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

 May 10, 2016

Best Answer 

 #2
avatar+24949 
+15

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:

 

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

 

laugh

 May 10, 2016
 #1
avatar
0

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

 May 10, 2016
 #2
avatar+24949 
+15
Best Answer

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:

 

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

 

laugh

heureka May 10, 2016
 #3
avatar
+5

BRILLIANT heureka!!. Congrats.

 May 10, 2016
 #4
avatar+111321 
+12

That is very nice, heureka  !!!

 

 

cool cool cool

 May 10, 2016
 #5
avatar+111321 
+5

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

 

 

cool cool cool

 May 10, 2016
 #6
avatar+24949 
+10

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

 

Hallo CPhill,

i did it by computing program. I wrote a algorithm for this in c++.

Here is the counting for all sums from 1 to 60:

 

   1 = 0
   2 = 0
   3 = 0
   4 = 0
   5 = 0
   6 = 0
   7 = 0
   8 = 0
   9 = 0
  10 = 1
  11 = 10
  12 = 55
  13 = 220
  14 = 715
  15 = 2002
  16 = 4995
  17 = 11340
  18 = 23760
  19 = 46420
  20 = 85228
  21 = 147940
  22 = 243925
  23 = 383470
  24 = 576565
  25 = 831204
  26 = 1151370
  27 = 1535040
  28 = 1972630
  29 = 2446300
  30 = 2930455
  31 = 3393610
  32 = 3801535
  33 = 4121260
  34 = 4325310
  35 = 4395456
  36 = 4325310
  37 = 4121260
  38 = 3801535
  39 = 3393610
  40 = 2930455
  41 = 2446300
  42 = 1972630
  43 = 1535040
  44 = 1151370
  45 = 831204
  46 = 576565
  47 = 383470
  48 = 243925
  49 = 147940
  50 = 85228
  51 = 46420
  52 = 23760
  53 = 11340
  54 = 4995
  55 = 2002
  56 = 715
  57 = 220
  58 = 55
  59 = 10
  60 = 1

 

laugh

heureka  May 11, 2016
 #7
avatar
0

 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

 May 12, 2016

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