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Question
Let X=((10C1)2+2(10C2)2+3(10C3)2)+...+10(10C10)2), where 10Cr, r∈{1,2,...,10} denote binomial coefficients. Then, the value of 11430X is
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Solution
X=n∑r=0r⋅(nCr)2 ; n=10,
⇒X=n∑r=0nr⋅r⋅(nCr)⋅(n−1Cr−1) ; n=10,
⇒X=n⋅n∑r=0(nCn−r)(n−1Cr−1) ; n=10,
⇒X=n⋅(2n−1Cn−1) ; n=10,
⇒X=10⋅19C9
⇒X1430=10⋅19C91430=646
Alternate solution:
X=((10C1)2+2(10C2)2+3(10C3)2)+...+10(10C10)2)
It can be rewritten as,
X=0(10C0)2+1(10C1)2+2(10C2)2+...+10(10C10)2X=10(10C10)2+9(10C9)2+8(10C8)2+...+0(10C0)2⇒2X=10[(10C0)2+(10C1)2+(10C2)2+…+(10C10)2]⇒X=5[20C10]∴X1430=646
⇒X=n∑r=0nr⋅r⋅(nCr)⋅(n−1Cr−1) ; n=10,
⇒X=n⋅n∑r=0(nCn−r)(n−1Cr−1) ; n=10,
⇒X=n⋅(2n−1Cn−1) ; n=10,
⇒X=10⋅19C9
⇒X1430=10⋅19C91430=646
Alternate solution:
X=((10C1)2+2(10C2)2+3(10C3)2)+...+10(10C10)2)
It can be rewritten as,
X=0(10C0)2+1(10C1)2+2(10C2)2+...+10(10C10)2X=10(10C10)2+9(10C9)2+8(10C8)2+...+0(10C0)2⇒2X=10[(10C0)2+(10C1)2+(10C2)2+…+(10C10)2]⇒X=5[20C10]∴X1430=646
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