question: Compute $\dbinom{16}{5}$.

answer: 4368

raw_answer: $\dbinom{16}{5}=\dfrac{16\times 15\times 14\times 13\times 12}{5\times 4\times 3\times 2\times 1}=\boxed{4368}.$


question: Determine the number of ways to arrange the letters of the word PROOF.

answer: 60

raw_answer: There are two O's and five total letters, so the answer is $\dfrac{5!}{2!} = \boxed{60}$.


question: 23 people attend a party. Each person shakes hands with at most 22 other people. What is the maximum possible number of handshakes, assuming that any two people can shake hands at most once?

answer: 253

raw_answer: Note that if each person shakes hands with every other person, then the number of handshakes is maximized. There are $\binom{23}{2} = \frac{(23)(22)}{2} = (23)(11) = 230+23 = \boxed{253}$ ways to choose two people to form a handshake.


question: James has 7 apples. 4 of them are red, and 3 of them are green. If he chooses 2 apples at random, what is the probability that both the apples he chooses are green?

answer: 1/7

raw_answer: There are $\binom{7}{2}=21$ total ways for James to choose 2 apples from 7, but only $\binom{3}{2}=3$ ways for him to choose 2 green apples. So, the probability that he chooses 2 green apples is $\frac{3}{21}=\boxed{\frac{1}{7}}$.


question: We are allowed to remove exactly one integer from the list $$-1,0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11,$$and then we choose two distinct integers at random from the remaining list.  What number should we remove if we wish to maximize the probability that the sum of the two chosen numbers is 10?

answer: 5

raw_answer:
For each integer $x$ in the list besides 5, the integer $10-x$ is also in the list.  So, for each of these integers, removing $x$ reduces the number of pairs of distinct integers whose sum is 10.  However, there is no other integer in list that can be added to 5 to give 10, so removing 5 from the list will not reduce the number of pairs of distinct integers whose sum is 10.

Since removing any integer besides 5 will reduce the number of pairs that add to 10, while removing 5 will leave the number of pairs that add to 10 unchanged, we have the highest probability of having a sum of 10 when we remove $\boxed{5}$.


question: The numbers 1 through 25 are written on 25 cards with one number on each card. Sara picks one of the 25 cards at random. What is the probability that the number on her card will be a multiple of 2 or 5? Express your answer as a common fraction.

answer: 3/5

raw_answer: There are $12$ even numbers and $5$ multiples of $5$ in the range $1$ to $25$.  However, we have double-counted $10$ and $20$, which are divisible by both $2$ and $5$.  So the number of good outcomes is $12+5-2=15$ and the probability is $\frac{15}{25}=\boxed{\frac{3}{5}}$.


question: A bag has 3 red marbles and 5 white marbles.  Two marbles are drawn from the bag and not replaced.  What is the probability that the first marble is red and the second marble is white?

answer: 15/56

raw_answer: The probability that the first is red is $\dfrac38$.  Now with 7 remaining, the probability that the second is white is $\dfrac57$. The answer is $\dfrac38 \times \dfrac57 = \boxed{\dfrac{15}{56}}$.


question: Find the largest prime divisor of 11! + 12!

answer: 13

raw_answer: Since $12! = 12 \cdot 11!$, we can examine the sum better by factoring $11!$ out of both parts: $$ 11! + 12! = 11! + 12 \cdot 11! = 11!(1 + 12) = 11! \cdot 13. $$Since no prime greater than 11 divides $11!$, $\boxed{13}$ is the largest prime factor of $11! + 12!$.


question:
These two spinners are divided into thirds and quarters, respectively. If each of these spinners is spun once, what is the probability that the product of the results of the two spins will be an even number? Express your answer as a common fraction.

[asy]

size(5cm,5cm);

draw(Circle((0,0),1));

draw(Circle((3,0),1));

draw((0,0)--(0,1));

draw((0,0)--(-0.9,-0.47));

draw((0,0)--(0.9,-0.47));

draw((2,0)--(4,0));

draw((3,1)--(3,-1));

label("$3$",(-0.5,0.3));

label("$4$",(0.5,0.3));

label("$5$",(0,-0.5));

label("$5$",(2.6,-0.4));

label("$6$",(2.6,0.4));

label("$7$",(3.4,0.4));

label("$8$",(3.4,-0.4));

draw((0,0)--(0.2,0.8),Arrow);

draw((3,0)--(3.2,0.8),Arrow);

[/asy]

answer: 2/3

raw_answer: We will subtract the probability that the product is odd from 1 to get the probability that the product is even. In order for the product to be odd, we must have both numbers be odd. There are $2\cdot2=4$ possibilities for this (a 3 or 5 is spun on the left spinner and a 5 or 7 on the right) out of a total of $3\cdot4=12$ possibilities, so the probability that the product is odd is $4/12=1/3$. The probability that the product is even is $1-1/3=\boxed{\frac{2}{3}}$.


question: No two students in Mrs. Vale's 26-student mathematics class have the same two initials. Each student's first name and last name begin with the same letter. If the letter ``Y'' is considered a vowel, what is the probability of randomly picking a student whose initials are vowels? Express your answer as a common fraction.

answer: 3/13

raw_answer: The students' initials are AA, BB, CC, $\cdots$, ZZ, representing all 26 letters. The vowels are A, E, I, O, U, and Y, which are 6 letters out of the possible 26. So the probability of picking a student whose initials are vowels is $\frac{6}{26}=\boxed{\frac{3}{13}}$.


question: What is the expected value of the roll of a standard 6-sided die?

answer: 3.5

raw_answer: Each outcome of rolling a 6-sided die has probability $\frac16$, and the possible outcomes are 1, 2, 3, 4, 5, and 6.  So the expected value is $$ \frac16(1) + \frac16(2) + \frac16(3) + \frac16(4) + \frac16(5) + \frac16(6) = \frac{21}{6} = \boxed{3.5}. $$


question: How many positive divisors of 30! are prime?

answer: 10

raw_answer: The only prime numbers that divide $30!$ are less than or equal to 30.  So 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 are primes that divide $30!$, and there are $\boxed{10}$ of these.


question: Marius is entering a wildlife photo contest, and wishes to arrange his seven snow leopards of different heights in a row. If the shortest two leopards have inferiority complexes and demand to be placed at the ends of the row, how many ways can he line up the leopards?

answer: 240

raw_answer:
There are two ways to arrange the shortest two leopards. For the five remaining leopards, there are $5!$ ways to arrange them.

Therefore, the answer is $2\times5!=\boxed{240\text{ ways.}}$


question: My school's math club has 6 boys and 8 girls.  I need to select a team to send to the state math competition.  We want 6 people on the team.  In how many ways can I select the team without restrictions?

answer: 3003

raw_answer: With no restrictions, we are merely picking 6 students out of 14.  This is $\binom{14}{6} = \boxed{3003}$.


question: Nathan will roll two six-sided dice. What is the probability that he will roll a number less than three on the first die and a number greater than three on the second die? Express your answer as a common fraction.

answer: 1/6

raw_answer: For the first die to be less than three, it must be a 1 or a 2, which occurs with probability $\frac{1}{3}$. For the second die to be greater than 3, it must be a 4 or a 5 or a 6, which occurs with probability $\frac{1}{2}$. The probability of both of these events occuring, as they are independent, is $\frac{1}{3} \cdot \frac{1}{2} = \boxed{\frac{1}{6}}$.


question: A Senate committee has 8 Republicans and 6 Democrats.  In how many ways can we form a subcommittee with 3 Republicans and 2 Democrats?

answer: 840

raw_answer: There are 8 Republicans and 3 spots for them, so there are $\binom{8}{3} = 56$ ways to choose the Republicans.  There are 6 Democrats and 2 spots for them, so there are $\binom{6}{2} = 15$ ways to choose the Democrats.  So there are $56 \times 15 = \boxed{840}$ ways to choose the subcommittee.


question: How many different positive, four-digit integers can be formed using the digits 2, 2, 9 and 9?

answer: 6

raw_answer: We could go ahead and count these directly, but instead we could count in general and then correct for overcounting. That is, if we had 4 distinct digits, there would be $4! = 24$ orderings. However, we must divide by 2! once for the repetition of the digit 2, and divide by 2! for the repetition of the digit 9 (this should make sense because if the repeated digit were different we would have twice as many orderings). So, our answer is $\frac{4!}{2!\cdot 2!} = 2 \cdot 3 = \boxed{6}$.


question: I won a trip for four to the Super Bowl.  I can bring three of my friends.  I have 8 friends.  In how many ways can I form my Super Bowl party?

answer: 56

raw_answer: Order does not matter, so it is a combination. Choosing $3$ out of $8$ is $\binom{8}{3}=\boxed{56}.$


question: Determine the number of ways to arrange the letters of the word MADAM.

answer: 30

raw_answer: First we count the arrangements if all the letters are unique, which is $5!$. Then since the M's and the A's are not unique, we divide by $2!$ twice for the arrangements of M's and the arrangements of A's, for an answer of $\dfrac{5!}{2! \times 2!} = \boxed{30}$.


question: A palindrome is a number that reads the same forwards and backwards, such as 3003. How many positive four-digit integers are palindromes?

answer: 90

raw_answer: Constructing palindromes requires that we choose the thousands digit (which defines the units digit) and the hundreds digit (which defines the tens digit). Since there are 9 choices for the thousands digit, and 10 choices for the hundreds digit, creating $9 \cdot 10 = \boxed{90}$ palindromes.