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.

Thank you Jim and Dan. Doug

P.S. 4 nucleotides, 20 amino acids == faces of smallest and largest Platonic solids.

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<r := 0
f[r_,r_] := r!
f[n_, r_] := f[n,r] = r*f[n-1,r] + r*f[n-1,r-1]

This gives:

f[64,21]                                                               
(* Out[16]= 1510109515792918244116781339315785081841294607960614956302330123544242628820336640000 *)

I was not able to convince RSolve or RecurrenceTable to do this but perhaps those might be used in a similar manner.

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

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.

Is there an online Mathematics journal that is peer reviewed where we could publish this number and how it was derived?

After that, we could drop the reference in Wikipedia in a few places so this question does not come up again or get answered using bad math.

I obviously have some reading to do, because I like to be able to derive the math I am using from scratch.