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# Write the value of (^{5}C_{1} + ^{5}C_{2} + ^{5}C_{3} + ^{5}C_{4} + ^{5}C_{5}).

Answer :

⇒ ^{5}C_{1} + ^{5}C_{2} + ^{5}C_{3} + ^{5}C_{4} + ^{5}C_{5} ⇒ ^{6}C_{2} + ^{6}C_{4} + 1 [As ^{5}C_{5} = 1] ⇒ 15 + 15 + 1 ⇒ 31

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Prove that the product of 2n consecutive negative integers is divisible by (2n)!

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