Here's the question you clicked on:
shubhamsrg
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
there are 4 options, though C(50,25) is an option but the ans given at the back is C(100,51)
sorry i actually need to prove it it should be C(2n,n-1) or C(2n,n+1) hence answer is correct ..
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^(n-1) hence from RHS we need the coefficient of x^(n-1) which is C(2n,n-1) here n=50 hence C(100,49) =C(100,100-49)=C(110,51)
a more genaralised result is C(50,0) * C(50,r) + C(50,1) * C(50,r+1) .....+C(50,n-r) * C(50,50) = C(2*50,50-r) though it can be generalised further but u can proceed as above...
i shall get back to you,,please gimme some time..
as i hardly remain online...
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) ?
here u take (1+x)^100 *(x+1)^200 =(1+x)^300
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
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
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
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 \[ \sum_{k=0}^{50}\binom{50}{k} \binom{50}{49-k}\] you can directly apply Vandemonde's identity to get \[ \binom{100}{49} \text{ or } \binom{100}{51}\]
Woops \[ \sum_{k=0}^{49}\binom{50}{k} \binom{50}{49-k} \]