in the binomial theorem how the n(n-1)/2! came please explain

Question

anonymous · Answer

there are a lot more terms than $\frac{n(n-1)}{2}$

amistre64 · Answer

lots of expanding binomial powers

amistre64 · Answer

a pattern develops, the coeffs relate to pascals triangle

anonymous · Answer

$\frac{n(n-1)}{2}$ is the number of ways you can choose 2 items from a set of $n$
commonly called "n choose 2" and written as $\binom{n}{2}$ or $_nC_2$

amistre64 · Answer

$$(a+b) = 1a+1b$$
$$(a+b)^2 = 1a^2+2ab+1b^2$$
$$(a+b)^3 = 1a^3+3a^2b+3ab^2+1b^3$$
$$(a+b)^4 = 1,4,6,4,1$$
$$(a+b)^5 = 1,5,10,10,5,1$$
etc ...

amistre64 · Answer

there is some induction method that then applies to form a proof is all

anonymous · Answer

you can also see it, if you have the patience, by actually computing this and not using the commutative law
it is messy, but it becomes very clear why the coefficients are "n choose k"

amistre64 · Answer

1
                   1  1
                 1  2  1
               1  3  3  1
  1       4        6        4        1
1 (1+4)  (4+6) (6+4)  (4+1) 1

etc ....

anonymous · Answer

for example 
$$(a+b)(a+b)(a+b)=(aa+ab+ba+bb)(a+b)$$
$$=aaa+aba+baa+abb+aab+abb+bab+bbb$$

amistre64 · Answer

messy indeed :)

anonymous · Answer

true, but it takes the mystery out of the $\binom{n}{k}$ part right? how many terms look like $a^2b$?   $aab,aba, baa$ the number of ways you can choose 2 out of three slots to put the $a$

anonymous · Answer

all of this is obscured somewhat by the rather useful commutative law
but if you were working non commutative system, things would be different