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
There are 40 choices for first place, 39 choices for second place (since the same project can't win both first and second), and 38 choices for third place. To count the total number of ways, we multiply these numbers together:
$40\times39\times38=59,\!280$
Therefore, there are 59,280 ways to determine the first, second, and third place winners. The answer is $\boxed{\textbf{(D)}\ 59,280}$.
A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn’t replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?
A. one over fifty-five
B. two over fifty-five
C. two over one hundred twenty-one
D. three over one hundred twenty-one
Let's first find the probability that Tyler chooses the letter I on his first draw. There are 11 tiles in the bag, and one of them is an I. Therefore, the probability of drawing an I on the first draw is $\frac{1}{11}$.
Now, assuming that Tyler did not replace the tile he drew on the first draw (meaning there are now only 10 tiles left in the bag), the probability that he chooses an I on the second draw is $\frac{1}{10}$.
To find the probability that he chooses an I both times, we need to multiply the probabilities of each event happening, since the events are independent:
$\frac{1}{11} \times \frac{1}{10} = \frac{1}{110}$
Therefore, the probability that Tyler chooses the letter I both times is $\boxed{\textbf{(A)}\ \frac{1}{55}}$.
A sandwich shop offers the following toppings. How many two-topping sandwiches can you make?
lettuce
tomato
bacon
cheese
mustard
A. 8
B. 10
C. 12
D. 20
To count the number of two-topping sandwiches, we need to count the number of ways to choose 2 of the 5 given toppings. We can use the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ to calculate this:
$$\binom{5}{2} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$$
Therefore, there are 10 two-topping sandwiches that can be made. The answer is $\boxed{\textbf{(B)}\ 10}$.
Hallie is trying to win the grand prize on a game show. Should she try her luck by spinning a wheel with 6 equal sections labeled from 1 to 6 and hope she gets a 5, or should she roll two number cubes and hope she gets the same number on both cubes? Explain
We can compare the probability of winning using each method.
First, consider spinning the wheel. Since there are 6 equal sections and only 1 of them is labeled with a 5, the probability of winning is $\frac{1}{6}$.
Second, consider rolling two number cubes. There are 6 possible outcomes for each roll of a single number cube (the numbers 1 through 6). Since we want the same number on both cubes, there is only 1 favorable outcome for each of these 6 outcomes (for example, if the first cube shows a 3, then the second cube must also show a 3). Therefore, the probability of winning by rolling two number cubes is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$.
Since $\frac{1}{6} < \frac{1}{36}$, Hallie should spin the wheel if she wants a better chance of winning the grand prize. Therefore, spinning the wheel is the better choice.
A bag contains 2 white marbles and 7 purple marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then a second marble is drawn.
a. What is the probability of selecting a purple marble and then a white marble?
b. What is the probability of selecting two white marbles?
c. Is there a greater chance of selecting two white marbles in a row or two purple marbles in a row? Show your work.
a. The probability of selecting a purple marble on the first draw is $\frac{7}{9}$, since there are 7 purple marbles out of 9 total marbles. Then, after one marble is drawn and not replaced, there are 8 marbles left in the bag, including 2 white marbles. Therefore, the probability of selecting a white marble on the second draw is $\frac{2}{8}$ (since there are 2 white marbles left out of 8 total marbles). By the multiplication principle, the probability of selecting a purple marble and then a white marble is:
$$\frac{7}{9} \times \frac{2}{8} = \frac{7 \times 2}{9 \times 8} = \frac{7}{36}$$
b. The probability of selecting a white marble on the first draw is $\frac{2}{9}$, and, assuming that the first marble drawn was white and not replaced, the probability of selecting another white marble on the second draw is $\frac{1}{8}$ (since there is now only 1 white marble left out of 8 total marbles). By the multiplication principle, the probability of selecting two white marbles is:
$$\frac{2}{9} \times \frac{1}{8} = \frac{1}{36}$$
c. The probability of selecting two white marbles in a row is $\frac{1}{36}$, and the probability of selecting two purple marbles in a row can be calculated by the same method as part (a), but with all of the numbers changed:
$$\frac{7}{9} \times \frac{6}{8} = \frac{21}{36}$$
Since $\frac{21}{36} > \frac{1}{36}$, there is a greater chance of selecting two purple marbles in a row.