I can do a binomial expansion using Pascal's Triangle, but...?

Question

what about a trinomial expansion using something similar ?

Answer 1

1.convert the trinomial into a binomial by bracketing two terms
2.use the binomial expansion keeping the bracketed terms as one block.
3.now expand the binomial brackets

Answer 2

Great question! It took me a while to derive the answer. Start with the general trinomial:

(a + b + c)^n

Every term in the expansion has the form:

m(a^d)(b^e)(c^f)

I originally wrote the entire derivation of the answer. but decided that might be excessive. The terms in the trinomial expansion are:

n!/(d!e!f!) *(a^d)(b^e)(c^f)

Just make sure that you choose every possible combination of d, e, and f so that d + e + f = n. Remember to include the cases where the exponents are 0 and that 0! =1.

For example:

(a + b + c)^3 = a^3 + (3!/(2!1!))a^2b + (3!/(2!1!))a^2c + (3!/(1!2!))ab^2 + (3!/(1!1!1!)abc + (3!/(1!2!))ac^2 + (3!/3!)b^3 + (3!/(2!1!))b^2c + (3!/(1!2!))bc^2 + (3!/3!)c^3

Simplifying:

(a + b + c)^3 = a^3 + 3a^2b + 3a^2c + 3ab^2 + 6abc + 3ac^2 + b^3 + 3b^2c + 3bc^2 + c^3

The real power of this system is it will work for not just trinomials but for any n-nomial to any power. If you go to a simple binomial, you will find that Pascal's triangle is equivalent to saying that any term of an order n binomial expansion can be written as:

n!/(i!j!) * a^i b^j

Once I saw this, I stopped using Pascal's triangle.

If you want the derivation of the formula, e-mail me.

Answer 3

Seeing Pretzel's post reminded me: there is such a thing as a multinomial distribution using a generalization of the binomial coeff.