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Q.
The sum ${ }^{10} C _3+{ }^{11} C _3+{ }^{12} C _3+\ldots \ldots \ldots . .+{ }^{20} C _3$ is equal to
Binomial Theorem
Solution:
${ }^{10} C _3+{ }^{11} C _3+{ }^{12} C _3+\ldots \ldots .+{ }^{20} C _3$
$={ }^{10} C _4+{ }^{10} C _3+{ }^{11} C _3+12^{ C }{ }_3+\ldots \ldots .+{ }^{20} C _3-{ }^{10} C _4$
$={ }^{11} C _4+{ }^{11} C _3+{ }^{12} C _3+\ldots \ldots . .+{ }^{20} C _3-{ }^{10} C _4$
$={ }^{21} C _4-{ }^{10} C _4={ }^{21} C _{17}-{ }^{10} C _6$.