Odd dolls

Python/Sage Code:

1. Define the Factorial:

   f = lambda n: reduce(lambda x,y: x*y,range(1,n+1))
   

2. Define c(n,r):

   c = lambda n,r: f(n)/(f(n-r)*f(r))
   

3. Find the sum of all the terms in 2014Choose1, 2014Choose3,.... 2014Choose 2013

   sum(map(lambda n: c(2014,n),range(1,2014,2)))
   

Returns

$940548665568669306251536958404757081311082661551059108231284930007667721\ 332629761111375451190481936317190934232258529504007879852739014927504122\ 966971722289472552871078398197765642478782382447586745769902600079317250\ 993878191213477456421077323500371729653731804466371042874405729266169033\ 000257306380965325571431702400684077894692739546431970516457689091514390\ 255260678803636290619279616342108306500134556458970287457296562420600045\ 913373517277196622966242525937013186214163166593344799391393518637733932\ 076446381883519208233258347987390537033939221566465437167851936340727978\ 540772997219818300612848648192$

Mathematica (half the world is already built-in, so one line kills it all)

Mod[Sum[Binomial[2014, 2 i - 1], {i, 1, 1007}], 1000]
= 192

Note: I solved it the hard way first.

The problem states choose the odd number of Dolls from 2014 dolls. This means we can choose 1, 3, 5 ... 2013 dolls out of 2014. Now, how many ways are there for choosing 1 doll $\binom{2014}{1}$ Similarly to choose 3 dolls , there are $\binom{2014}{3}$ ways and so on. So total number of ways is given by, $\binom{2014}{1} + \binom{2014}{3} + \binom{2014}{5} + \dots + \binom{2014}{2013}$ We know that, $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n$ This consists of equal number of odd and even terms so sum of only odd terms is $2^n/2$ In our case this is $\boxed{2^{2013}}$ Finally we have to calculate $2^{2013} \% 1000$ which comes out to be $\boxed{192}$