Q4. - Algebra 1

Question

Eduardo has 20 $\mathit{C}\mathit{D}\mathit{s}$. He wants to choose 3 of them at random to take on a road trip. How many different ways can he do this if the order is not important?

 $\mathit{F}$ 60

$\mathit{G}$ 84

$\mathit{H}$ 1,140

$\mathit{J}$ 6,840

Step-by-Step Solution

Answer

There are 1,140 different ways in which Eduardo can choose 3 $CDs$ out of 20 if the order is not important.

Step 1. Combination formula.

The number of combinations of$n$objects taken$r$at a time is the quotient of$n!$and $(n-r)!r!$, that is,$C(n,r)=\frac{n!}{(n-r)!r!}$.

Step 2. Substitution.

Substitute 20 for $n$ and 3 for $r$ into combination formula.

$C(20,3)=\frac{20!}{(20-3)!3!}$

Step 3. Simplify.

Simplify the above obtained expression.

$\begin{array}{l}C(20,3)=\frac{20!}{17!3!}\\ =1140\end{array}$

Therefore, there are 1140 different ways in which Eduardo can choose 3 $CDs$ out of 20 if the order is not important. Option $H$ is the required answer.