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  • The value of <img alt="\begin{aligned} & \left(\begin{array}{c} 30 \\ 0 \end{array}\right)\left(\begin{array}{l} 30 \\ 10 \end{array}\right)-\left(\begin{array}{c} 30 \\ 1 \end{array}\righ

The value of \begin{aligned} & \left(\begin{array}{c} 30 \\ 0 \end{array}\right)\left(\begin{array}{l} 30 \\ 10 \end{array}\right)-\left(\begin{array}{c} 30 \\ 1 \end{array}\right)\left(\begin{array}{l} 30 \\ 11 \end{array}\right)+ \left(\begin{array}{c} 30 \\ 2 \end{array}\right)\left(\begin{array}{c} 30 \\ 12 \end{array}\right)+\ldots \ldots+\left(\begin{array}{l} 30 \\ 20 \end{array}\right)\left(\begin{array}{l} 30 \\ 30 \end{array}\right) \end{aligned}

Option: 1

{ }^{60} \mathrm{C}_{20}


Option: 2

{ }^{30} \mathrm{C}_{10}


Option: 3

{ }^{60} \mathrm{C}_{30}


Option: 4

{ }^{40} \mathrm{C}_{30}


Answers (1)

best_answer

\\(1-x)^{30}={ }^{30} C_0 x^0-{ }^{30} C_1 x^1+{ }^{30} C_2 x^2 +\ldots \ldots+(-1)^{30}{ }^{30} C_{30} x^{30} \\\\ (x+1)^{30}={ }^{30} C_0 x^{30}+{ }^{30} C_1 x^{29}+{ }^{30} C_2 x^{28} +\ldots \ldots+{ }^{30} C_{10} x^{20}+\ldots .+{ }^{30} C_{30} x^0Multiplying (i) and (ii) and equating the coefficient of x^{20} on both sides, we get required sum = coefficient of x^{20} in \left(1-x^2\right)^{30}={ }^{30} C_{10}
 

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Gaurav

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