Counting Subsets of a Set: Combinations Lecture 31 Section 6.4 Wed, Mar 21, 2007

r-Combinations An r-combination of a set of n elements is a subset of r of the n elements. The order of the elements does not matter. The 3-combinations of the set {a, b, c, d, e} are {a, b, c}, {a, b, d}, {a, b, e}, {a, c, d}, {a, c, e}, {a, d, e}, {b, c, d}, {b, c, e}, {b, d, e}, {c, d, e}.

Counting r-Combinations Theorem: The number of r-combinations of a set of n elements is Examples: C(4, 2) = (4  3)/(2  1) = 6. C(10, 3) = (10  9  8)/(3  2  1) = 120. C(1000, 2) = (1000  999)/(2  1) =

Some Useful Facts C(n, 0) = 1 for all n  0. C(n, 1) = n for all n  1. Notice that C(n, r) = C(n, n – r). For example, C(100, 99) = C(100, 1) = 100/1 = 100. Therefore, C(n, n) = 1 for all n  0. C(n, n – 1) = n for all n  1.

Another Useful Fact The TI-83 will calculate C(n, r). Enter n. Select MATH > PRB > nCr. Enter r. Press ENTER. The value of C(n, r) appears.

Counting r-Combinations Proof of the theorem (by induction on n). Base case: Let n = 0. Then r = 0 and there is only one 0-combination, the null set. Also, 0!/(0!0!) = 1. So the statement is true when n = 0.

Counting r-Combinations Inductive case: Suppose that the statement is true when n = k, for some integer k  0. Consider a set of k + 1 elements. If r = 0, then there is only one 0-combination, the null set, and

Counting r-Combinations If r = k + 1, then there is only one k- combination, the entire set, and So let r be any number between 0 and k + 1 (0 < r < k + 1). Select an arbitrary element a from the set.

Counting r-Combinations For each r-combination of the k + 1 elements, a is either a member or not a member. We will count the r-combinations for which a is a member and then count the r- combinations for which a is not a member.

Counting r-Combinations Case 1: a is not a member of the combination: The r elements come from the remaining k elements. By the inductive hypothesis, there are such sets.

Counting r-Combinations Case 2: a is a member of the combination: The other r – 1 elements in the subset come from the k remaining elements in the set. By the inductive hypothesis, there are such sets.

Counting r-Combinations Therefore, the number of r-combinations of k + 1 elements is A “little algebra” shows that this equals

Counting r-Combinations Therefore, the statement is true when n = k + 1. Thus, the statement is true for all n  1.

Example: Counting r- Combinations Recently I needed to find the distribution of averages of 10 numbers selected at random from a set of 19 numbers. I wrote a C++ program to use brute force to calculate the distribution. It is much easier to write the program if the sampling is done with replacement.

Example: Counting r- Combinations Sampling with replacement, there are possible samples = The program took 21.2 seconds to compute the distribution using 7 instead of 10 numbers. How long would it take using 10 numbers?

Example: Counting r- Combinations How many possibilities are there if we sample without replacement? How long would it take to calculate the distribution?

Example: Counting r- Combinations How can that be determined? Can a computer program make the determination by brute force (exhaustive checking) within a reasonable amount of time? There are C(48, 4) = 194,580 possible choices. A computer can do the math really fast, in say one second.

Lotto South In Lotto South, a player chooses 6 numbers from 1 to 49. Then the state chooses at random 6 numbers from 1 to 49. The player wins according to how many of his numbers match the ones the state chooses. See the Lotto South web page.Lotto South web page

Lotto South There are C(49, 6) = 13,983,816 possible choices. Match all 6 numbers There is only 1 winning combination. Probability of winning is 1/ =

Lotto South Match 5 of 6 numbers There are 6 winning numbers and 43 losing numbers. Player chooses 5 winning numbers and 1 losing numbers. Number of ways is C(6, 5)  C(43, 1) = 258. Probability is

Lotto South Match 4 of 6 numbers Player chooses 4 winning numbers and 2 losing numbers. Number of ways is C(6, 4)  C(43, 2) = Probability is

Lotto South Match 3 of 6 numbers Player chooses 3 winning numbers and 3 losing numbers. Number of ways is C(6, 3)  C(43, 3) = Probability is

Lotto South Match 2 of 6 numbers Player chooses 2 winning numbers and 4 losing numbers. Number of ways is C(6, 2)  C(43, 4) = Probability is

Lotto South Match 1 of 6 numbers Player chooses 1 winning numbers and 5 losing numbers. Number of ways is C(6, 1)  C(43, 5) = Probability is

Lotto South Match 0 of 6 numbers Player chooses 6 losing numbers. Number of ways is C(43, 6) = Probability is

Lotto South Note also that the sum of these integers is Note also that the lottery pays out a prize only if the player matches 3 or more numbers. Match 3 – win $5. Match 4 – win $75. Match 5 – win $1000. Match 6 – win millions.

Lotto South Given that a lottery player wins a prize, what is the probability that he won the $5 prize? P(he won $5, given that he won) = P(match 3)/P(match 3, 4, 5, or 6) = / =

Example Theorem (The Vandermonde convolution): For all integers n  0 and for all integers r with 0  r  n, Proof: See p. 362, Sec. 6.6, Ex. 18.

Another Lottery In the previous lottery, the probability of winning a cash prize is Suppose that the prize for matching 2 numbers is… another lottery ticket! Then what is the probability of winning a cash prize?

Lotto South What is the average prize value of a ticket? Multiply each prize value by its probability and then add up the products: $10,000,000  = $1000  = $75  = $5  = $0  =

Lotto South The total is $0.8945, or cents (assuming that the big prize is ten million dollars). A ticket costs $1.00. How large must the grand prize be to make the average value of a ticket more than $1.00?

Another Lottery What is the average prize value if matching 2 numbers wins another lottery ticket?

Permutations of Sets with Repeated Elements Theorem: Suppose a set contains n 1 indistinguishable elements of one type, n 2 indistinguishable elements of another type, and so on, through k types, where n 1 + n 2 + … + n k = n. Then the number of (distinguishable) permutations of the n elements is n!/(n 1 !n 2 !…n k !).

Proof of Theorem Proof: Rather than consider permutations per se, consider the choices of where to put the different types of element. There are C(n, n 1 ) choices of where to place the elements of the first type.

Proof of Theorem Proof: Then there are C(n – n 1, n 2 ) choices of where to place the elements of the second type. Then there are C(n – n 1 – n 2, n 3 ) choices of where to place the elements of the third type. And so on.

Proof, continued Therefore, the total number of choices, and hence permutations, is C(n, n 1 )  C(n – n 1, n 2 )  C(n – n 1 – n 2, n 3 ) … C(n – n 1 – n 2 – … – n k – 1, n k ) = …(some algebra)… = n!/(n 1 !n 2 !…n k !).

Example How many different numbers can be formed by permuting the digits of the number ? 6!/(3!2!1!) = 720/(6  2  1) = 60.

Example How many permutations are there of the letters in the word MISSISSIPPI? How many for VIRGINIA? How many for INDIVISIBILITY?

Poker Hands Two of a kind. Two pairs. Three of a kind. Straight. Flush. Full house. Four of a kind. Straight flush. Royal flush.