AoPS:

AoPS: Introduction to Probability and Counting

Chapter 3 Correcting for Overcounting

Permutations with Repeated Elements Let’s start with a problem we should (hopefully) already know how to do. Problem 3.1: How many possible distinct arrangements are there of the letters in the word DOG?

Problem 3.1 How many possible distinct arrangements are there of the letters in the word DOG? We could list them or we should know by now that there are 3 ways to pick the 1st, 2 ways to pick the 2nd, and 1 way to pick the last letter, for a total of 3! = 6 ways. This is just a basic permutation problem like in Chapter 1.

Problem 3.2 How many possible distinct arrangements are there of the letters in the word BALL?

Solution 3.2 How many possible distinct arrangements are there of the letters in the word BALL? The Bogus Solution: 4! possibilities. That’s because BAL1L2 and BAL2L1 are the same once the subscripts are removed, so we have over-counted and need to correct this.

Solution 3.2 How many possible distinct arrangements are there of the letters in the word BALL? The Bogus Solution is 4! Possibilities. That’s because BAL1L2 and BAL2L1 are the same once the subscripts are removed, so we have over-counted and need to correct this. So we have to divide the # of arrangements by 2! because the L’s can be arranged 2 different ways: 4! / 2! = 12

Problem 3.3 How many distinct arrangements are there of TATTER?

Solution 3.3 TATTER: pretend that all the T’s are different (T1, T2, and T3). Then are are 6! possibilities. The 3 T’s can be arranged in 3! different ways, which leads to overcounting. The solution: 6! / 3! = 120. This is called strategic overcounting.

Problem 3.4 One more twist: How many distinct arrangements are there of PAPA?

Solution 3.4 One more twist: How many distinct arrangements are there of PAPA? Simple: because the P’s repeat twice and the A’s repeat twice, the answer is 4! / (2! X 2!) = 6 ways

Exercise 3.1 Compute the # of distinct arrangements of the letters in the word EDGE.

Solution 3.1 Compute the # of distinct arrangements of the letters in the word EDGE. 4! / 2! = 12 ways

Exercise 3.2 For each of the following words, determine the # of ways to arrange the letters of the word. • WAR • THAT • CEASE • ALABAMA • MISSISSIPPI

Solution 3.2 For each of the following words, determine the # of ways to arrange the letters of the word. • WAR = 3! = 6 • THAT = 4! / 2! = 12 • CEASE = 5! / 2! = 60 • ALABAMA = 7! / 4! = 210 • MISSISSIPPI = 11! / (4! X 4! X 2!) = 34,650

Problem 3.3 I have 5 books, two of which are identical copies of the same math book (and all of the rest of the books are different.) In how many ways can I arrange them on the shelf?

Solution 3.3 I have 5 books, two of which are identical copies of the same math book (and all of the rest of the books are different.) In how many ways can I arrange them on the shelf? 5! / 2! = 60

Problem 3.4 Thereare 8 pens along a wall in the pound. The pound has to allocate 4 pens to dogs, 3 to cats, and one to roosters. In how many ways can the pound make the allocation?

Solution 3.4 Thereare 8 pens along a wall in the pound. The pound has to allocate 4 pens to dogs, 3 to cats, and one to roosters. In how many ways can the pound make the allocation? 8! / (4! X 3!) = 280

Counting Pairs of Items A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during the an 8- person round-robin tournament?

Counting Pairs of Items A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during the an 8- person round-robin tournament? Bogus Solution: Each of the 8 players plays 7 games, or 8 x 7 = 56 total games played.

Counting Pairs of Items A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during the an 8- person round-robin tournament? Alice plays Bob and Bob plays Alice, but it’s the same game, so 8 x 7 = 28. 2

Counting Pairs of Items Another way to look at the problem: Alice plays 7 matches. Bob also plays 7 matches, but the one with Alice has already been counted, leaving Bob with 6 more to play. The next player, Carol, has already played A & B, so Carol has 5 more matches to play, and so on down the line, to the last player.

Counting Pairs of Items Another way to look at the problem: Alice plays 7 matches. Bob also plays 7 matches, but the one with Alice has already been counted, leaving Bob with 6 more to play. The next player, Carol, has already played A & B, so Carol has 5 more matches to play, and so on down the line, to the last player. That means we have a total of 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 games total played

Counting Pairs of Items That means we have a total of 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 games total played. This is a classic example of counting pairs of objects.

Problem 3.5 • Compute the sum 1 + 2 + 3 + 4 of the 1st 4 positive integers. • Compute the sum of 1 + 2 + 3 + 4 + 5 of the 1st 5 positive integers. • Compute the sum of 1 + 2 + 3 + … + 9 + 10 of the 1st 10 positive integers. • Find a formula for the sum of the 1stn positive integers. • Compute the sum of 1 + 2 + 3 + … + 100 of the 1st 100 positive integers.

Solution 3.5 • Compute the sum 1 + 2 + 3 + 4 of the 1st 4 positive integers. 10 • Compute the sum of 1 + 2 + 3 + 4 + 5 of the 1st 5 positive integers. 15 • Compute the sum of 1 + 2 + 3 + … + 9 + 10 of the 1st 10 positive integers. 55 • Find a formula for the sum of the 1stn positive integers. [n (n + 1)] / 2 • Compute the sum of 1 + 2 + 3 + … + 100 of the 1st 100 positive integers. 5050

Problem 3.6 A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during a n-person round-robin tennis tournament, where n > 2 is a positive integer?

Solution 3.6 A round-robin tennis tournament consists of each player playing every other player exactly once. How many matches will be held during a n-person round-robin tennis tournament, where n > 2 is a positive integer? Each of the n players must play every other player, so each player must play n – 1 matches which gives the preliminary count n (n – 1) matches.

Solution 3.6 Each of the n players must play every other player, so each player must play n – 1 matches which gives the preliminary count n (n – 1) matches. But this counts each match twice, so divide by 2 to get n (n – 1) 2

Problem 3.7 A convex polygon is a polygon in which every interior angles is less than 180°. A diagonal of a convex polygon is a line segment which connects 2 non-adjacent vertices. Find a formula for the # of diagonals of a convex polygon with n sides, where n is any positive integer greater than 2.

Solution 3.7 A polygon with n sides has n vertices. A diagonal corresponds to a pair of vertices. By similar reasoning to Prob. 3.6, there are n (n – 1) pairs 2 of vertices.

Solution 3.7 A polygon with n sides has n vertices. A diagonal corresponds to a pair of vertices. By similar reasoning to Prob. 3.6, there are n (n – 1) pairs 2 of vertices. However n of these pairs correspond to edges of the polygons rather than diagonals, so subtract these from the count: the # of diagonals n (n – 1) - n 2

Solution 3.7 Simplifying this expression n (n – 1) - n 2 leads to the following: n (n – 3) 2 If you’re interested in the math behind this transformation, I can show it to you; all you have to do is ask!

Exercise 3.3.1 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents?

Solution 3.3.1 A club has 15 members and needs to choose 2 members to be co-presidents. In how many ways can the club choose its co-presidents? If the co-president positions are unique, there are 15 choices for the 1st and 14 choices for the 2nd. However, since the positions are identical, we must divide by 2! (the # of arrangements of co- presidents) = (15 x 14) / 2! = 105ways.

Exercise 3.3.2 I have twenty balls numbered 1 through 20 in a bin. In how many ways can I select 2 balls is the order in which I draw them doesn’t matter?

Solution 3.3.2 I have twenty balls numbered 1 through 20 in a bin. In how many ways can I select 2 balls is the order in which I draw them doesn’t matter? This is like the previous problem, so we get (20 x 19) / 2! = 190

Exercise 3.3.3 A sports conference has 14 teams in two divisions of 7. How many games are in a complete season for the conference if each team must play every other team in its own division twice and every team in the other division once?

Solution 3.3.3 A sports conference has 14 teams in two divisions of 7. How many games are in a complete season for the conference if each team must play every other team in its own division twice and every team in the other division once? Each team plays 6 other teams in its division twice and the other 7 teams once, for a total of 6 x 2 + 7 = 19 games for each team.

Solution 3.3.3 Each team plays 6 other teams in its division twice and the other 7 teams once, for a total of 6 x 2 + 7 = 19 games for each team. There are 14 teams total, which gives a preliminary count of 19 x 14 = 266 games, but we must divide by 2 because we counted each game twice. The answer is (19 x 14) / 2 = 133 games.

Exercise 3.3.4 • Find a formula for the sum of the 1stn even integers: 2 + 4 + 6 + … + 2n. • Find a formula for the sum of the 1stn odd integers: 1 + 3 + 5 + … + (2n – 1).

Solution 3.3.4 • Find a formula for the sum of the 1stn even integers: 2 + 4 + 6 + … + 2n. Let S = 2 + 4 + 6 + … + 2n. ( S had n terms.) Another expression for S is 2n + …+ 6 +4 + 2, so adding these together we obtain the equation 2S = (2n + 2) + (2n + 2) + … + (2n + 2) .

Solution 3.3.4 • Find a formula for the sum of the 1stn even integers: 2 + 4 + 6 + … + 2n. Let S = 2 + 4 + 6 + … + 2n. ( S had n terms.) Another expression for S is 2n + …+ 6 +4 + 2, so adding these together we obtain the equation 2S = (2n + 2) + (2n + 2) + … + (2n + 2) . This sum has n terms, so 2S = n (2n + 2) = 2n (n + 1), and S = n (n+ 1)

Solution 3.3.4 Let S = 2 + 4 + 6 + … + 2n. ( S had n terms.) Another expression for S is 2n + …+ 6 +4 + 2, so adding these together we obtain the equation 2S = (2n + 2) + (2n + 2) + … + (2n + 2) . This sum has n terms, so 2S = n (2n + 2) = 2n (n + 1), and S = n (n+ 1) OR, S = 2(1 + 2 + 3 +…+n) = 2(n(n + 1)) = 2 n (n + 1).

Solution 3.3.4 (b) Find a formula for the sum of the 1stn odd integers: 1 + 3 + 5 + … + (2n – 1). Let S = 1 + 3 + 5 + … + (2n – 1). (S has n terms.) Another expression for S = (2n – 1) + … + 3 + 1, so adding these together we get the equation 2S = 2n + 2n + … + 2n.

Solution 3.3.4 (b) Find a formula for the sum of the 1stn odd integers: 1 + 3 + 5 + … + (2n – 1). Let S = 1 + 3 + 5 + … + (2n – 1). (S has n terms.) Another expression for S = (2n – 1) + … + 3 + 1, so adding these together we get the equation 2S = 2n + 2n + … + 2n. This sum has n terms, so 2S = n (2n) = 2n2 so S = n2.

Exercise 3.3.5 How many interior diagonals does an icosahedron have? ( An icosahedron is a 3-dimentional figure with 20 triangular faces and 12 vertices, with 5 faces meeting at each vertex. An interior diagonal is a segment connecting two vertices which do not lie on a common face.)

Solution 3.3.5 How many interior diagonals does an icosahedron have? There are 12 vertices in the icosahedron, so there are potentially 11 other vertices to which we could extend a diagonal. However 5 of these 11 points are connected to the original point by an edge, so they are not connected by internal diagonals.

Solution 3.3.5 There are 12 vertices in the icosahedron, so there are potentially 11 other vertices to which we could extend a diagonal. However 5 of these 11 points are connected to the original point by an edge, so they are not connected by internal diagonals. So each vertex is connected to 6 other points by interior diagonals. This gives the preliminary count of 12 x 6 = 72 interior diagonals. However, they were counted twice, so divide by 2 = 36 diag.

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