Section 2.4 Combinatorics

Transcription

1 Section 2.4 Combinatorics Counting is so commonplace we don t give it a second thought, yet as the reader will see can be one of the most challenging. For very small sets, we simply count the elements like 1, 2, 3,. But for many sets, either because they are large or because their structure is complex, the task can be daunting so we apply some standard tools of counting. Although Venn diagrams and tree structures often are useful for counting complex sets, this section focuses on a basic rule, the multiplication principle, which although simple, has far reaching consequences. In its most basic form, it seems ridiculously simple; for example, how many dots are there in the diagram in Figure 1? Count the Dots Figure 1 We suspect you counted the 10 dots in the first row and then multiplied by 3. If this is true, then you used the multiplication principle. As a small step up in complexity chain, suppose you count the paths from A to C in Figure 2. No doubt this problem didn t stump you either, getting 4 5 = 20. Again you used the multiplication principle. How Many Paths from A to C? Figure 2

2 Now let s try the multiplication principle on a more interesting problem. How many paths are there in the road system in Figure 3, starting at the upper left and moving to the lower right, always moving to the right and down? Note in the middle all paths must go through a gap. Still think counting is easy? Did you find all 2450 paths? Counting Paths Figure 3 Don t worry we will count the paths in this network shortly; we just need a little more ammunition. Permutations A permutation is simply a rearrangement of objects. If you are standing next to someone and you change places, you have a new permutation. Definition: A permutation of r elements in a set of n elements is an arrangement of the r elements.

3 We are not so much interested in a single permutation of elements but the number of permutations there are in a given set. To find this we use, of course, the multiplication principle. Example 1: Arrangements of a Set with 3 Elements Find the number of permutations of size r = 1, 2 and 3 in the set {a, b, c}. Solution: The permutation of 1, 2 and 3 elements from {1, 2, 3} are easily found and are listed in Table 1. r =1 r = 2 r = 3 a ab abc b ac acb c ba bac bc bca ca cab cb cba Permutations of Size r in {a, b, c} Factorial When counting sets, one often encounters the product of consecutive integers from 1 to n. This product is called n factorial and denoted by n!= n (n 1) (n 2) 2 1 It is convenient to define 0! = 1. Theorem 1: The number of permutations of r elements in a set of size n, denoted P(n, r ), is n! P(n, r) = = n (n 1) (n 2) (n r +1) (n - r)!

4 Proof: The proof is a straightforward application of the multiplication principle. Choosing r elements from a set of size n, we see (think of ordering them from left to right) the first element can be selected n ways the second element can be selected n 1 ways (since we have already chosen 1) the third element can be selected n 2 ways (since now there are n 2 left. the r th element can be selected n r +1 ways Hence, by the multiplication principle P(n, r ) = n (n 1) (n 2) (n r +1). Margin Note: Permutations: Order matters! It is useful to note that P(n, n) = n (n 1) (n 2) 2 1 = n! Example 2. Typical permutations are as follows. How many ways can 10 people line up for a picture? P(10,10) =10!= 3,628,800. How many ways can 5 runners finish a race? P(5,5) = 5!=120. How many ways can first 3 place runners finish when there are 10 people in the race? P(10,3) = = 720.

5 Combinations Combinations are essentially permutations where order doesn t matter. So to count combinations we count the number of permutations (or arrangements) of r elements in a set of size n, then divide by r!, the number of ways of ordering the selected items. For example there are P(10,3) = 720 ways of doing 3 errands from a list of 10 errands. If we don t care about the order which we do the errands then we divide 720 by 3!= 6, getting C(10,3) =120, which is the number of combinations of 3 things taken from a set of 10 things. This discussion leads us to the following definition. Definition: A combination is a selection of elements taken from a set. Or simply a subset from a given set. We are more interested in finding the number of combinations rather than focusing on the individual combinations themselves. Example 2: Find all combinations (i.e. subsets) of size r =1, 2 and 3 from the set {a, b, c}. Solution: It is a simple matter to enumerate the combinations listed Table 2. r =1 r = 2 r = 3 {a} {a, b} {a, b, c} {b} {a, c} {c} {b, a} Combinations of Different Sizes of {a, b, c}. Table 2 Note: Note that the combination {a, b } is the same as the combination {b, a}. They are simply sets. Note how this contrasts with permutations where the permutation ab is not the same as ba.

6 Theorem 2: The number of combinations (subsets) of size r taken from a set of n size n, denoted by C(n, r ) or is r n n! C(n, r ) = = r r!(n -r)! Proof: The proof is based on the multiplication principle. To find the number of subsets of size r, we note that the number of permutations (arrangements) of size r is P(n, r) = n!. We now divide P(n, r) by r!, which is the number of (n - r)! permutations of the r elements, getting P(n, r) C(n, r ) = r! = n! r!(n -r)!. n Recall: The numbers C(n, r ) or are called binomial coefficients because r they are the coefficients in the binomial expansion n ( a + b) = n k = 0 n k a n It helps in thinking about combinations to read as "n choose r " r since it is the number of ways one can choose r items from a set 4 of n items. For example = 6 is read as 4 choose 2 is 6 meaning the number 2 of ways of choosing 2 things from 4 things is 6. n k b k Margin Note: Combinations: Order does not matter Example 3. How many ways can 10 people play a game with five members on each side?

7 Solution: A simple way to approach problem this is to think of the ways you can select your four teammates from 9 players, which is nine choose four, or 9 9! = = 4 4!5! = 126 Example 4. You flip a penny 10 times. The number of ways you get Pascal s Identity Verify Pascal s triangle n r = n 1 r + n 1 r 1 Proof: The algebraic proof is straightforward:

8 n n! = = r r!(n - r)! (n -1)! n r!(n - r)! = (n -1)! (n - r) r!(n - r)! (n -1)! r + r!(n - r)! = (n -1)! r!(n - r -1)! + (n - 1)! (r -1)!(n - r)! n 1 n 1 = + r r 1 It produces the well-known Pascal triangle. Pascal s Triangle Figure 4 Example 5. How many ways can the digits 0,1, 2,, 9 be arranged in a decreasing sequence of 5 digits? Solution: After selecting any five (distinct) digits, arrange them in decreasing order. The number of decreasing digits is the same as the number of subsets of size 5. That is

9 C(10, 5) = 10! 5!5! = 252. Example 6 (Going to the Movies). Three boys and two girls are going to a movie. How many ways can they sit next to each other under the following conditions? a) No boys sit next to each other. b) The two girls sit next to each other. Solution: a) The only way they can sit is boy-girl-boy-girl-boy. But the boys can be permuted 3!= 6 ways and the girls 2!= 2 ways, so the total number of arrangements is 3!2!=12. b) First think of the two girls are a single girl so you have 4 persons, 3 boys and 1 girl. Hence there are 4!= 24 ways to permute the two girls among the 3 boys. But, for each of these arrangements, we can permute the two girls 2!= 2 ways, and so the total number of arrangements is 4!2!= 48. Example 7 (Finding Paths). How many paths are there from Start to End in the road system in Figure 5, always moving to the right and down? Counting Paths Figure 5

10 Solution: Since all paths pass through the one-point gap, we divide the problem into two parts; finding the paths from Start to the gap, then finding the paths from the gap to End, and then multiply them together. From Start to the gap, note that we travel a total of eight blocks, four blocks to the right and four blocks down. Labeling each block as R or D, depending whether the move to the right or down, all paths can be written {x, x, x, x, x, x, x, x}, where four of the x ' s are R and four are D. The total number of paths is the number of ways you can select 4 D s (or R s) from a set of size 8, which is 8 choose 4 or C(8,4) = 70. Similarly, the number of paths from the gap to End is C(7,3) = 35. Hence, the total number of paths is = 2,450.

11 Problems 1. Compute the following. a) P(5,3) b) P(4,1) c) P(30,2) d) C(4,1) e) C(10,8) 7 f) 2 9 g) 2 h) (a + b) 6 i) (a + b) 10 Permutations and Combinations 2. (Tree Representation of Permutations) Draw a tree diagram showing all permutations of size 2 from the letters abcd. How many permutations are there? 3. (Distinguishable Permutations) The number of distinguishable permuations of 5! the word TOOTS is. Since there are five letters in the word one writes 5! in 2!2! the numerator. However, we cannot distinguish the 2 Ts and the 2 Os in the word, hence we divide 2!2!. Find the number of distinguishable permutations in the following words. a) TO b) TWO c) TOO d) TOOT e) SNOOT f) DALLAS g) TENNESSEE h) MISSISSIPPI i) ILLINOIS

12 4. (Going to the Movies) Four girls and four boys are going to a movie. How many ways can they be seated if no two girls sit next to each other? 5. (Baseball Season) A baseball league consists of 9 teams. How many games will be played over the course of a year if each team plays every other team exactly 20 times? 6. (Hmmmmmmmm) How many 2-element sets of the set {x Ν 1 n 100} such that the sum of the 2 elements is even? 7. (Picky People) How many ways can 8 people sit next to each other at a movie if a certain 2 of them refuse to sit next to each other? 8. (One Committee) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members? 9. (Two Committees) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members and an entertainment committee of 4 members if no member of the society can serve on both committees? 10. (Three Committees) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members, an entertainment committee of 3 members, and a welcoming committee of 2 members if no member of the society can serve on more than one committees? 11. (Serving on More than One Committee) How many ways can the Snail Darter Society, who has 25 members, elect an executive committee of 2 members, an entertainment committee of 3 members, and a welcoming committee of 2 members if members can serve on more than one committee?

13 12. (Counting Poker Hands) A hand of 5 cards is dealt from a deck of 52 to a poker player. a) How many total hands can be dealt? b) How many ways can a player be dealt a royal flush (A, K, Q, J,10) of the same suit? c)how many ways can a player be dealt a straight flush (five consecutive cards of the same suit, but not a royal flush)? d) How many ways can a player be dealt four of a kind (four cards alike)? e) How many ways can a player be dealt a full house (three cards alike and a pair)? 13. (Choosing Sides) How many ways can eight players choose sides to play four-against-four? 14. (Flipping Coins) Suppose you flip a penny 10 times. a) How many possible outcomes or sequences of heads and tails are possible? b) How many possible outcomes have 1 head? c) How many possible outcomes have 5 heads? d) If the probably of getting 5 heads is the number of ways of getting 5 heads, divided by the total number of outcomes, what is the probability of getting 5 heads and 5 tails? 15. (Counting Softball Teams). A college softball team is taking 25 players on a road trip. The traveling squad consists of 3 catchers, 6 pitchers, 8 infielders, and 6 outfielders. Assuming each player can only play her own position, how many different teams can the coach put on the field?