1. Counting Subsets of a Set: Combinations Lecture 28 Section 6.4 Wed, Mar 2, 2005

2. 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}.

3. Counting r-Combinations Theorem: The number of r-combinations of a set of n elements is C(n, r) = n!/[r!(n – r)!]. 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.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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.

9. 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.

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

11. Example: Lotto South In Lotto South (Loot the 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

12. Example: Virginia Lottery 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 (it ain’t gonna happen)

13. Example: Virginia Lottery 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 (forget about it).

14. Example: Virginia Lottery 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 (nope).

15. Example: Virginia Lottery Match 3 of 6 numbers Player chooses 3 winning numbers and 3 losing numbers. Number of ways is C(6, 3)  C(43, 3) = 246820. Probability is 0.01765 (it just might happen.)

16. Example: Virginia Lottery Match 2 of 6 numbers Player chooses 2 winning numbers and 4 losing numbers. Number of ways is C(6, 2)  C(43, 4) = 1851150. Probability is 0.1324 (it will happen).

17. Example: Virginia Lottery Match 1 of 6 numbers Player chooses 1 winning numbers and 5 losing numbers. Number of ways is C(6, 1)  C(43, 5) = 3011652. Probability is 0.4130 (it will happen).

18. Example: Virginia Lottery Match 0 of 6 numbers Player chooses 6 losing numbers. Number of ways is C(43, 6) = 2760681. Probability is 0.4360 (it will happen).

19. Example: Virginia Lottery 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.

20. Example: Virginia Lottery 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.

21. 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.

22. 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 !).

23. 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.

24. 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.

25. 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 !).

26. 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.

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