Re: Probability Problem - Sets

On Feb 4, 7:00 pm, Ray Vickson <RGVick...@xxxxxxx> wrote:
On Feb 4, 3:21 pm, trittium <matt.melchi...@xxxxxxxxx> wrote:

Thank you for your quick reply. Forgive me for my ignorance, but I
have a few additional questions. In your polynomial, (x + x^2 + x^3 +
x^5)^10, the exponents on the x's within the parens are the numbers in
the set (1,2,3 and 5). The external exponent is what I called "x",
the number of times choosing from the set.

No, quasi was using a different 'x' from yours.

Er, that was me.

In answer the above question, I noticed that the
last few coefficients were

.... 55, 10, 10, 0, 1]

and I knew the last one corresponded to x^50, so
I just counted backwards.

1 combination adding to 50
0 combinations adding to 49
10 combinations adding to 48
10 combinations adding to 47
55 combinations adding to 46, etc.

- Randy
.

Relevant Pages

  • Re: Probability Problem - Sets
    ... The external exponent is what I called "x", ... last few coefficients were ... I just counted backwards. ...
    (sci.math)
  • Re: Probability Problem - Sets
    ... The external exponent is what I called "x", ... If we want to extract all the coefficients in correct order, ... you convolve the coefficients in Matlab and find your list of ... could arrive at if I selected from the set 10 times with replacement. ...
    (sci.math)