Eric has two identical number cubes. The faces of each cube are numbered 1–6. Eric will roll both of the number cubes at the same time. What is the probability that both of the numbers showing face-up will be multiples of 3?
A. two-ninths
B. one-third
C. start fraction 1 over 6 end fraction
D. one-ninth
A multiple of 3 is either a 3 or a 6. There are two such faces on each cube, so there are $2\cdot2=4$ outcomes where both numbers showing face-up are multiples of 3. There are $6\cdot6=36$ total possible outcomes (since each cube has 6 faces, and there are 2 cubes), so the probability is $\dfrac{4}{36}=\boxed{\textbf{(D)}\ \frac19}$.
How many different arrangements can be made with the letters from the word SPACE?
A. 24
B. 32
C. 60
D. 120
There are 5 letters in the word SPACE, so there are 5 choices for which letter is first, then 4 choices for which letter is second, then 3 choices for which letter is third, then 2 choices for which letter is fourth, and 1 choice for which letter is last. Therefore, there are $5\times4\times3\times2\times1=\boxed{\textbf{(D)}\ 120}$ different arrangements.
ou want to draft a four-player tennis team. There are eight players to choose from. How many different teams can you form?
A. 70
B. 64
C. 48
D. 336
There are 8 choices for the first slot, then 7 remaining choices for the second slot, then 6 remaining choices for the third slot, and finally 5 remaining choices for the fourth slot. However, the order in which we select the players does not matter, so we have counted each team 4! times (since there are 4! ways to order the four players we have selected). Therefore, we must divide our count by 4!, to get \[\dfrac{8\cdot7\cdot6\cdot5}{4!}=2\cdot7\cdot6\cdot5=\boxed{\textbf{(A)}\ 70}.\]
Write the number of permutations in factorial form. Then simplify. How many different ways can you and six of your friends sit in your assigned seats in math class?
A. 6!; 120
B. 6!; 720
C. 7!; 2,520
D. 7!; 5,040
There are 7 people (you and 6 friends) who will sit in the seats. There are 7 choices for the first seat, 6 for the second, 5 for the third, and so on, down to 1 choice for the last seat. Therefore, the number of ways that the group of 7 can be seated is \[7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 7!\] So the answer is $\boxed{\textbf{(C)}\ 7!; 2,520}$.
Below are the results of tossing a number cube 8 times. Find the experimental probability of tossing an even number.
3, 4, 1, 2, 5, 1, 6, 5
A. one-fourth
B. Start Fraction 3 over 8 End Fraction
C. The fraction is one-half.
D. start fraction 5 over 8 end fraction
An even number will be thrown whenever a 2, 4, or 6 is thrown. In the list of throws given, there are three such numbers, so the experimental probability is $\dfrac{3}{8} = \boxed{\textbf{(B)}\ \frac38}$.
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