1. Section 4.3: Permutations and Combinations Def: A permutation of a set of distinct objects is an ordered arrangement of these objects. Ex: {a, b, c, d} is a set of 4 distinct objects. Here are some permutations of this set: abcd, dcba, abdc, cadb, etc. Def: An r-permutation of a set of n distinct elements is an ordered arrangement of r of the n objects. Ex: {a, b, c, d} is a set of 4 distinct objects. cad is a 3-permutation of this set. b is a 1-permutation of this set. This set has no 5-permutations.

2. Notation: We use P(n, r) to denote the number of r-permutations of a set with n distinct elements. Remark: Note that the number of permutations of a set with n elements is P(n, n) = n!. Theorem: Let n, r  N with r  n. Then P(n, r) = n(n – 1)(n – 2) … (n – (r – 1)) = n!/(n – r)! Proof: Let n, r  N with r  n. This is a simple application of the product rule. To choose an r-permutation of a set with n elements, we must choose a first element (there are n to choose from) then we must choose a second element (there are n – 1 left to choose from, …, and finally we must choose an rth element (there are n – (r – 1) left to choose from).  Ex: There are P(5, 5) = 5! = 120 permutations of a set with 5 elements.

3. Ex: 100 people enter a contest. How many different ways are there to choose a first, second, and third place winner. Since abc must be in the permutation, we can think of it as one entity. Namely, we reduce the problem to finding the number of permutations of the set {abc, d, e, f, g, h}. This is P(6, 6) = 6! = 720. This is the number of 3-permutations of a set with 100 elements. That is, P(100, 3) = 100!/(100 – 3)! = 100*99*98 = 970200. So there are 970200 different ways to select winners. Ex: How many permutations of the set {a, b, c, d, e, f, g, h} contain the string abc? Sometimes we don’t care about the order that elements are chosen. We only care about the number of different ways to choose r things from a set with n elements. [Choose a subset].

4. Def: An r-combination of a set of n distinct objects is an unordered arrangement of r of the n objects. In other words, an r-combination of a set with n distinct elements is a subset with of cardinality r. Ex: {a, b, c, d} is a set of 4 distinct objects. Here are some 3- combinations of this set: {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}. In fact these are all of the 3-combinations of this set. Notation: C(n, r) denotes the number of r-combinations of a set with n distinct elements. This is also denoted ( n r ). Theorem: Let n, r  N with r  n. Then C(n, r) = n!/[r!(n – r)!]. Proof: Note that P(n, r) = C(n, r)*P(r, r). So C(n, r) = P(n, r)/P(r, r) = [n!/(n – r)!]/[r!/(r – r)!] = n!/[r!(n – r)!]. 

5. Theorem: Let n, r  N with r  n. Then C(n, r) = C(n, n – r). Proof1: C(n, r) = n!/[r!(n – r)!] by the previous theorem. C(n, n – r) = n!/[(n – r)!(n – (n – r)!] = n!/[(n – r)!r!] as well.  Proof2: Let S be a set with n elements. Then C(n, r) is the number of subsets of this set with r elements. C(n, r) is also the number of ways to choose r elements from a set with n elements. Now C(n, n – r) is the number of ways to choose n – r elements from a set with n elements. We could consider this as a way of constructing a subset with r elements. The n – r elements we choose are the ones we are excluding. So C(n, r) = C(n, n – r) since they are both represent the number of ways to choose a subset of r elements from a set with n elements. 

6. Ex: How many ways are there to select five players from a 10 member tennis team to play in a match? This is C(10, 5) = 10!/[5!5!] = (10*9*8*7*6)/(5*4*3*2*1) = 252. Ex: What if the players in the above example have to be seeded as 1, 2, 3, 4, and 5 to play the matches? Now we care about the order we choose the 5 players in. So this is P(10, 5) = 10!/5! = 10*9*8*7*6 = 30240 Ex: How many bit strings of length n contain exactly r 1s? We must choose r of the n bit positions to put the 1s in. So this is C(n, r). Ex: So there are C(5, 3) = 5!/[3!2!] = (5*4)/2 = 10 bit strings of length 5 that contain exactly three 1s.

7. Reminders: –Test3 is scheduled on Friday, April 25 (in 3 weeks) Email was sent that included a proposed schedule –HW 8 is due Wednesday, April 9 in class Return Test 2: –Avg: 70, Median: 74 –http://www.cs.virginia.edu/~cmt5n/cs202/test2/test2key.dochttp://www.cs.virginia.edu/~cmt5n/cs202/test2/test2key.doc –< 50 scores are headed in the direction of failing –Be sure to review material that gave you trouble –Lower right hand corner has current letter grade –Turn in Homework assignments on time 25% of 20 is 5 full points on final grade Equivalent to 33 pts on a test, 14 pts on final