Talk:Multinomial theorem

message: :(x_1+x_2+\cdots+x_m+x_{m+1})^n = (x_1+x_2+\cdots+(x_m+x_{m+1}))^n 
:   = \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}(x_m+x_{m+1})^K
by the induction hypothesis.  Applying the binomial theorem to the last factor,
: = \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}\sum_{k_m+k_{m+1}=K}{K\choose k_m,k_{m+1}}x_m^{k_m}x_{m+1}^{k_{m+1}}
: = \sum_{k_1+k_2+\cdots+k_{m-1}+k_m+k_{m+1}=n}{n\choose k_1,k_2,\ldots,k_{m-1},k_m,k_{m+1}} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}x_m^{k_m}x_{m+1}^{k_{m+1}}
which completes the induction.  The last step follows because
:{n\choose k_1,k_2,\ldots,k_{m-1},K}{K\choose k_m,k_{m+1}} = {n\choose k_1,k_2,\ldots,k_{m-1},k_m,k_{m+1}},
as can easily be seen by writing the three coefficients using factorials as follows:
: \frac{n!}{k_1! k_2! \cdots k_{m-1}!K!} \frac{K!}{k_m! k_{m+1}!}=\frac{n!}{k_1! k_2! \cdots k_{m+1}!}
RobotMacheen (talk) 15:05, 16 November 2008 (UTC)
thread_title:  proof 1 

message: Now perform the inductive step for m and fixed n. For m = 1, both sides equal x_1^n. Similar to before, we suppose the multinomial theorem holds for m and fixed n.  Then
:(x_1+x_2+\cdots+x_m+x_{m+1})^n = ((x_1+x_2+\cdots+x_m)+(x_{m+1})))^n 
Applying binomial theorem to the two terms inside the parentheses to get
: = \sum_{j=1}^n{n \choose j}(x_1+\cdots+x_m)^{n-j}x_{m+1}^j
Then we apply the multinomial theorem for an exponent of n-j and rearrange
: = \sum_{j=1}^n\sum_{|\alpha|=n-j}{n \choose j}{n-j \choose \alpha}x^{\alpha}x_{m+1}^j.
: = \sum_{j=1}^n\sum_{|\alpha|=n-j}\frac{n!}{j! \alpha_1!\cdots\alpha_m!}x_1^{\alpha_1} \cdots x_m^{\alpha_m} x_{m+1}^j
So now we can define \beta=(\alpha, j), i.e. \beta_{m+1}=j and |\beta| = j +|\alpha| =j + n - j = n. With this, the above sum simplifies to
:(x_1+\cdots+x_{m+1})^n = \sum_{|\beta|=n}{n \choose \beta}x^{\beta}
for \beta = (\beta_1,\cdots,\beta_{m+1}), thus concluding our proof of the theorem.
RobotMacheen (talk) 15:05, 16 November 2008 (UTC)
thread_title:  proof 2 

message: I am used to seeing and writing the multinomial coefficient without commas,
:{n \choose k_1 \; k_2 \; \ldots \; k_m}
rather than with commas, 
:{n \choose k_1, k_2, \ldots, k_m} \,.
Should we delete the commas from the article?  Quantling (talk) 17:27, 25 June 2010 (UTC)
replies: text: I see that the United States National Institute of Standards and Technology uses commas, e.g., here.  I guess we should keep the commas.  Quantling (talk) 19:49, 30 August 2010 (UTC)
thread_title:  Commas appropriate? 

message: Speaking as a mathematician who has previously seen neither the multinomial theorem nor the multinomial coefficient, I found the beginning of the Theorem section very confusing.  I subsequently hunted all over Wikipedia and some other web sites to find an explanation of the notation for the multinomial coefficient, only to finally find it defined further down in this article!  The Theorem section needed some immediate explanation of what is in the formula.  Consequently I inserted a brief definition of the multinomial coefficient immediately after the first appearance of the formula, refering to the more thorough description below it.  I also re-worded some of the text to make the section easier to understand.  Other than that, the section was easy to follow and understand, but a first-time reader should not have to already understand a concept before reading about it. — Anita5192 (talk) 07:00, 9 November 2011 (UTC)
replies: text: Joel, I like the way you subsequently cleaned up parts of this article.  I think this is much more readable now.  Nice work! — Anita5192 (talk) 02:50, 23 November 2011 (UTC)
replies: text:  Thanks -- I thought your comments were a helpful guide (and the change you made was certainly necessary).  --Joel B. Lewis (talk) 03:13, 23 November 2011 (UTC)
thread_title:  Multinomial coefficient definition 

message: Theorem : The number of non-negative integral solution of the equation x_1+x_2+x_3+\cdot\cdot\cdot+x_k=n is {n+k-1 \choose k-1}
Proof : 
Let Equation-1 :=x_1+x_2+x_3+\cdot\cdot\cdot+x_k=n
Let f(y_1,y_2,y_3,\cdot\cdot\cdot,y_k)=(y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n
∴ Each term of f(y_1,y_2,y_3,\cdot\cdot\cdot,y_k) will be of the form \displaystyle\sum_{x_1+x_2+x_3+\cdot\cdot\cdot+x_k=n} {\frac{n!}{x_1! x_2! x_3! \cdot\cdot\cdot x_k!}\ y_1^{x_1}} y_2^{x_2} y_3^{x_3} \cdot\cdot\cdot y_k^{x_k}=\displaystyle\sum_{Equation-1} {\frac{n!}{x_1! x_2! x_3! \cdot\cdot\cdot x_k!}\ y_1^{x_1}} y_2^{x_2} y_3^{x_3} \cdot\cdot\cdot y_k^{x_k} Where\ x_i\geq0\ for\ all\ i\in[0,k]
# For each unique non-negative solution of Equation-1, there exist a unique monomial term of degree n of (y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n.
# For each unique monomial term of (y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n, there exists an unique non-negative solution of Equation-1
# No unique non-negative solution of Equation-1 yields two different monomial terms of (y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n.
# No unique monomial term of (y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n yields two different non-negative solution of Equation-1.
Thus, by  bijection , \#terms\ of\ g(x)=\#non-negative\ integral\ solutions\ of\ Equation-1
Thus, to prove the theorem, it is sufficient to prove that number of terms of f(x) is {n+k-1 \choose k-1}
Now, we shall proceed by Mathematical Induction.
* Base Case: (y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n for k=1 is (y_1)^n=y_1^n, which has 1={n+1-1 \choose 1-1}\ terms terms.
Also, for k=2, (y_1+y_2+y_3+\cdot\cdot\cdot+y_k)^n is (x_1+x_2)^n=\displaystyle\sum_{i=0}^n {{n \choose i}x^i y^{n-i}}, which has \displaystyle\sum_{i=0}^n 1\ terms\ of\ the\ form\ {n \choose i}x^iy^{n-i}. \displaystyle\sum_{i=0} ^n 1=(n+1)={n+2-1  \choose 2-1}
* Induction Hypothesis: Let for some k=r, (x_1+x_2+x_3+\cdot\cdot\cdot+x_r)^n\ \ has\ {n+r-1 \choose r-1}\ terms
* Inductive Step: We have to prove that for k=r+1 too, (x_1+x_2+x_3+\cdot\cdot\cdot+x_{r+1})^{n}\ \ has\ {n+(r+1)-1 \choose (r+1)-1}\ terms,\ i.e.\ {n+r \choose r}\ terms
(x_1+x_2+x_3+\cdot\cdot\cdot+x_{r+1})^{n}=\left[(x_1+x_2+x_3+\cdot\cdot\cdot+x_r)+(x_{r+1})\right]^n=\displaystyle\sum_{i=0}^n {{n \choose i}(x_1+x_2+x_3+\cdot\cdot\cdot+x_r)^iy^{n-i}}
For\ each\ unique\ i,\ {n \choose i}(x_1+x_2+x_3+\cdot\cdot\cdot+x_r)^ix^{n-1}\ has\ {i+r-1 \choose r-1}\ unique\ terms.\left[ From\ Induction\ Hypothesis\right]
Thus, Number\ of\ terms\ of\ (x_1+x_2+x_3+\cdot\cdot\cdot+x_{r+1})^{n}=\displaystyle\sum_{i=0}^n {{n \choose i}(x_1+x_2+x_3+\cdot\cdot\cdot+x_r)^iy^{n-i}} \  is\ \displaystyle\sum_{i=0}^n{i+r-1 \choose r-1}\ terms.
\begin{align}
\displaystyle\sum_{i=0}^n{i+r-1 \choose r-1}={r-1 \choose r-1}+{r \choose r-1}+{r+1 \choose r-1}+{r+2 \choose r-1}+\cdot\cdot\cdot \\
+{n+r-2 \choose r-1}+{n+r-1 \choose r-1}=\left[{r \choose r}+{r \choose r-1}\right]+{r+1 \choose r-1}+{r+2 \choose r-1}+\cdot\cdot\cdot \\
+{n+r-2 \choose r-1}+{n+r-1 \choose r-1}=\left[{r+1 \choose r}+{r+1 \choose r-1}\right]+{r+2 \choose r-1}+\cdot\cdot\cdot \\
+{n+r-2 \choose r-1}+{n+r-1 \choose r-1}=\cdot\cdot\cdot \\
=\left[{n+r-2 \choose r}+{n+r-2 \choose r-1}\right]+{n+r-1 \choose r-1}={n+r-1 \choose r}+{n+r-1 \choose r-1}={n+r \choose r}\ \left[ Thus,\ proved \right]
\end{align}
: This is a theorem about the multichoose function, not about multinomial coefficients.  It is covered in our articles on multisets and stars and bars (combinatorics). --JBL (talk) 18:42, 2 December 2020 (UTC)
thread_title:  Multinomial Theorem of 2ⁿᵈ kind