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I failed to simplify this expression on my exams, not sure if it was even needed, but i wonder if it is possible to simplify $$\sum_{k=0}^{299} \binom{1000}{k}$$ by using $\binom{n}{k}=\binom{n}{n-k}$ and $\sum_{k=0}^n \binom{n}{k}=2^n$, etc.

user1078292
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    It's unlikely it can be simplified but it can be approximated fairly well by a normal distribution. If it was $499$ instead of $299$ then it could be made substantially simpler. – Erick Wong Sep 01 '16 at 19:10
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    What he said. Basically, apart from symmetry and total sum being $2^k$, very little exact can be said about sums of elements from a row of Pascal's triangle. – Arthur Sep 01 '16 at 19:16
  • Evaluated on a computer we get 403621297224372179587174778525490662939513429874229602901374585037574482875428641087322949044551375327065506507441402376473867715785996303960372112340768255477294037470591404017799106792622675867069713194644599674908017109185154253594250702645503751375904561165672 which is a very large number (which has no simple prime factorization). Putting two lines under the expression you got should be good enough imo. – Winther Sep 01 '16 at 19:47
  • Large deviations estimate suggest this would be of the order of $$10^{292}\approx2^{970}$$ – Did Sep 01 '16 at 20:43
  • Judging from the complexity of the duplicate answer, i guess i was not supposed to even try simplify, on that probabilities exam. Initial question was to estimate the probability that at least 300 odd digits and 300 even digits exist in a random 1000 digit number. – user1078292 Sep 01 '16 at 21:19

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