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I have some problem understanding this Exercise/problem. What is summand ? I have searched for it, but nothing concrete came up.

Problem: Look at the multinomial theorem. How many summands are there in $(x+y+z)^7$ and in $(w+x+2y+z)^9$ ?

Can someone explain to me what it is(summand) and how to get the solution.

The solution should be 36 for the first one and 220 for the last one.

Hanne
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  • Do yo know the general formula to solve the number of integer solutions to $x_1+x_2+x_3+\dots+x_n = N$? – Frank Vel Nov 20 '14 at 20:57

4 Answers4

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The number of summands is determined by $\binom{n+k-1}{k-1}$ or $\binom{n+k-1}{n}$ where $n$ is the exponent and $k$ is the number of unknowns ($x,y,z...$)

A summand for $(x+y+z)^7$ might be $210x^3y^2z^2$ or $x^7y^0z^0$ (or just written $x^7$)

Frank Vel
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A summand is an expression in a sum.

In $a+b$, $a$ and $b$ are the summands.

Henrik supports the community
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  • Ahh..okay I understood that one, thanks. But how do I find this summands in $(x+y+z)^7$ for example ? – Hanne Nov 20 '14 at 16:07
  • Basically you would expand it. You'll probably see a pattern, before you've gotten to 36 (I hope you'll see on before you get to 220 in the second case). As a hint: What will all the summands look like? – Henrik supports the community Nov 20 '14 at 16:15
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If you were to multiply out the expressions and combine like terms, you'd arrive at the number of summands.

The number of summands in the first is the number of triples $(a,b,c)$, with $0 \leq a,b,c \in \mathbb{Z} \leq 7$ and $a+b+c=7$. (Do you see why?)

The number of summands in the second is the number of quadruples $(a,b,c,d)$, with $0 \leq a,b,c,d \in \mathbb{Z} \leq 9$ and $a+b+c+d=9$. (Do you see why?)

EDIT: As an example, let's do $(a+b+c)^3$.

Multiplied out (I used Wolfram Alpha), it's:

$$a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3b^2c + 3b^2a + 3c^2a + 3c^2b + 6abc.$$

Now, let's look at the exponents of $a,b,c$ on each summand, in order:

$$(3,0,0), (0,3,0), (0,0,3), (2,1,0), (2,0,1), (0,2,1), (1,2,0), (1,0,2), (0,1,2), (1,1,1).$$

I listed all of the triples $(a,b,c)$ for which $0 \leq a,b,c \in \mathbb{Z} \leq 3$ and $a+b+c=3$.

Hopefully this makes it a bit clearer.

John
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  • I don`t see it, is there a more expanded or prof based version of how you can show it to me ? So that it is easier to see. – Hanne Nov 20 '14 at 16:34
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It's evident that the number of summands in the expansion of $(a+b)^n$ is $(n+1)$. By substituting $a=x, b=y+z$, and $n=7$, we can observe that each term in the expanded expression $(x+y+z)^7$ contains $(y+z)^n$, with $n$ ranging from $0$ to $7$. Additionally, each term is associated with a unique power of $x$ ranging from $7$ to $0$. This ensures that no two terms in the partial expansion can be directly combined. Given that the number of summands in the expansion of $(y+z)^n$ is also $(n+1)$, when $n$ ranges from $0$ to $7$, the total number of summands in the partial expansion is $1+2+3+4+5+6+7+8=36$.

Ricky
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Socrates 9ja
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