InterviewSolution
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InterviewSolution
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If `sum_(r = 0)^(25) {""^(50)C_(r)* ""^(50-r)C_(25 - r)} = K(""^(50)C_(25)),` then, K is equal toA. `2^(24)`B. `2^(25) - 1`C. `2^(25)`D. `(25)^(2)` |
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Answer» Correct Answer - C Given, `overset(25)underset(r = 0)sum {""^(50)C_(r)*""^(50-r)C_(25 - r)} = K " "^(50)C_(25)` `rArr overset(25)underset(r = 0)sum ((50!)/(r!(50-r)!) xx ((50 - r)!)/((25 - r)!25!)) = K " "^(50)C_(25)` `rArr" "overset(25)underset(r = 0)sum ((50!)/(25!25!) xx (25!)/(r!(25 - r)!)) = K" "^(50)C_(25)` [on multiplying `25!` in numerator and denominator.] `rArr" "^(50)C_(25)overset(25)underset(r = 0)sum ""^(25)C_(r) = K" "^(50)C_(25)" " [because ""^(50)C_(25) = (50!)/(25!25!)]` `rArr" "K = overset(25)underset(r = 0)sum ""^(25)C_(r) = 2^(25)` `[because ""^(n)C_(0)+ ""^(n)C_(1) + ""^(n)C_(2) + .....+ ""^(n)C_(n) = 2^(n)]` `rArr" " K = 2^(25)` |
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