 1 1 2 3 3 1 2 2 2 3 2 2 2 2 2 Total: 7 ways Note how much easier it is if I am not considering combinations/permutations here. Now all of these possible ways of getting 10 for the first set exist for each one of the ways in the second set. In other words: 1 1 1 1 6 + 1st 7 ways 1 1 1 2 5 + 1st 7 ways 1 1 1 3 4 + 1st 7 ways 1 1 2 2 4 + 1st 7 ways 1 1 2 3 3 + 1st 7 ways 1 2 2 2 3 + 1st 7 ways 2 2 2 2 2 + 1st 7 ways Thus the grand total is: 7^2 = 49 Not sure about this answer, but I am still thinking about it. Please let me know if there are any glaring errors.
anonymous
  • anonymous
Ok apparently there is overcounting in this scheme: 22222|11233 = 22213|12223 12223|11134 = 11233|11224 So far.... Sorry man I was always weak when it came to raw counting principle problems :(
anonymous
  • anonymous
Does anyone know how to solve this with the hokey stick identity? Because ive been researching and that seems to be the most efficient, but i dont know how to get around it....
anonymous
  • anonymous
@Loser66 @ganeshie8 are you guy here? can you please help?
Loser66
  • Loser66
I think : the only 1 way to get 20 when all of them are 2, right? now, if 1 of them is 1 then another must be 3 to get 20. And we have 10C2 cases of that.
Loser66
  • Loser66
if 2 of them are 1, then another pair will be 3 to get 20, again, we have 10 C4 for this case
Loser66
  • Loser66
if 3 of them are 1, then another three must be 3, we have 10 C6 for this
Loser66
  • Loser66
if 4 of them are 1, then another 4 must be 3, we have 10 C8 for this and the last one is 5 of them are 1 , another 5 must be 3, we have 10 C5 for this Add them up

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