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In how many ways can an Interview team select one or more candidates from the Group I of 16 different candidates,the Group II of 5 different candidates, and the Group III of 6 different candidates?

Option: 1

715


Option: 2

2^{27}


Option: 3

2^{27}-1


Option: 4

490


Answers (1)

best_answer

Note the following:

  • The formula for the combination for the selection of the \mathrm{x} items from the \mathrm{y}different items is \mathrm{^{y}C_{x}=\frac{y!}{x!\left ( y-x \right )!}}

  • The combination for the selection of the one or more items from the \mathrm{n}different items is \mathrm{=2^{n}-1}

From the available data, the following is evident.

  • The combination for the selection of the one or more different candidates from the Group I is \mathrm{=\left ( ^{4}C_{0}+^{4}C_{0}+............+^{16}C_{16} \right )}

  • The combination for the selection of the one or more different candidates from the Group II is \mathrm{=\left ( ^{5}C_{0}+^{5}C_{1}+............+^{5}C_{5} \right )}

  • The combination for the selection of the one or more different candidates from the Group III is\mathrm{=\left ( ^{6}C_{0}+............+^{6}C_{6} \right )}

Therefore, the required combination for selection of none or more candidates

\begin{aligned} & =\left({ }^4 C_0+{ }^4 C_1+\ldots \ldots \ldots+{ }^{16} C_{14}\right)\left({ }^5 C_0+{ }^5 C_1+\ldots \ldots+{ }^5 C_5\right)\left({ }^6 C_0+\ldots \ldots \ldots .+{ }^6 C_6\right) \\ & =2^{16} \times 2^5 \times 2^6 \\ & =2^{27} \end{aligned}

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Divya Prakash Singh

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