Lots of choices
Calculate 10C1+11C2+12C3+...+20C11 Sure, you could use a calculator, but where's the fun in that? PS you may use a calculator to calculate your final answer after you change this whole equation into just 2 terms
Clue (Don't read this if you want a challenge): add 10C0=1 to the equation
The answer is 352715.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
First, add 10C0=1 to the equation. So it becomes 10C0+10C1+11C2+12C3+...+20C11. Then note that for any positive integers x and y, xCy+xC(y+1)=(x+1)C(y+1). The simplest proof of this is pascal's triangle. xCy and xC(y+1) are terms side-by-side, and the term below them, their sum, is (x+1)C(y+1). So 10C0+10C1=11C1, 11C1+11C2=12C2, etc until 20C10+20C11=21C11. It's a telescopic sum. So that entire equation becomes 21C11-1. Then, use a calculator and you get the answer ^_^