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Mathematics 8 Online
OpenStudy (anonymous):

What is the multinomial theorem in probability ? I did not understand anything.

OpenStudy (anonymous):

The multinomial theorem is a statement about expanding a polynomial when it is raised to an arbitrary power.

OpenStudy (anonymous):

It looks really complex.. Can you explain it?

OpenStudy (anonymous):

do you know the binomial theorem? remember it's like:$$(x+y)^n=\sum_{k=0}^n\binom{n}kx^{n-k}y^k$$

OpenStudy (anonymous):

the idea is that the multinomial theorem extends this to \((x+y+z)^n\) and even more -- past mere binomials.

OpenStudy (anonymous):

I did not know the summation form of the binomial theorem. What does the bracket with the n above and k below mean ?

OpenStudy (ybarrap):

I think of the multinomial theorem as an extension of the binormial theorem. Where in binomial, you choose from only 2 different types of things, like heads or tails, or 1 or 0, in mulinomial, you can choose from more than two, like the numbers on a die or anything else that has more than 2 possibilities.

OpenStudy (anonymous):

@Tushi it is called the binomial coefficient and is usually read '\(n\)-choose-\(k\)' -- it is the way to pick a \(k\)-combination from \(n\) objects, or the number of subsets of size \(k\) to pick from a finite set of size \(n\)

OpenStudy (anonymous):

What is the expansion of binomial theorem ? Is binomial coefficient attached to every term ?

OpenStudy (anonymous):

What is the expansion of binomial theorem ? Is binomial coefficient attached to every term ?

OpenStudy (anonymous):

each term has a binomial coefficient, yes... it tells the number of ways to partition \(n\) into \(k\) and \(n-k\), i.e. \(x^ky^{n-k}\)

OpenStudy (anonymous):

In the expansion of (a+b)^2, is the binomial coefficient attached to every term?

OpenStudy (ybarrap):

When you expand (a+b)^2, the binomial coefficients indicate the number of ways that the particular terms of x and y and their corresponding powers work out to be. @oldrin.bataku expressed it succinctly and precisely.

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