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Question

If n is a positive integer, then find the value of n+2Cnn+3Cn+1n+4Cn+2n+3Cn+1n+4Cn+2n+5Cn+3n+4Cn+2n+5Cn+3n+6Cn+4.

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Solution

∣ ∣ ∣∣ ∣ ∣n+2Cnn+3Cn+1n+4Cn+2n+3Cn+1n+4Cn+2n+5Cn+3n+4Cn+2n+5Cn+3n+6Cn+4∣ ∣ ∣∣ ∣ ∣

we know that
nCr+nCr1=n+1Crn+1CrnCr1=nCr1
Applying R3R3R2 and R2R2R1

∣ ∣ ∣∣ ∣ ∣n+2Cnn+3Cn+1n+4Cn+2n+2Cn+1n+3Cn+2n+4Cn+3n+3Cn+2n+4Cn+3n+5Cn+4∣ ∣ ∣∣ ∣ ∣ (from equation - 1)

Applying R3R3R2
∣ ∣ ∣∣ ∣ ∣n+2Cnn+3Cn+1n+4Cn+2n+2Cn+1n+3Cn+2n+4Cn+3n+2Cn+2n+3Cn+3n+4Cn+4∣ ∣ ∣∣ ∣ ∣ (from equation - 1)
∣ ∣∣ ∣n+2C2n+3C2n+4C2n+2n+3n+4111∣ ∣∣ ∣ (nCr=nCnr)
Expanding and simplifying gives
=|(n+2)(n+3)|=1

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