The value of 15 C 0 2 - 15 C 1 2 - 15 C 2 2 - - 15 C 15 2 is

The correct option is C 0 ^15C^2_0 ^15C^2_1+ ^15C^2_2... ^15C^2_7+ ^15C^2_8... ^15C^2_15 Pairing the terms. ( ^15C^2_0 ^15C^2_15)+( ^15C^2_1+ ^15C^2 ...

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Standard XII

Mathematics

Implicit Differentiation

The value of ...

Question

The value of ${^{15} C_{0}}^{2} - {^{15} C_{1}}^{2} + {^{15} C_{2}}^{2} - \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot - {^{15} C_{15}}^{2}$ is

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-15

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Solution

The correct option is C 0
$^{15} C_{0}^{2} -^{15} C_{1}^{2} +^{15} C_{2}^{2} . . . -^{15} C_{7}^{2} +^{15} C_{8}^{2} . . .^{15} C_{15}^{2}$
Pairing the terms.
$(^{15} C_{0}^{2} -^{15} C_{15}^{2}) + (-^{15} C_{1}^{2} +^{15} C_{14}^{2}) + . . . (-^{15} C_{7}^{2} +^{15} C_{8}^{2})$
By using the property of binomial coefficient.
$^{n} C_{r} =^{n} C_{n - r}$
$(^{15} C_{0}^{2} -^{15} C_{0}^{2}) + (-^{15} C_{1}^{2} +^{15} C_{1}^{2}) + . . . (-^{15} C_{7}^{2} +^{15} C_{7}^{2})$
$= 0$

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The value of 15C02−15C12+15C22−⋅⋅⋅⋅⋅⋅⋅⋅−15C152 is

The correct option is C 015C20−15C21+15C22...−15C27+15C28...15C215Pairing the terms.(15C20−15C215)+(−15C21+15C214)+...(−15C27+15C28)By using the property of binomial coefficient.nCr=nCn−r(15C20−15C20)+(−15C21+15C21)+...(−15C27+15C27)=0

The value of ${^{15} C_{0}}^{2} - {^{15} C_{1}}^{2} + {^{15} C_{2}}^{2} - \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot - {^{15} C_{15}}^{2}$ is