<p>Can any math whizes here prove why the binomial coefficients of each term in a (x+b)^n polynomial is equal to tthe number of combinations of the power of one of the terms (either x or b) out of n? Or is this fact taken for granted because it is not necessary to know why?</p>
<p>Just a hint:
(x+b)^5= (x+b) (x+b) (x+b) (x+b) (x+b)
How many ways can you get, say, x^3, after expanding?
You have to chose any three (x+b)'s out of 5 to multiply just x's out of them. It's 5C3.
The remaining two (x+b)'s add factors (b<em>b) to (x</em>x*x), but they are determined after selecting 3 (x+b)'s for x^3.
(x+b)^5 = 5C5 * x^5 * b^0 + 5C4 * x^4 * b^1 + 5C3 * x^3 * b^2 +
5C2 * x^2 * b^3 + 5C1 * x^1 * b^4 + 5C0 * x^0 * b^5.
It would help to see the mechanics of it if you do (x+1)^5 manually.</p>
<p>Thanks for the advice gcf101. I multiplied out (x+b)^5 and I see the pattern, which is (x+b)^5 is essentially x(x+b)^4 + b(x+b)^4 and so on and so on. So X has to be multiplied by every combination of multiples of previous x and y factors. </p>
<p>So (x+b)^5 is equal to the sum of all possible multiples of factors which are the different combination of the x's and y's. So that means xy^4 must be the number of combinations where there are 4 y out of 5 factors, etc.</p>
<p>Binomial theorom </p>
<p>(x+y)^a</p>
<p>The are a+1 terms in the expantion.
The sum of the exponents of x and y is equal to a.
The y exponent is one less than the number of the term (i.e. term 5, the y exponent is 4)
The coeficient of each term is equal to (a/either exponent) (either x's exponent or y's) I don't mean divided by like (4/2) would mean the top number multiplyed by the number 1 less than it over and over (like factorical 5<em>4</em>3<em>2) and the bottem number does the same thing until it hits 1 so in this case (4</em>3)/(2<em>1) or (9/6) for examlple is (9</em>8<em>7</em>6<em>5</em>4)/(6<em>5</em>4<em>3</em>2*1) </p>
<p>For example (x+y)^3=x^3+3x^2y+3xy^2+y^3</p>
<p>I appreciate the input stix, but my original question was about why the combinations were the coefficients themselves.</p>
<p>il bandito, if you are familiar with math induction, you can prove binomial in a more abstract and precise way.
If math induction is new to you, it'll take you 10min to grasp it and get totally awed.
(x+b)^1 = 1C1 * x^1 + 1C0 b^1</p>
<p>Now, instead of dealing with "n" and '"n+1", let's use our favorite "5".
Assuming our formula for (x+b)^5 is correct ,we'll prove it for (x+b)^6.</p>
<p>(x+b)^6 = (x+b)^5 * (x+b).
Let's zero in on just one term, say x^4 * b^2.
We are getting it only from
(5C4 * x^4 * b + 5C3 * x^3 * b^2) * (x+b).
Leaving only relevant:
5C4 * x^4 * b * b + 5C3 * x^3 * b^2 * x =
x^4 * b^2 * (5C4 + 5C3).
OK, now just
(5C4 + 5C3) = 5!/4! + (5<em>4</em>3)/3! =
((5<em>4</em>3)/3!) * ((2<em>1)/4 + 1) = ((5</em>4<em>3)/3!) * (6/4) =
(6</em>5<em>4</em>3)/4! = 6C4.
So (x+b)^6 = ... + 6C4 * x^4 * b^2 + ...
Looks very familiar! Right, just like a typical term in (x+b)^5.
All the other terms can be derived the same way.
End of proof.
Basically the same technique is used in a generic proof with "n" and "n+1".</p>
<p>Pascal triangle is a beautiful visual for this:
Second line in it is
1 1
or
1C1 1C0
5th line:
1 5 10 10 5 1
6th line:
1 6 15 20 15 6 1
Outer terms in line "6" are "1"s, or 6C6 and 6C1.
3rd term in line "6" is a sum of 2nd and 3rd terms from line "5":
5C4 + 5C3 = 6C4 = 15 - coefficient in the 3rd term of (x+b)^6:
6C4 * x^4 * b^2.
Wow.</p>
<p>But wait! No, don't really.
Barron's goes as far as binominal coefficients for (x+b)^t with "t" being a fraction or negative.
As much as I value Barron's, this is utterly ridiculous.</p>
<p>Hey, people at large: do you remember anything truly difficult on binominal coefficients ever on real math 2c? I've never heard of any big'ns.</p>
<p>Dear Bandito</p>
<p>It is a direct result of Pascal's triangle. I recommend that you check the internet on Pascal's triangle and I am sure you will get the answer to your question.</p>
<p>Rusen Meylani.</p>