Understanding Combinations: Counting Subsets of a Set

Counting Subsets of a Set: Combinations Lecture 28 Section 6.4 Thu, Mar 30, 2006

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) = 499500.

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 (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 • General 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 (k + 1)!/(0!(k + 1)!) = 1. • If r = k + 1, then there is only one k-combination, the entire set, and (k + 1)!/((k + 1)!0!) = 1.

Counting r-Combinations • So let r be any number between 0 and k + 1 (0 < r < k + 1). • Select an arbitrary element a from the set. • 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 • r-combinations for which a is not a member: • The r elements come from the remaining k elements. • Therefore, by the inductive hypothesis, there are k!/(r!(k – r)!) such sets. • r-combinations for which a is a member: • The other r – 1 elements in the subset come from the k remaining elements in the set. • By the inductive hypothesis, there are k!/((r – 1)!(k – (r – 1))!) such sets.

Counting r-Combinations • Therefore, the number of r-combinations of k + 1 elements is • A little algebra shows that this equals • Thus, the statement is true when n = k + 1, etc.

Example: Counting r-Combinations • In Math 121, I used to collect 48 daily homework assignments. • Some assignments count more than others. • I drop the “lowest” 4 homework grades. • Which should be dropped: 0 out of 30 or 15 out of 40? • I drop the 4 grades that hurt the student’s average the most.

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.

Example: Counting r-Combinations • What if I dropped 6 grades? • There would be C(48, 6) = 12,271,512 possible choices. • Over 60 times as many. • This would require about 60 secs = 1 min.

Example: Counting r-Combinations • What if I dropped 12 grades? • There would be C(48, 12) = over 69 billion choices! • More than 350,000 as many! • This would require almost 350,000 sec = over 4 days.

Example: Counting r-Combinations • What if I dropped 44 grades!!!??? • This must involve unimaginably many possibilities! • How many would it be?

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 • There are C(49, 6) = 13,983,816 possible choices. • Match all 6 numbers • There is only 1 winning combination. • Probability of winning is 1/13983816 = 0.00000007151.

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 0.00001845.

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) = 13545. • Probability is 0.0009686.

Lotto South • Match 0 of 6 numbers • Player chooses 6 losing numbers. • Number of ways is C(43, 6) = 2760681. • Probability is 0.4360.

Lotto South • Note also that the sum of these integers is 13983816. • 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) = 0.01765/0.01864 = 0.9469.

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 0.018637545. • 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  0.00000007151 = 0.7151 • $1000  0.00001845 = 0.0185 • $75  0.0009686 = 0.0726 • $5  0.01765 = 0.0883 • $0  0.9814 = 0.0000

Lotto South • The total is $0.8945, or 89.45 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 n1 indistinguishable elements of one type, n2 indistinguishable elements of another type, and so on, through k types, where n1 + n2 + … + nk = n. Then the number of (distinguishable) permutations of the n elements is n!/(n1!n2!…nk!).

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, n1) choices of where to place the elements of the first type.

Proof of Theorem • Proof: • Then there are C(n – n1, n2) choices of where to place the elements of the second type. • Then there are C(n – n1 – n2, n3) 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, n1) C(n – n1, n2) C(n – n1 – n2, n3) … C(n – n1 – n2 – … – nk – 1, nk) = …(some algebra)… = n!/(n1!n2!…nk!).

Example • How many different numbers can be formed by permuting the digits of the number 444556? • 6!/(3!2!1!) = 720/(6  2  1) = 60.

Example • How many permutations are there of the letters in the word MISSISSIPPI? • 11!/(4!4!2!1!) = 34650. • How many for VIRGINIA? • How many for INDIVISIBILITY?

Understanding Combinations: Counting Subsets of a Set