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

We have 10 standard 6-sided dice, all different colors.

In how many ways can we roll them to get a sum of 20?

1.

The number of ways of partitioning 20 into exactly 10 distinct parts with  the 6 numbers {1,2,3,4,5,6} we have:

30 partitions of 20 into ten distinct parts:

The 30 distinct partitions with the sum 20 are:

   1.) {2,2,2,2,2,2,2,2,2,2}
   2.) {1,2,2,2,2,2,2,2,2,3}
   3.) {1,1,2,2,2,2,2,2,3,3}
   4.) {1,1,2,2,2,2,2,2,2,4}
   5.) {1,1,1,2,2,2,2,3,3,3} 
   6.) {1,1,1,2,2,2,2,2,3,4}
   7.) {1,1,1,2,2,2,2,2,2,5}
   8.) {1,1,1,1,2,2,3,3,3,3}
   9.) {1,1,1,1,2,2,2,3,3,4}
  10.) {1,1,1,1,2,2,2,2,4,4}
  11.) {1,1,1,1,2,2,2,2,3,5}
  12.) {1,1,1,1,2,2,2,2,2,6}
  13.) {1,1,1,1,1,3,3,3,3,3}
  14.) {1,1,1,1,1,2,3,3,3,4}
  15.) {1,1,1,1,1,2,2,3,4,4}
  16.) {1,1,1,1,1,2,2,3,3,5}
  17.) {1,1,1,1,1,2,2,2,4,5}
  18.) {1,1,1,1,1,2,2,2,3,6}
  19.) {1,1,1,1,1,1,3,3,4,4}
  20.) {1,1,1,1,1,1,3,3,3,5}
  21.) {1,1,1,1,1,1,2,4,4,4}
  22.) {1,1,1,1,1,1,2,3,4,5}
  23.) {1,1,1,1,1,1,2,3,3,6}
  24.) {1,1,1,1,1,1,2,2,5,5}
  25.) {1,1,1,1,1,1,2,2,4,6}
  26.) {1,1,1,1,1,1,1,4,4,5}
  27.) {1,1,1,1,1,1,1,3,5,5}
  28.) {1,1,1,1,1,1,1,3,4,6}
  29.) {1,1,1,1,1,1,1,2,5,6}
  30.) {1,1,1,1,1,1,1,1,6,6}

2.

In how many ways can we roll them to get a sum of 20?
We need the permutations of each partitions and the sum is the answer.

\(\begin{array}{rcll} 1.)& \{2,2,2,2,2,2,2,2,2,2\} & \frac{10!}{10!} &=& 1 \\ 2.)& \{1,2,2,2,2,2,2,2,2,3\} & \frac{10!}{1!8!1!} &=& 90 \\ 3.)& \{1,1,2,2,2,2,2,2,3,3\} & \frac{10!}{2!6!2!} &=& 1260 \\ 4.)& \{1,1,2,2,2,2,2,2,2,4\} & \frac{10!}{2!6!1!} &=& 360 \\ 5.)& \{1,1,1,2,2,2,2,3,3,3\} & \frac{10!}{3!4!3!} &=& 4200 \\ 6.)& \{1,1,1,2,2,2,2,2,3,4\} & \frac{10!}{3!5!1!1!} &=& 5040 \\ 7.)& \{1,1,1,2,2,2,2,2,2,5\} & \frac{10!}{3!6!1!} &=& 840 \\ 8.)& \{1,1,1,1,2,2,3,3,3,3\} & \frac{10!}{4!2!4!} &=& 3150 \\ 9.)& \{1,1,1,1,2,2,2,3,3,4\} & \frac{10!}{4!3!2!1!} &=& 12600 \\ 10.)& \{1,1,1,1,2,2,2,2,4,4\} & \frac{10!}{4!4!2!} &=& 3150 \\ 11.)& \{1,1,1,1,2,2,2,2,3,5\} & \frac{10!}{4!4!1!1!} &=& 6300 \\ 12.)& \{1,1,1,1,2,2,2,2,2,6\} & \frac{10!}{4!5!1!} &=& 1260 \\ 13.)& \{1,1,1,1,1,3,3,3,3,3\} & \frac{10!}{5!5!} &=& 252 \\ 14.)& \{1,1,1,1,1,2,3,3,3,4\} & \frac{10!}{5!1!3!1!} &=& 5040 \\ 15.)& \{1,1,1,1,1,2,2,3,4,4\} & \frac{10!}{5!2!1!2!} &=& 7560 \\ 16.)& \{1,1,1,1,1,2,2,3,3,5\} & \frac{10!}{5!2!2!1!} &=& 7560 \\ 17.)& \{1,1,1,1,1,2,2,2,4,5\} & \frac{10!}{5!3!1!1!} &=& 5040 \\ 18.)& \{1,1,1,1,1,2,2,2,3,6\} & \frac{10!}{5!3!1!1!} &=& 5040 \\ 19.)& \{1,1,1,1,1,1,3,3,4,4\} & \frac{10!}{6!2!2!} &=& 1260 \\ 20.)& \{1,1,1,1,1,1,3,3,3,5\} & \frac{10!}{6!3!1!} &=& 840 \\ 21.)& \{1,1,1,1,1,1,2,4,4,4\} & \frac{10!}{6!1!3!} &=& 840 \\ 22.)& \{1,1,1,1,1,1,2,3,4,5\} & \frac{10!}{6!1!1!1!1!} &=& 5040 \\ 23.)& \{1,1,1,1,1,1,2,3,3,6\} & \frac{10!}{6!1!2!1!} &=& 2520 \\ 24.)& \{1,1,1,1,1,1,2,2,5,5\} & \frac{10!}{6!2!2!} &=& 1260 \\ 25.)& \{1,1,1,1,1,1,2,2,4,6\} & \frac{10!}{6!2!1!1!} &=& 2520 \\ 26.)& \{1,1,1,1,1,1,1,4,4,5\} & \frac{10!}{7!2!1!} &=& 360 \\ 27.)& \{1,1,1,1,1,1,1,3,5,5\} & \frac{10!}{7!1!2!} &=& 360 \\ 28.)& \{1,1,1,1,1,1,1,3,4,6\} & \frac{10!}{7!1!1!1!} &=& 720 \\ 29.)& \{1,1,1,1,1,1,1,2,5,6\} & \frac{10!}{7!1!1!1!} &=& 720 \\ 30.)& \{1,1,1,1,1,1,1,1,6,6\} & \frac{10!}{8!2!} &=& 45 \\ \hline && \text{sum} &=& 85228 \end{array} \)

3.

The probability is:

\(\begin{array}{rcll} \frac{ 85228 } { 6^{10} } &=& \frac{ 85228 } { 60466176 } \\ &=& 0.00140951530 \end{array}\)

The EXACT probability is:85,228 / 6^10=0.00140951529662......Your task is to figure out how to arrive at these numbers!!!.

BRILLIANT heureka!!. Congrats.

That is very nice, heureka  !!!

One question, heureka.......did you you find the number of partitions by "trial and error"  or did you use some algorithm for this???

 Here is an alternative solution to this problem:
 Expand the generating function
(1/6x+1/6x^2+1/6x^3 + 1/6x^4+1/6x^5+1/6x^6)^10
 (by using Wolfram Alpha, for example)
and find the  x^20  term, which is 21,307/15,116,544=85,228/60,466,176  in this case. Each coefficient gives the probability of getting a sum of that power of x.

EXPANDED FORM:
x^60/60466176+(5 x^59)/30233088+(55 x^58)/60466176+(55 x^57)/15116544+(715 x^56)/60466176+(1001 x^55)/30233088+(185 x^54)/2239488+(35 x^53)/186624+(55 x^52)/139968+(11605 x^51)/15116544+(21307 x^50)/15116544+(36985 x^49)/15116544+(243925 x^48)/60466176+(191735 x^47)/30233088+(576565 x^46)/60466176+(23089 x^45)/1679616+(63965 x^44)/3359232+(2665 x^43)/104976+(986315 x^42)/30233088+(611575 x^41)/15116544+(2930455 x^40)/60466176+(1696805 x^39)/30233088+(3801535 x^38)/60466176+(1030315 x^37)/15116544+(240295 x^36)/3359232+(7631 x^35)/104976+(240295 x^34)/3359232+(1030315 x^33)/15116544+(3801535 x^32)/60466176+(1696805 x^31)/30233088+(2930455 x^30)/60466176+(611575 x^29)/15116544+(986315 x^28)/30233088+(2665 x^27)/104976+(63965 x^26)/3359232+(23089 x^25)/1679616+(576565 x^24)/60466176+(191735 x^23)/30233088+(243925 x^22)/60466176+(36985 x^21)/15116544+(21307 x^20)/15116544+(11605 x^19)/15116544+(55 x^18)/139968+(35 x^17)/186624+(185 x^16)/2239488+(1001 x^15)/30233088+(715 x^14)/60466176+(55 x^13)/15116544+(55 x^12)/60466176+(5 x^11)/30233088+x^10/60466176