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
The captain can be any one of the 20 team members. The coach must then select a co-captain from the remaining 19 team members. That gives us $20\cdot19 = 380$ ways to select the captain and co-captain. Therefore, 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
You must select one of three different lengths, one of five different colors, and one of two different styles. Therefore, by the Multiplication Principle, the total number of choices is $3\cdot5\cdot2=\boxed{\textbf{(D)}\ 32}$.
There are 25 people competing in a race. In how many ways can they finish in first and second place?
A. 49
B. 400
C. 600
D. 625
There are 25 choices for first place. After the first place finisher has been selected, there are 24 remaining people who can finish in second place. Therefore, there are $25\cdot24=\boxed{\textbf{(D)}\ 625}$ ways for the race to end.
A bag contains tiles with the letters A-R-I-T-H-M-E-T-I-C. Amelia chooses a tile without looking and doesn’t replace it. She chooses a second tile without looking. What is the probability that she will choose the letter I both times?
A. start fraction 1 over 25 end fraction
B. Start Fraction 1 over 45 End Fraction
C. start fraction 2 over 45 end fraction
D. start fraction 2 over 55 end fraction