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  • In how many ways can a collection of one or more ornaments be displayed from 21 golden ornaments, 25 silver ornaments, 16 diamond ornaments and 19 artificial stone ornaments?<div clas

In how many ways can a collection of one or more ornaments be displayed from 21 golden ornaments, 25 silver ornaments, 16 diamond ornaments and 19 artificial stone ornaments?

Option: 1

2^4\left(2^2+1\right)\left(2^3+3\right)\left(2^3+5\right)\left(2^4+1\right)

 


Option: 2

2^4\left(2^2+1\right)\left(2^3+3\right)\left(2^3+5\right)\left(2^4+1\right)-1


Option: 3

2^4\left(2^2+1\right)\left(2^3+5\right)\left(2^4+1\right)-1


Option: 4

\left(2^2+1\right)\left(2^3+3\right)\left(2^3+5\right)\left(2^4+1\right)-1


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!(y-x)!}}

  • The combination for the selection of the none or more items from the \mathrm{n} identical items is \mathrm{=n+1}

From the available data, the following is evident.

  • The combination for the selection of the none or more golden ornaments is =21+1=22

  • The combination for the selection of the none or more silver ornaments is =25+1=26

  • The combination for the selection of the none or more diamond ornaments is =16+1=17

  • The combination for the selection of the none or more artificial stone ornaments is =19+1=20

Therefore, the required combination is

\begin{aligned} & =22 \times 26 \times 17 \times 20-1 \\ & =(2 \times 11) \times(2 \times 13) \times 17 \times\left(2^2 \times 5\right)-1 \\ & =2^4 \times 5 \times 11 \times 13 \times 17-1 \\ & =2^4\left(2^2+1\right)\left(2^3+3\right)\left(2^3+5\right)\left(2^4+1\right)-1 \end{aligned}

Posted by

sudhir.kumar

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