User talk:Aprogressivist/Nwoddice

[edit] Copy/paste

$C_k^n = {n \choose k} = \frac{n!}{k!(n-k)!}$ .

Is it possibly to simplify the problem by setting 8 = 1 suxx, 9 = 1 suxx, 10 = 1.3R suxx?
Can one construct P(x,n) from P(x-1,n), P(x,n-1), etc.?

P(x,n) will generally achieve x successes in some combination of the following forms:

S-Form: s^xf^n-x

...

T-Form: f^x-1t^x-1s

P(0,1): f

P(1,1): s

P(2,1): ts

etc.

P(0,2): ff = f^2

P(1,2): fs + sf = 2fs

P(2,2): fts + tsf + ss = 2fts + s^2

P(3,2): ftts + ttsf + sts + tss = 2ft^2s + 2ts^2

P(4,2): 2ft^3s + 2st^2s = 2ft^3s + 2t^2s^2

etc.

P(0,3) = f^3

P(1,3) = 3ffs = 3f^2s

P(2,3) = 3ffts + 3fss = 3f^2ts + 3fs^2

P(3,3)= 3fftts + 6fsts + sss = 3f^2t^2s + 6fts^2 + s^3

...

$\ P(0,6) = f^6$

$P(1,6) = C_1^6 f^5s$

$P(2,6) = C_1^6 f^5ts + C_2^6 f^4s^2$

$P(3,6) = C_1^6f^5t^2s + C_2^6 C_1^2 f^4ts^2 + C_3^6 f^3s^3$