Understanding part of the multinomial formula

Widening this out, the binomial will provide the total number of ways of doing k things with n outcomes.
  • #1

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.
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  • #3
Vital said:
What is the mathematical meaning of this process?
One technicality to settle is whether the multinomial formula has a "mathematical" proof. Combinatorics and probability are topics that have long history. They were investigated before the modern standards of mathematics were established. In introductory courses, it is traditional (and perhaps wise) not to teach topics in a precise way and to rely on non-mathematical arguments that appeal to our intuitions about applying mathematics to particular situations.

For such an intuitive argument, imagine that we have a list of ##n## distinct names and spreadsheet with ##n## columns. We write down all possible permutations of the names by filling out ##n!## lines of the spreadsheet.

Next we arbitrarily stipulate that the the names in first ##n_1## columns will correspond to the selection of ##n_1## items to be in "set 1", the names in the next ##n_2## columns will correspond to the selection of items to be in "set 2" and the last ##n_3## columns will correspond to the selection of items in "set 3".

Although we don't know how many different sets are defined in this manner, we can imagine making a list of the different triplets of sets by tedious clerical work. We create a second spreadsheet with ##n## columns. We look at each line of the first spreadsheet. If the line defines sets that are not listed on the second spreadsheet, we copy it to the second spreadsheet. If the line lists sets that are already on the second spreadsheet, we don't copy it. (The order in which names are listed in a set doesn't matter. We don't copy lines that define sets that are already listed when the lines merely list names in a different order.)

When we are finished, the second spread sheet contains one line for each distinct "way" of picking items to be in the sets. (You will find that in discussing combinatorial problems, the word "way" is often used. It has different meanings depending on what problem is considered.)

For each line on the second spreadsheet, how many lines on the first spreadsheet represent the same assignment of names to sets? Can you see intuitively that each line on the second spreadsheet is one of ##n_1! n_2! n_3!## lines that represent assigning names that create the same 3 sets?

The number of lines ##W## on the second spreadsheet is the answer for the numbers of "ways" to do what we want.

Each of the ##W## lines is one of ##n_1! n_2! n_3!## lines in the first spreadsheet that define the same "way". There are ##n!## lines on the first spreadsheet. So ##(W)( n_1! n_2! n_3!) = n!## and ##W = \frac{n!}{n_1!n_2!n_3!}##.
  • #4
One way to approach multinomial is to do binomial first and expand on it. For example assume we have two possibilities then out of n tries, there are ##\binom{n}{k}=\frac{n!}{k!(n-k)!}## ways of getting k outcomes for the first (or second) possibility. Next assume the first possibility has two cases within the k possibilities, then the first case has ##\binom{k}{j}=\frac{k!}{j!(k-j)!}## ways of getting j outcomes. Combining we have ##\frac{n!}{j!m!i!}## where m=k-j and i=n-k. Note i+j+m=n for the trinomial.