The tutorial for this set of problems can be found here. The answers are at the end of this article.
Questions
1. A bookstore has a selection of 10 books for free. However, any customer is allowed only 3 books. How many ways can a customer choose 3 books?
2. A car race has 8 contestants. A record keeper writes down the car names as they cross the finish line. How many possible ways could the record turn out?
3. Your house has 10 (different) paintings that you are packing. You have 2 burlap bags that can hold 3 paintings and a box that can hold 4 paintings. How many ways can you pack the paintings?
4. You are planning a vacation for 6 days. You can either spend 2 days each at 3 destinations or 3 days each at 2 destinations. There are 12 destinations in your surroundings. a:How many different vacation plans are there? b:You can't make up your mind so you write down all possibilities on slips of paper, put them in a (big) box, and draw one. What is the probability that drawn vacation plan only includes 2 destinations?
5. In a variation of chess, the 16 pieces are randomly arranged in the two back rows of the board. A chess set has 8 pawns, 2 rooks, 2 knights, 2 bishops, a queen, and a king.
- a. What is the probability of getting the standard setup?
- b. What is the probability of getting the pawns all in the front row?
6. Suppose you are playing 5 card poker.
- a. What is the probability of getting a 4 of a kind?
- b. What is the probability of getting a 2 pair (and nothing better)?
7. A fun-size bag of M&M's has 15 M&M's. There are 5 red, 3 green, 3 blue, 2 yellow, 1 brown, and 1 orange.
- a. You randomly take 3 M&M's. What is the probability that all 3 are red?
- b. You randomly take 3 M&M's. What is the probability that 1 is red, 1 is green, and 1 is blue?
- c. You randomly take a subset of at least size 3 from the M&M's. What is the probability that the subset only contains red, green, and blue M&M's?
- d. You randomly take a subset of at least size 3 from the M&M's. What is the probability that all 3 green M&M's are included in the subset?
8. A movie theater has 9 showings in a day. The theater will be playing 3 new movies. There will be 5 showings of one movie, 3 showings of another, and just one showing of the third.
- a. How many ways are there to organize the showings?
- b. Suppose the manager enjoys making completely arbitrary decisions for the showings. What is the probability that no movie is played in consecutive showings?
Answers
1. This is equivalent to taking a subset of size 3 without order from a set of size 10. So the number of choices is,
$ {10 \choose 3} = \frac{10!}{3! 7!} = 120 $
2. This is equivalent to finding all permutations of size 8 which is $ 8! = 40320 $
3. This is equivalent to partitioning a set of size 10 into 2 subsets of size 3 and a subset of size 4 where two of the subsets are indistinguishable. So the number of ways is,
$ \frac{1}{2!} {10 \choose 3,3,4} = \frac{10!}{2! 3! 3! 4!} = 2100 $
4.a. If we only go to two destinations, this is equivalent to taking a subset of size 2 with order from a set of size 12. So the number of options with two destinations is,
$ P^{12}_2 = \frac{12!}{10!} = 132 $
If we go to three destinations, in the same way, the number of options is,
$ P^{12}_3 = \frac{12!}{9!} = 1320 $
So the total number of vacation plans is $ 132 + 1320 = 1452 $
4.b. The sample set S is all vacation plans and the event A is all vacation plans with only 2 destinations.
$ P(A) = \frac{|A|}{|S|} = \frac{132}{1452} = \frac{1}{11} $
5.a. The sample set S is all random setups and the event A is that the standard setup so $ |A|=1 $.
The size of S is the same as the number of partitions of a set of size 16 into subsets of size 8, 2, 2, 2, 1, and 1. So,
$ |S| = \frac{16!}{8! 2! 2! 2! 1! 1!} = 64864800 $
$ P(A) = \frac{|A|}{|S|} = \frac{1}{64864800} $
5.b. I will solve this two different ways.
First, we could lay out the 16 pieces and then choose 8 pieces for the front row. Let the sample set S be the choices for the front row. Let A be the choices that have the all pawns in the front row, so $ |A|=1 $.
$ |S| = { 16 \choose 8 } = \frac{16!}{8! 8!} = 12870 $
$ P(A) = \frac{|A|}{|S|} = \frac{1}{12870} $
Secondly, we could define S as in part a. So the size of A would be the number of ways to organize the back row with non-pawn pieces which is the number of partitions of a set of size 8 into subsets of size 2, 2, 2, 1, and 1. So,
$ |A| = {8 \choose 2,2,2,1,1} = \frac{8!}{2! 2! 2! 1! 1!} = 5040 $
$ P(A) = \frac{|A|}{|S|} = \frac{5040}{64864800} = \frac{1}{12870} $
