**This is not arranged in order of difficulty. If you are stuck, skip the problem to find more that you can do. 0. What three-digit integer is equal to the sum of the factorials of its digits? - $\boxed{145}=1!+4!+5!=1+24+120$ 1. The set $S=\{1,2,3, \ldots, 49,50\}$ contains the first 50 positive integers. After the multiples of 2 and the multiples of 3 are removed, how many integers remain in the set $S$ ? - $1+8(2)=\boxed{17}$ 2. What is the probability that a random arrangement of the letters in the word 'SEVEN' will have both E's next to each other? - $\frac{4!\cdot2!}{5!}=\boxed{\frac{2}{5}}$ 3. There is a machine with 8 toys in it that each cost between 25 cents and 2 dollars, with each toy being 25 cents more expensive than the next most expensive one. Each time Sam presses the big red button on the machine, the machine randomly selects one of the remaining toys and gives Sam the option to buy it. If Sam has enough money, he will buy the toy, the red button will light up again, and he can repeat the process. If Sam has 8 quarters and a ten dollar bill and the machine only accepts quarters, what is the probability that Sam has to get change for the 10 dollar bill before he can buy his favorite toy- the one that costs $\$ 1.75$ ? Express your answer as a common fraction. - $1-\left(\frac18 +\frac{1}{8\cdot7}\right)=\boxed{\frac67}$ 4. Amaretta's birthday is July 27, and her brother Enzo's birthday is September 3. Every year, Amaretta and Enzo celebrate by eating cake every day from Amaretta's birthday through Enzo's birthday (including both birthdays). If they did this for the first time in 2008, how many cake-eating days will they have observed by the end of 2016? - $9(39)=\boxed{351}$ 5. How many 4-digit positive integers exist that satisfy the following conditions: (A) Each of the first two digits must be 1, 4, or 5, and (B) the last two digits cannot be the same digit, and (C) each of the last two digits must be 5, 7, or 8? - $3^2\cdot3\cdot2=\boxed{54}$ 6. How many numbers can be expressed as the sum of two or more distinct elements of the set $\{0,1,2,4,8,16\}$? - $2^5-1=\boxed{31}$ 7. The sequence $2,3,5,6,7,10,11, \ldots$ contains all the positive integers from least to greatest that are neither squares nor cubes. What is the $400^{\text {th }}$ term of the sequence? - $400+20+7-2=\boxed{425}$ 8. I have a spinner that lands on 1 with a probability of $\frac{1}{10}, 2$ with a probability of $\frac{2}{10}, 3$ with a probability of $\frac{3}{10}$, and 4 with a probability of $\frac{4}{10}$. If Phil and Sarah both spin the spinner, what is the probability that they get the same number? - $\left(\frac{1}{10}\right)^2+\left(\frac{2}{10}\right)^2+\left(\frac{3}{10}\right)^2+\left(\frac{4}{10}\right)^2=\frac{30}{100}=\boxed{\frac{3}{10}}$ 9. A student must choose a program of four courses from a list of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and *at least* one mathematics course. In how many ways can this program be chosen? - English is fixed, but there are ${5\choose3}=10$ ways to choose the other 3 classes. ${3\choose3}=1$ of those ways does not contain any math. $10-1=\boxed9$ 10. In how many ways can the digits of 45,520 be arranged to form a 5-digit number? (Remember, numbers cannot begin with 0) - $(5!\underset{\text{ no leading zero }}{-4!)}\underset{\text{ the 5's are identical }}{\div2!}=2\cdot4!=48$ 11. Four politicians and three lawyers attend a party. Each politician shakes hands exactly once with everyone, and each lawyer shakes hands exactly once with each politician. How many handshakes take place? - Total possible handshakes $-$ lawyers' missed handshakes$\frac{7\cdot6}{2}-\frac{3\cdot2}{2}=21-3=\boxed{18}$ 12. Each of eight houses on a street is painted brown, yellow or white. Each house is painted only one color and each color is used on at least one house. No two colors are used to paint the same number of houses. In how many ways could the eight houses on the street be painted? - 1+2+5, 1+3+4, 2 imply $\set{x,y,z}=\set{1,2,5} \text{ or } \set{1,3,4}$ where $x,y,z$ are the numbers of houses with the same color - There is no way to leave out 1, because 2+2+4 contains two 2's. - $\frac{8 !}{1 ! \cdot 2 ! \cdot 5 !}$ ways to shuffle first case - $\frac{8 !}{1 ! \cdot 3 ! \cdot 4 !}$ ways to shuffle the second case - $3!$ ways to assign color to the monochromatic groups - $\begin{aligned}3!\left(\frac{8 !}{1 ! \cdot 2 ! \cdot 5 !}+\frac{8 !}{1 ! \cdot 3 ! \cdot 4 !}\right)&=3!\left(\frac{3\cdot8 !}{1 ! \cdot 3 ! \cdot 5 !}+\frac{5\cdot 8 !}{1 ! \cdot 3 ! \cdot 5 !}\right)\\ &=\cancel{3!}\left(\frac{8\cdot8 !}{1 ! \cdot \cancel{3 !} \cdot 5 !}\right)\\ &= 8\cdot8\cdot7\cdot6=64\cdot42=\boxed{2688} \end{aligned}$ 13. Mr. Brennan has 7 boys and 4 girls in his statistics class. In how many ways can he pick 3 boys and 2 girls to give a group presentation tomorrow? (The order in which the boys and girls are chosen does not matter.) - ${7\choose3}\cdot{4\choose2}=\frac{7!}{3!4!}\cdot\frac{4!}{2!2!}=7\cdot6\cdot5=\boxed{210}$ 14. Our basketball team has 10 players. We need to divide into two teams of 5 for an intra-squad scrimmage. In how many ways can we do this without restrictions? - Ways to pick first team ${10\choose5}=\frac{10\cdot9\cdot8\cdot7\cdot\cancel6}{5\cdot4\cdot\cancel{3\cdot2}}=4\cdot63=252$ - For each team of 5, there is a unique complement, so in the 252 scrimmages counted, there is double counting because A vs B is the same scrimmage as B vs A. So $252 \div 2=\boxed{126}$ 15. Our basketball team has 10 players, including Steve and Danny. We need to divide into two teams of 5 for an intra-squad scrimmage. In how many ways can we do this if Steve and Danny insist on playing on the same team? - Find the number of 3-men teams that can play alongside steve and danny. ${8\choose3}=\boxed{56}$ 16. Gretchen has eight socks, two of each color: magenta, cyan, black, and white. She randomly draws four socks. What is the probability that she has exactly one pair of socks with the same color? - Find all possible combinations $8\choose4$ - For one pair to occur, there must be 3 of 4 colors present $4\choose3$ - Any of those 3 can be the pair $3\choose1$ - $\frac{{4\choose3}\cdot{3\choose1}}{8\choose4}=\frac{48}{70}=\boxed{\frac{24}{35}}$