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shubhamsrg

  • 3 years ago

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

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  1. shubhamsrg
    • 3 years ago
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    there are 4 options, though C(50,25) is an option but the ans given at the back is C(100,51)

  2. matricked
    • 3 years ago
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    sorry i actually need to prove it it should be C(2n,n-1) or C(2n,n+1) hence answer is correct ..

  3. matricked
    • 3 years ago
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    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)

  4. matricked
    • 3 years ago
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    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...

  5. shubhamsrg
    • 3 years ago
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    i shall get back to you,,please gimme some time..

  6. matricked
    • 3 years ago
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    if possible sure ..

  7. matricked
    • 3 years ago
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    as i hardly remain online...

  8. shubhamsrg
    • 3 years ago
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    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) ?

  9. matricked
    • 3 years ago
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    here u take (1+x)^100 *(x+1)^200 =(1+x)^300

  10. shubhamsrg
    • 3 years ago
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    okay,,following..

  11. shubhamsrg
    • 3 years ago
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    sir ? hmm..?

  12. matricked
    • 3 years ago
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    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

  13. KKJ
    • 3 years ago
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    nCr = nC(n-r)

  14. experimentX
    • 3 years ago
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    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

  15. KKJ
    • 3 years ago
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    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

  16. experimentX
    • 3 years ago
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    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 \[ \binom{100}{49} \text{ or } \binom{100}{51}\]

  17. experimentX
    • 3 years ago
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    Woops \[ \sum_{k=0}^{49}\binom{50}{k} \binom{50}{49-k} \]

  18. shubhamsrg
    • 3 years ago
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    thank you :)

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