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The formula for the number of monomials in variables $w,x,y,\ldots,z$ of degree $m$ (where e.g. $x^iy^jz^k$ degree $m=i+j+k$) is $$\binom{m+n-1}{n-1},$$ where $m$ is degree of monomial and $n$ is the number of distinct variables in that monomial(e.g. in $x^2y^3z^4~, m=9,n=3$).

My book describes that the formula is same as that of number of ways of placing identical objects into distinct boxes as these two situations are similar.

But I can't find how are they similar... Kindly help me with this...

user26857
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coool
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4 Answers4

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A monomial of degree $m$ in $n$ variables is

$$x_1^{a_1}x_2^{a_2}\ldots x_n^{a_n}$$

The number of monomials in $n$ variables of degree $m$ equals the number of solutions in natural numbers $a_1$, $\ldots$, $a_n$ of the equation

$$a_1 + \ldots + a_n = m$$

Arrange $m + n-1$ white balls in a row. Color $n-1$ of them in black. There are $m$ white balls left and the black balls will separate them into $n$ groups. Hence you get a writing of $m$ as a sum of natural numbers.

There are $\binom{m+n-1}{n-1}$ ways to choose the balls to be colored.

orangeskid
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The boxes are your $n$ distinct variables, and the identical objects are the $m$ places (factors) you have available in a monomial of degree$~m$. Each factor is equal to one of the variables, and thereby "places an object in the box" for that variable; in the end one only cares how many objects have been put in each box (the power of that variable), since the order of factors in a monomial is irrelevant.

Marc van Leeuwen
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A monomial is an algebraic expression consisting of 1 term. So it can be expressed in the form: $a_1^{b_1}a_2^{b_2}a_3^{b_3}...a_n^{b_n}$

Where $b_1+...+b_n=m$

We then simply have to distribute the m exponents amongst the n possible bases ${a_1, a_2...,a_n}$. This is done in $\binom{m+n-1}{n-1}$, if you use stars and bars.

Sherlock Holmes
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  • The first sentence is not quite precise; that description would also allow a scalar coefficient to be included in a monomial. Some people indeed define monomials to allow that, but for this problem it is clearly not intended (the number of monomials would become infinite). – Marc van Leeuwen Oct 02 '14 at 08:06
  • @MarcvanLeeuwen Thanks for the correction, I didn't know that. – Sherlock Holmes Oct 02 '14 at 08:11
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Consider monomials in $x,y,z$. They are of the form $x^ay^bz^c$. If this is of order $m$, then $a+b+c=m$, and of course $a,b,c\geq0$. Thus to make a monomial, you have to "divide the $m$ into three boxes corresponding to $x$, $y$ and $z$" by choosing the numbers $a$, $b$ and $c$.

The same argument works with any number of variables, but this three variable case should give you the idea. The number of boxes is the number of variables $n$, and the number of objects to place in the boxes is the degree $m$.

Joonas Ilmavirta
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