Oops!

how do I compute the last three digits of 2014C1+2014C3+2014C5+...+2014C2013 by hand?

Consider the question: In how many ways can you select things from a collection of 2014 things?

This is equivalent to 2014C1 + 2014C2 + 2014C3 + .. 2014C2014

For each doll you have, 2 choices. To select it or not, so there are 2^2014 ways.

Now, for only odd number of dolls:

Half of those cases we selected odd number of dolls

This is equivalent to asking in how many cases you are choosing the odd terms.

Clearly the answer is 2^2013

bobbym wrote:

]]>Algebra is Combinatorics, Combinatorics is Algebra

Olinguito wrote:

The tune of the present-day Austrian national anthem was written by him.

I meant the German national anthem, not Austrian. Haydn himself was Austrian.

]]>]]>An editor is someone who separates the wheat from the chaff and then prints the chaff.

But there's nothing wrong with local matters being decided locally in any democracy, so let's hope this is good for all parts of the UK.

Bob

]]>What reason is that?

I was able to sign up now. The website looks promising. I love the navigator and the explanations and overviews are great. I didn't see any examples or practice problems. It's also not as easy as some other sites to navigate but the website is only in it's better phase.

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