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  • The society enlists names of 21 adult male, 22 adult females and 20 children to give 25 free movie tickets to the elected ones. Already 12 members of the society withdrew their name while

The society enlists names of 21 adult male, 22 adult females and 20 children to give 25 free movie tickets to the elected ones. Already 12 members of the society withdrew their name while another 14 are already confirmed to get the free movie ticket. In how many ways can the members be selected now for the free movie tickets?

Option: 1

\frac{37!}{11!26!}


Option: 2

\frac{37!}{18!17!}


Option: 3

\frac{35!}{19!17!}


Option: 4

Cannot be determined


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 \mathrm{r} items from the \mathrm{n} different items with \mathrm{k} particular things always included and \mathrm{h} particular things always excluded is \mathrm{=^{n-k-h}C_{r-k}}

Since12 members are withdrawn and another 14 are already confirmed to get the free movie ticket, the following is evident.

  • The number from which the restricted combination is to be made is \mathrm{=n-k-h=\left ( 21+22+20 \right )-14-12=37}.

  • The number with which the restricted combination is to be made is \mathrm{=r-k=25-14=11}

Therefore, the required restricted combination is

\mathrm{=^{n-k-h}C_{r-k}}

\mathrm{=^{37}C_{11}}

\mathrm{=\frac{37!}{11!26!}}

Posted by

Shailly goel

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