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shubhamsrg
Group Title
Binomial question:
simplify :
C(50,0) * C(50,1) + C(50,1) * C(50,2) .....+C(50,49) * C(50,50)
C(n,r) has the usual meaning i.e. n(combination)r
 one year ago
 one year ago
shubhamsrg Group Title
Binomial question: simplify : C(50,0) * C(50,1) + C(50,1) * C(50,2) .....+C(50,49) * C(50,50) C(n,r) has the usual meaning i.e. n(combination)r
 one year ago
 one year ago

This Question is Closed

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
there are 4 options, though C(50,25) is an option but the ans given at the back is C(100,51)
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
sorry i actually need to prove it it should be C(2n,n1) or C(2n,n+1) hence answer is correct ..
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
first we use (1+x)^n (x+1)^n =(1+x)^(2n) i am actually very lazy in writing ... can u check that the LHS of ur question is just the coefficient of x^(n1) hence from RHS we need the coefficient of x^(n1) which is C(2n,n1) here n=50 hence C(100,49) =C(100,10049)=C(110,51)
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
a more genaralised result is C(50,0) * C(50,r) + C(50,1) * C(50,r+1) .....+C(50,nr) * C(50,50) = C(2*50,50r) though it can be generalised further but u can proceed as above...
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i shall get back to you,,please gimme some time..
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
if possible sure ..
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
as i hardly remain online...
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
well,,i tried but i couldnt understand the concept really sir,,but i rote the formulla ! :P next ques was C(50,0) ^2 + C(50,1) ^2. .....(C(50,50) ^2 so according to the formulla,,r=0 and on plugging in that, we get the right ans i.e. C(100,50) but what about this kind ques : C(100,0)* C(200,150) + C(100,1) * C(200,151) ..... C(100,50) * C(200,200) ?
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
here u take (1+x)^100 *(x+1)^200 =(1+x)^300
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
okay,,following..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
sir ? hmm..?
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.1
is LHS the coefficient of x^50 then from RHS u need the coefficient of x^50 ( if i am wrong just find the proper power of x) hence coefficient of X^50 in RHS is C(300,50) or C(300,150) is the ans correct
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
C(50,0) * C(50,1) + C(50,1) * C(50,2) .....+C(50,49) * C(50,50) C(50,0) * C(50,49) + C(50,1) * C(50,48) .....+C(50,49) * C(50,0) http://en.wikipedia.org/wiki/Vandermonde's_identity
 one year ago

KKJ Group TitleBest ResponseYou've already chosen the best response.0
C(50,0) * C(50,1) = C(50,49) * C(50,50) C(50,1) * C(50,2) = C(50,48) * C(50,49) C(50,2) * C(50,3) = C(50,47) * C(50,48) etc
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
let \( r=49\) C(50,0) = C(50,50) C(50,1) = C(50,49) C(50,2) = C(50,48) ... the expression is like \[ \sum_{k=0}^{50}\binom{50}{k} \binom{50}{49k}\] you can directly apply Vandemonde's identity to get \[ \binom{100}{49} \text{ or } \binom{100}{51}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
Woops \[ \sum_{k=0}^{49}\binom{50}{k} \binom{50}{49k} \]
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
thank you :)
 one year ago
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