# Mathematica function for # possible arrangements of items in bins?

Posted 7 years ago
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 You have a hat with an infinite number of marbles with 10 color categories - equiprobable.You have 50 (numbered) bins to fill, so order matters.How many possible arrangements are there?Here is where I get stuck: What if we require that all of the 10 colors be used at least once?
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Posted 7 years ago
 This isn't a Mathematica question, but...consider recasting your bins as digit places and your colored marbles as numbers from 0 through 9. So bin#1 is the ones place, bin#2 the tens place, etc. And red=0, yellow=1,...,violet=9.Your (first) question is thus equivalent to this: how many base 10 numbers are there of 50 digits?the requirement that all colors get used at least once is equivalent to asking how many 50 digit numbers there are that use every digit between 0 and 9. For that, you might start with the answer to the first question and subtract off however many numbers there are that DO NOT use all 10 digits. So then ask: how many use all but the first digit? All but the second? And so on. Now you have to account for double-subtracting since the ones that use all but the first overlap those that use all but the second (they are the ones that use neither first nor second). From here one develops an inclusion-exclusion method.If I wrote the computation correctly, it is this: s1 = Sum[(-1)^j*Binomial[10, j]*j^50, {j, 10}] (* Out[90]= 94910235292763223632455966616702223835130977516800 *) N[s1/10^50] (* Out[93]= 0.949102352928 *) So around 95% of the 50 digit numbers use all ten digits.
Posted 7 years ago
 For generality use the (-1)^ndigits as @Jim Baldwin mentioned. In this example it amounts to taking the absolute value.
Posted 7 years ago
 Taking the absolute value works just fine in this case but I wanted to assure you that it is not just a matter of convenience. One can write the appropriate general equation using the inclusion/exclusion principle (inclusion/exclusion principle wiki) just as Daniel Lichtblau described. For whatever it's worth below is more of a step-by-step use of that principle for a "less numerous" example than the original question. (* Determine the number of permutations of length n with up to d digits that contain all d digits. *) (* Set values that only require a few steps *) d = 4; n = 8; (* Total number of permutations *) s0 = d^n (* Now subtract off the number of permutations that leave off one number and use just the remaining number of digits. There are d = Binomial[d,1] possible digits to remove and (d-1)^n permutations of the remaining d-1 digits. Those permutations consist of the remaining d-1 digits. *) s1 = s0 - Binomial[d, 1] (d - 1)^n (* We've now subtracted off too many times the number of permutations that had two digits left off. There are Binomial[d,2] possible pairs of digits to leave off and (d-2)^n permutations of the remaining digits. So add those back in. *) s2 = s1 + Binomial[d, 2] (d - 2)^n (* We've now add added in too many times the number of permutations that had three digits left off. There are Binomial[d,3] possible triplets of digits and there are (d-3)^n permutations of the remaining digits. So subtract those out. *) s3 = s2 - Binomial[d, 3] (d - 3)^n (* We are now done for this example after considering removing up to 3 digits because we only had 4 digits to start with as all permutations have to have at least one digit represented. *) (* If we did this in general, we see that because Binomial[d,j] == Binomial[d,d-j], the answer is any of the equivalent forms below... *) d^n + Sum[Binomial[d, j] (d - j)^n (-1)^j, {j,d - 1}] (* Direct from the steps above *) Sum[Binomial[d, j] (d - j)^n (-1)^j, {j, 0, d - 1}] Sum[Binomial[d, j] j^n (-1)^(d - j), {j, d}] (-1)^d*Sum[(-1)^j*Binomial[d, j]*j^n, {j, d}] Abs[Sum[(-1)^j*Binomial[d, j]*j^n, {j, d}]] 
Posted 7 years ago
 Another approach is to code a recursion in two variables. If we have n places and r digits, and each has to be used at least once, then call the number of possibilities f(n,r). The rules are as follows.If there is but one digit, we only have one possibile value for a given n (every digit is the one that's allowable).If we have fewer places than digits there are no possibilities that fulfill the requirements.If number digits = number of places, call it r, then there are r! possibilities; each is a permutation of the available digits. (I didn't check whether this rule is really needed; maybe not.)for other n, r the recurrence is that we have f(n,r-1) ways to fill out the first n-1 positions using all r digits, times r possibilities for the last, PLUS we have f(n-1,r-1) ways of filling out the first n-1 places using only r-1 digits, and the last pace is then forced on us, but multiply by the r choices for which digit gets omitted until the end.In code this is as follows. f[n_,1] = 1 f[n_,r_] /; n
Posted 7 years ago
 Douglas, Consider this to be the genetic code, and for once and all we are going to state how many possible codes there are! 64 bins for the codons. 21 items, for the 20 amino acids plus one stop codon. Does Nature know left from right? Should you divide your count by 2?
Posted 7 years ago
 Douglas (way up there),Sorry, I did not see your question. I probably thought it was me!Yes, Nature does know left from right. DNA or codons are given 5' and 3' ends based on the sugars in the sugar-phosphate backbone. In proteins, these ends correlate with the amino and carboxy termini of the first and final amino acids, respectively.Doug
Posted 7 years ago
 Daniel, Thank you. I'll try it. Doug
Posted 7 years ago
 You can generalize from Daniel's formula to the following: (* Number of digits and number of draws *) ndigits = 3; n = 4; (* Total number of arrangements *) ndigits^n (* Number of arrangements with all digits *) (-1)^ndigits*Sum[(-1)^j*Binomial[ndigits, j]*j^n, {j, ndigits}] (* If you want to see all of the possible arrangements to check on the formula *) digits = Table[i, {i, 0, ndigits - 1}] If[ndigits^n < 500, Tuples[digits, n]] `The multiplier (-1)^ndigits is needed when there is an odd number of digits.
Posted 7 years ago
 Consider this to be the genetic code, and for once and all we are going to state how many possible codes there are! 64 bins for the codons. 21 items, for the 20 amino acids plus one stop codon.Am I suppose to take an absolute value if these equations produce negative numbers?Dougs1 = Sum[(-1)^jBinomial[21, j]j^64, {j, 21}]N[s1/21^64]-151010951579291824411678133931578508184129460796061495630233012354424\ 2628820336640000-0.360557
Posted 7 years ago
 Thank you Jim and Dan. DougP.S. 4 nucleotides, 20 amino acids == faces of smallest and largest Platonic solids.