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## Main Question or Discussion Point

Hello.

I am trying to decipher the formula, making sure I understand what exactly is going on in each part of the expression. I will be grateful for your guidance, corrections and help.

Below I show the formula and the example for only 3 possible outcomes (in general it would be k)

\(p = \frac{n!}{n_{1}!n_{2}!n_{3}!} p1^{n1}p2^{n2}p3^{n3}\)

where n = 12, n1 = 7, n2 = 2, n3 = 3, p1 = 0.4, p2 = 0.35, p3 = 0.25

(oh, tried to use LaTex here, but it didn't work - how to correct it to display the math formulae?)

p = ( n! / [n1! n2! n3!] ) p1^n1 p2^n2 p3^n3

where n = 12, n1 = 7, n2 = 2, n3 = 3, p1 = 0.4, p2 = 0.35, p3 = 0.25

Looking at each fraction of this formula:

1) n! = 12! produces the number of all possible combinations, in this case all possible combinations given 12 outcomes (for the first outcome we can choose out of 12, for the second out of 11, for the third out of 10, etc till we have 0 possible outcomes)

2) same logic applies to each of n1! = 7! , n2!, = 2!, n3! = 3!

3) but then I am not sure I understand what we get by dividing n! / n_{1}! n_{2}! n_{3}!

As I see it, we are eliminating each of n1! = 7! , n2!, = 2!, n3! = 3! out of the n! = 12! outcomes.

If so, what does it give us and why are we doing it? What is the mathematical meaning of this process?

Thank you very much.

I am trying to decipher the formula, making sure I understand what exactly is going on in each part of the expression. I will be grateful for your guidance, corrections and help.

Below I show the formula and the example for only 3 possible outcomes (in general it would be k)

\(p = \frac{n!}{n_{1}!n_{2}!n_{3}!} p1^{n1}p2^{n2}p3^{n3}\)

where n = 12, n1 = 7, n2 = 2, n3 = 3, p1 = 0.4, p2 = 0.35, p3 = 0.25

(oh, tried to use LaTex here, but it didn't work - how to correct it to display the math formulae?)

p = ( n! / [n1! n2! n3!] ) p1^n1 p2^n2 p3^n3

where n = 12, n1 = 7, n2 = 2, n3 = 3, p1 = 0.4, p2 = 0.35, p3 = 0.25

Looking at each fraction of this formula:

1) n! = 12! produces the number of all possible combinations, in this case all possible combinations given 12 outcomes (for the first outcome we can choose out of 12, for the second out of 11, for the third out of 10, etc till we have 0 possible outcomes)

2) same logic applies to each of n1! = 7! , n2!, = 2!, n3! = 3!

3) but then I am not sure I understand what we get by dividing n! / n_{1}! n_{2}! n_{3}!

As I see it, we are eliminating each of n1! = 7! , n2!, = 2!, n3! = 3! out of the n! = 12! outcomes.

If so, what does it give us and why are we doing it? What is the mathematical meaning of this process?

Thank you very much.