How do I find the number of terms in (x + 5y + 3z + w)^6?

Answer 1

Here's how we can solve the problem logically, without doing all the multiplication.

First of all, the coefficients aren't relevant to the problem; we don't care what the terms are specifically, just how many of them there are.

Let's create two new temporary variables: let a = x + 5y, and b = 3z + w. That reduces the problem, for the moment, to (a + b)^6.

Now, any time you take a binomial to a power -- such as (a+b)^n -- there will be n+1 terms in the final expansion: (a+b)^2 has 3 terms, (a+b)^3 has 4 terms, etc.

So in (a+b)^6 there will be seven terms, which will have various coefficients but generally be of the forms a^6, a^5b, a^4b^2, a^3b^3, a^2b^2, ab^5, and b^6. Note that, in each term, the sum of the powers is 6.

Now, if we reinstate the original variables, substituting back in for a and b, we can consider each of those seven terms.

a^6 becomes (x+5y)^6, which will have 7 terms.
a^5b becomes (x+5y)^5 (3z+w), which will have 6 * 2 = 12 terms.
a^4b^2 becomes (x+5y)^4 (3z+w)^2, which will have 5 *3 = 15 terms.
a^3b^3 becomes (x+5y)^3 (3z+w)^3, which will have 4 * 4 = 16 terms.
a^2b^4 becomes (x+5y)^2 (3z+w)^4, which will have 3 *5 = 15 terms.
ab^5 becomes (x+5y) (3z+w)^5, which will have 2 *6 = 12 terms.
And, finally, b^6 will have 7 terms.

7+12+15+16+15+12+7 = 84.

Hope that made sense.

Answer 2

The multinomial theorem says the terms will be of the form :
C(x)^n_1(5y)^n_2(3z)n_3(w)^n_4

where n_1+n_2+n_3+n_4 = 6 and the coefficient C = 6!/n_1!n_2!n_3!n_4!

For example there will be a (x)(5y)^2(3z)^2(w) term with a coefficient of 6!/1!2!2!1! = 180.

Answer 3

There are 84 terms

Unless i miscounted

i recommend going to www.quickmath.com, click on Expand under Algebra, then type in your problem, and it will give you the full expansion form of your problem.

Answer 4

what number - x?y?z?or w?..... or are you looking for what their total is?