Nauseated
Dec 25, 2014

#2**0 **

A Down vote was an available option on the forum’s previous operating system. Posts with negative points are a carryover from there.

Some of my** great **troll posts have negative points. We trolls consider them **points for excellent work.** The best ones Melody made into Scooby snacks

‘tis a pity it’s not an option anymore. **Those were the days** . . . . . ..

http://web2.0calc.com/questions/note-on-the-avatar-feature-if-you-have-a-unicorn-you-lose-24-points-a-day

Nauseated
Mar 14, 2016

#16**+5 **

**Errata:**

Corrected to indicate exponent in term of general expansion

\( - \binom{6}{5}g^{(y)} + 1 \)

Corrected odd term subtraction

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

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Wolfram alpha link for

Sum[Binomial[6, i] (-1)^i (6 - i)^11, {i, 0, 10}]

https://www.wolframalpha.com/input/?i=sum+%28binom%286,i%29*%28-1%29^i*%286-i%29^11%29+from+i%3D0+to+10

.

Nauseated
Mar 2, 2016

#15**+5 **

Your logic is excellent, Melody.

This helps to explain the formula referenced here.

http://web2.0calc.com/questions/this-is-a-continuation-of-an-earlier-counting-question-that-nauseated-and-geno-answered

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

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