Small question about binomial theorem

[ozone]

I was trying to make sense of the equation attached below which was on the wikipedia site.

However I'm not entirely sure how to make use of the "n choose 0" , "n choose 1", etc. statements that in front of each term in of the expansion. I roughly know how the expansion should look intuitively but I was hoping I could find a greater understanding.

Thank you.

 

[Muphrid]

Remember that binomial coefficients can be expressed in terms of factorials.

\binom{n}{m} = \frac{n!}{m!(n-m)!}

For example,

\binom{6}{2} = \frac{6!}{2!4!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(4 \times 3 \times 2 \times 1)} = \frac{6 \times 5}{2} = 15

You see that the 4! on the bottom canceled all but two factors on the top. One of the terms can always be used to get a lot of cancellations in this way.

 

[Ray Vickson]

I was trying to make sense of the equation attached below which was on the wikipedia site.

However I'm not entirely sure how to make use of the "n choose 0" , "n choose 1", etc. statements that in front of each term in of the expansion. I roughly know how the expansion should look intuitively but I was hoping I could find a greater understanding.

Thank you.

{n \choose m} \equiv \frac{n!}{m! (n-m)!} = \frac{n(n-1)...(n-m+1)}{m!},
so
{n \choose 0} = 1, \: {n \choose 1} = n, \; {n \choose 2} = \frac{n(n-1)}{2}, \; \cdots, {n \choose n} = 1.<br />
In practice it is often easier to get them recursively from
{n \choose 0} = 1, \: {n \choose 1} = n, \\<br /> {n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}, \; 1 \leq k \leq n.

RGV

 

[ozone]

Thank you it is much clearer now.