We know that the relationship between set with n Cardinality named A and Binomial Coefficient is all about subsets of the set A.
Binomial Coefficients describes Cardinality of subsets for A set.
But what about Trinomial Coefficient?
If cardinality of A equals 4.

We can wright down the Trinomial Coefficients in one line as we do normally for Binomial Coefficients!
But it is easy to see whats going on, if we will wright those coefficients not in one line, But in one plane.


There is 15 number which defines cardinality of 15 different sets.
My question is: What are those sets?!
My question is: What is the relationship between those unknown sets and set named by A?
All this means that we have a set A with 4 elements and we must find out 15 sets with their own elements.


$$n\choose{a,b,c}$$ is the number of partitions of a set of $n$ elements into three subsets, one of size $a$, one of size $b$, and one of size $c$.

  • 2
    $\begingroup$ Should be "ordered partitions" in case some of $a,b,c$ are equal. Worth noting that the binomial coefficient is also order partitions into two subsets... $\endgroup$ – Thomas Andrews Mar 18 '13 at 12:50
  • $\begingroup$ by the question I don't mean to calculate those Coefficient $\endgroup$ – IremadzeArchil19910311 Mar 18 '13 at 12:55
  • $\begingroup$ @IremadzeArchil19910311 It's not clear what you mean by that comment. Gerry did not calculate, he gave you a "counting" meaning for the trinomial coefficient. $\endgroup$ – Thomas Andrews Mar 18 '13 at 13:28
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    $\begingroup$ @IremadzeArchil19910311: In case it helps make it concrete for you, note that for $n=4$ there are $15$ triples of nonnegative integers $(a, b, c)$ such that $a + b + c = 4$. These are: $$004 \quad 013 \quad 022 \quad 031 \quad 040$$ $$103 \quad 112 \quad 121 \quad 130$$ $$202 \quad 211 \quad 220$$ $$301 \quad 310$$ $$400$$. For instance, you can read the 6 in the top row of your triangle as the cardinality of the (set of) partitions of (say) ${P, Q, R, S}$ into three sets of sizes $0, 2, 2$. $\endgroup$ – ShreevatsaR Mar 18 '13 at 13:37

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