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JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Hard
1 of 1
Analyzing +4 | -1
(300)(3010)βˆ’(301)(3011)+...+(3020)(3030)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ 0 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ {10} \end{array}} \right) - \left( {\begin{array}{cccccccccccccccccccc} {30}\\ 1 \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ {11} \end{array}} \right) + ... + \left( {\begin{array}{cccccccccccccccccccc} {30}\\ {20} \end{array}} \right)\left( {\begin{array}{cccccccccccccccccccc} {30}\\ {30} \end{array}} \right) is equal to
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Correct option: C
β€…β€Š30C10{\;^{30}}{C_{10}}
Option C is correct because the given expression is the coefficient of x20 in (1-x2)30, which simplifies to ^{30}C_{10}.

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