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Question

If ∣ ∣ ∣111mC1m+3C1m+6C1mC2m+3C2m+6C2∣ ∣ ∣ = 2α3β5γ then α+β+γ is equal to

A
3
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B
5
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C
7
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D
0
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Solution

The correct option is A 3
∣ ∣ ∣111mC1m+3C1m+6C1mC2m+3C2m+6C2∣ ∣ ∣=2α3β5γ
∣ ∣ ∣ ∣111mm+3m+6m(m1)2(m+3)(m+2)2(m+6)(m+5)2∣ ∣ ∣ ∣=2α3β5γ
Consider,∣ ∣ ∣ ∣111mm+3m+6m(m1)2(m+3)(m+2)2(m+6)(m+5)2∣ ∣ ∣ ∣
C1C1C2,C2C2C3
∣ ∣ ∣ ∣00133m+63(m+1)3(m+4)(m+6)(m+5)2∣ ∣ ∣ ∣
=32∣ ∣ ∣ ∣00111m+6(m+1)(m+4)(m+6)(m+5)2∣ ∣ ∣ ∣
=32[m+4m1]=33
So, 203350=2α3β5γ
On comparing,
α=γ=0,β=3
So,α+β+γ=3

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