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  • The names of 18 men, 16 women and 20 children participated in a lottery. But 11 among them got disqualified. The lottery will be awarded to 15 random participants. In how many ways can th

The names of 18 men, 16 women and 20 children participated in a lottery. But 11 among them got disqualified. The lottery will be awarded to 15 random participants. In how many ways can the participants be selected now?

Option: 1

=\frac{31!}{7!39!}


Option: 2

\frac{43!}{4!39!}


Option: 3

\frac{43!}{7!24!}


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!\left ( y-x \right )!}}
  • The restricted combination for the selection of the \mathrm{r} items from the \mathrm{n}different items with \mathrm{k} particular things always excluded is =^{\mathrm{n}-\mathrm{k}} \mathrm{C}_{\mathrm{r}}

Since 15 participants must never be included in the team, the following is evident.

  • The number from which the restricted combination is to be made is \mathrm{=n-k=(18+16+20)-11=43.}
  • The number with which the restricted combination is to be made is \mathrm{=r-k=15-11=4}

Therefore, the required restricted combination is

\begin{aligned} & ={ }^{\mathrm{n}-\mathrm{k}} \mathrm{C}_{\mathrm{r}} \\ & ={ }^{43} \mathrm{C}_4 \\ & =\frac{43 !}{4 ! 39 !} \end{aligned}

Posted by

himanshu.meshram

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