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Question

Prove that nr=03r nCr=4n.

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Solution

nr=03rnCr=4n[30.nCr+31nC1+32nC2+..........+3nnCn][1+3n+9.n(n1)2!+........3n.1][(1+x)n=1+nx+n(n1)2!x2+.....x2](1+3)n

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