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I'm not understanding the method of using multinomial theorem in combinatorics problems.

For example, suppose we want to distribute $17$ identical oranges among $4$ children such that each child gets at least $1$ orange, how many ways can u distribute the oranges?

The solution in the book says the max no of oranges a boy can get is $17$, so the required no of ways is

Coefficient of $x^{20}$ in $[ x + x^2 + x^3 + \ldots + x^{17} ]^4$

If I am not right, this is a multinomial expansion, but what is the intuition behind this step?

PS : I don't have a very high level of knowledge of math, so a simple intuitive explanation will be highly appreciated

kimtahe6
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user34304
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2 Answers2

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$x^{20}$ should be $x^{17}$ in your question.

$$(x+x^2+\cdots+x^{17})^{4}$$ $$=(x+x^2+\cdots+x^{17})\times (x+x^2+\cdots+x^{17})\times (x+x^2+\cdots+x^{17})\times (x+x^2+\cdots+x^{17})$$

So, you choose $x^A, x^B, x^C, x^D$ from each $()$ from the left to the right.

Then, what you need is the number of a set $(A,B,C,D)$ such that $$A+B+C+D=17\ \text{and}\ A,B,C,D\ge 1$$

Note that the latter is already satisfied.

Each represents the number of oranges which a child get. Then, imagine when you find the coefficient of $x^{17}$. You'll choose every pattern of $(A,B,C,D)$ such that $A+B+C+D=17$.

Hence, the coefficient of $x^{17}$ represents the ways you can distribute the oranges. I hope this is helpful.

mathlove
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  • I didn't understand the part about choosing x^a, x^b.... From the brackets. Can u please elucidate? – user34304 Dec 26 '13 at 06:36
  • What exactly do the terms in the brackets - x, x^2 .... Indicate? – user34304 Dec 26 '13 at 06:37
  • Suppose that you are trying to get the coefficient of $x^{17}$. Then, you'll choose one term in one bracket. For example, $x^{3}\times x^{4}\times x^{5}\times x^{5}=x^{17}$. This represents that you distribute the oranges to each child as $3, 4,5,5$. – mathlove Dec 26 '13 at 06:39
  • $i$ is more important than $x^i$. In one bracket, there are 17 choices as $1,2,\cdots,17$. But four numbers has to sum up to 17. – mathlove Dec 26 '13 at 06:40
  • One last question, sir - why exactly are we using only multinomial theorem, does any other method work – user34304 Dec 26 '13 at 07:04
  • Yes. First suppose that you give one orange to each child. So now you have $17-4=13$ oranges. By the way, imagine that you have 13 circles and 3 short lines. Then, you arrange these in a row and count the number of circles in each section between lines. The number of circles of each section represents the number of oranges. The number of permutation is $(13+3)!/(3!13!)$. Since you've already given one orange to each child, this is the answer, which is the same as user4140's answer. – mathlove Dec 26 '13 at 07:13
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First off: if a kid gets 17 oranges then the other kids die of hunger.And that one kid gets too much energy and becomes annoying.

Here is how I would approach the problem. First give 1 orange to each kid. You now have 13 oranges which you can distribute however you like.

Now use stars and bars to see in how many ways you can distribute the remaining 13 oranges.(see this answer)

Here there are 3 bars and 13 stars. So there are $\binom{16}{3}$ ways to distribute the oranges.

Hope this helps.

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Asinomás
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