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Question

In how many ways can a football team of $$11$$ players be selected from $$16$$ players? How many of these will Include $$2$$ particular players?


A
4638,2020
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B
4658,2001
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C
4368,2002
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D
None of these
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Solution

The correct option is C $$4368,2002$$
$$11$$ players can be selected of $$6$$ in $$\ ^{16}C_{11}$$ ways
$$\Rightarrow \dfrac{16!}{11!(16-11)!}=4368$$ ways

Including $$2$$ particular players

if two particular players is to included selection will be $$\ ^{14}C_{9}$$

$$\Rightarrow \dfrac{14!}{9!(14-9)!}\Rightarrow \dfrac{14\times 13\times 12\times 11\times 10\times 9!}{9!5\times 4\times 3\times 2}\Rightarrow 2002$$

Mathematics

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