# Nauseated

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### Congratulations Alan on Your Moderator’s Star!

Nauseated  May 3, 2015
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### Merry Christmas and Happy Holidays Melody, CPhill , and . . . ah . . . ah . . . Whoever the third answerer is. Alan too.

Nauseated  Dec 25, 2014
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### Rosala!! Please! Please come out from the Taj Mahal and post something to save us all . . . from the fiddling, whittling, twiddling drivel o

Nauseated  Nov 22, 2014
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This helps to explain the formula referenced here.

$$\displaystyle \sum \limits_{i=0}^{N-1} *(-1)^i*\binom{N}{i}* (N-i)^k\\ \text {Current problem}\\ \displaystyle \sum \limits_{i=0}^{10} *(-1)^i*\binom{N}{i}* (N-i)^{11} = 129230640$$

This formula is a modification of the formula to generate Stirling numbers of the second kind.

$$\displaystyle \text{Sterling number -2}^{nd} {~type:~} \dfrac{1}{N!}\sum \limits_{i=0}^{N-1} *(-1)^i*\binom{N}{i}* (N-i)^{k}$$

The modification is to multiply by N!.

Wolfram input

StirlingS2[11, 6] = 179487

Multiplying by 6! Returns value above.

StirlingS2[11, 6]*6! = 129320640

Stirling numbers of the second kind generate solutions to bąll and box distributions where the bąlls are unique and the boxes are identical.

This die rolling problem is equivalent to a bąll (die roll) and box (die face) problem where both the bąlls and boxes are unique, requiring an increase in the counts of solutions by a factor of 6!

General form:

$$\dfrac{11!}{(B_1!)*(B_2!)*(B_3!)*(B_4!)*( B_5!)* (B_6!)} = \text{unique distributions}\\ \text{where } {B_1+B_2+B_3+ B_4+ B_5+B_6 =11} \text{ with } {B_i} \ne 0$$

Expanded this by using a generating function for polynomials

[g^y - 1]^6 (The minus 1 correlates to having no empty box).

The general polynomial expansion is

$$g^{(6y)} - \binom{6}{1}g^{(5y)} + \binom{6}{2}g^{(4y)} - \binom{6}{3}g^{(3y)} + \binom{6}{4}g^{(2y)} - \binom{6}{5}{g(y)} + 1$$

$$\text{Now in terms of }~ y^{11} \text{ for the 11 rolls of the die.}$$

$$\left(\dfrac{6^{11}}{11!}\right) + 6 \left(\dfrac{5^{11}}{11!}\right) + 15 \left(\dfrac{4^{11}}{11!}\right) + 20 \left(\dfrac{3^{11}}{11!}\right) + 15 \left(\dfrac{2^{11}}{11!}\right) +6\left(\dfrac{1^{11}}{11!}\right)$$

$$\text{Factor out the } \left(\dfrac{1}{11!}\right)\\ \left(\dfrac{1}{11!}\right) [(6^{11} - 6(5^{11}) + 15(4^{11}) - 20(3^{11}) + 15(2^{11}) - 6(1^{11})]\\ = \left(\dfrac{1}{11!}\right) \left(129230640\right) \\$$

Note the similarity to the modification of the sterling number formula, where (6!) is factored out.

$$\dfrac{129230640}{11^6} = 0.0.3562064186099 \\ \text { (Probability of rolling at least one of each number (1-6) for 11 rolls of a fair die)}$$

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A Markov matrix may also be used to solve this type of problem.

Nauseated Mar 2, 2016