Jason wants to choose 9 players for his track team. There are 12 players to choose from. How many different teams can Jason make?
A. 21
B. 112
C. 220
D. 306
This is a combination problem, since the order in which the players are chosen does not matter. We want to choose a team of 9 players from a pool of 12, which can be written as "12 choose 9" or $\binom{12}{9}$. Using the formula $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, we get:
$\binom{12}{9} = \frac{12!}{9!3!} = \frac{12\times11\times10}{3\times2\times1} = 220$
Therefore, there are 220 different teams that Jason can make. The answer is $\boxed{\textbf{(C)}\ 220}$.
Write the number of permutations in factorial form. Then simplify. How many different ways can you and four of your friends sit in the backseat of a limousine?
A. 4!; 24
B. 4!; 120
C. 5!; 120
D. 5!; 720
The number of permutations of $n$ objects is $n!$, which is read as "$n$ factorial." In this problem, we want to find the number of ways that five people can sit in the backseat of a limousine, which is the number of permutations of five objects. Therefore, the answer is either $\textbf{(C)}$ or $\textbf{(D)}$.
To determine which one is correct, we can calculate $5!$:
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
Therefore, the answer is $\boxed{\textbf{(C)}\ 5!; 120}$.
Below are the results of tossing a number cube 7 times. Find the experimental probability of tossing an even number.
6 4 3 2 5 3 3
A. start fraction 3 over 7 end fraction
B. start fraction 2 over 7 end fraction
C. one-half
D. start fraction 4 over 7 end fraction
We need to count how many of the outcomes are even numbers and divide by the total number of outcomes. Out of the 7 outcomes listed, exactly 2 of them (the 6 and the 2) are even. Therefore, the experimental probability of tossing an even number is:
$\frac{\text{number of even outcomes}}{\text{total number of outcomes}} = \frac{2}{7}$
The answer is $\boxed{\textbf{(B) } \frac{2}{7}}$.
How many different ways can a coach select the captain and co-captain of a team from a group of 20 people?
A. 40
B. 160
C. 380
D. 420
The coach needs to choose 2 people out of a group of 20, which can be done in $\binom{20}{2}$ ways. Using the formula $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, we get:
$\binom{20}{2} = \frac{20!}{2!18!} = \frac{20\times19}{2\times1} = 190$
Therefore, there are 190 different ways for the coach to choose a captain and co-captain. The answer is $\boxed{\textbf{(C)}\ 380}$.
When buying a new dress, you have a choice of 3 different lengths, 5 different colors, and 2 different styles. How many choices are there for one dress?
A. 13
B. 15
C. 30
D. 32
Using the multiplication principle, the number of choices is the product of the number of choices for each category. That is:
$3 \times 5 \times 2 = 30$
Therefore, there are 30 choices for one dress. The answer is $\boxed{\textbf{(C)}\ 30}$.
There are 40 entries in the science fair. How many ways can the entries finish in first, second, and third place?
A. 240
B. 1,480
C. 12,640
D. 59,280