Q22. - Algebra 2

Question

Determine whether each situation involves a permutation or a combination. Then find the number of possibilities.

 

The winner and first, second, and third runners-up in a contest with 10 finalists.

Step-by-Step Solution

Answer

The situation is a permutation and the number of possibilities is 5040. 

Step 1. Given Information.

Given to choose the winner and first, second, and third runners-up in a contest with 10 finalists. It is to be determined if the situation involves a permutation or a combination and then the number of possibilities are to be calculated.

Step 2. Calculation .

A permutation is when n objects are available and r are to be picked and arranged in a certain order and the number of permutations is given by $P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$

A combination is when n objects are available and r are to be picked without arrangement and the number of combinations is given by $C\left(n,r\right)=\frac{n!}{\left(n-r\right)!r!}$

Here, the order of choosing does affect the final outcome i.e., every arrangement is different. Hence the finalists are to be arranged i.e., the given situation is a permutation.

 

Here n=10,r=4

Plugging the values:

 $\begin{array}{l}P\left(10,4\right)=\frac{10!}{\left(10-4\right)!}\\ P\left(10,4\right)=\frac{10!}{6!}\\ P\left(10,4\right)=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6!}{6!}\\ P\left(10,4\right)=10\cdot 9\cdot 8\cdot 7\\ P\left(10,4\right)=5040\end{array}$

Step 3. Conclusion .

Hence, the given situation is a permutation with 5040 possibilities.