Math 3790 Steven Noble November 22nd
Basic ideas of combinations and permutations Simple Addition. If there are a varieties of soup and b varieties of salad then there are a + b possible ways to order a meal of soup or salad (but not soup and salad).
Basic ideas of combinations and permutations Simple Addition. If there are a varieties of soup and b varieties of salad then there are a + b possible ways to order a meal of soup or salad (but not soup and salad). This is special case of the idea that if you have two finite sets A and B are disjoint (ie A B = ) then A B = A + B.
Basic ideas of combinations and permutations Simple Addition. If there are a varieties of soup and b varieties of salad then there are a + b possible ways to order a meal of soup or salad (but not soup and salad). This is special case of the idea that if you have two finite sets A and B are disjoint (ie A B = ) then A B = A + B. Simple Multiplication. If there are a varieties of soup and b varieties of salad, then there are a b possible ways to order a meal of soup and salad.
Basic ideas of combinations and permutations Simple Addition. If there are a varieties of soup and b varieties of salad then there are a + b possible ways to order a meal of soup or salad (but not soup and salad). This is special case of the idea that if you have two finite sets A and B are disjoint (ie A B = ) then A B = A + B. Simple Multiplication. If there are a varieties of soup and b varieties of salad, then there are a b possible ways to order a meal of soup and salad. This is special case of the idea that if you have two finite sets A and B then A B = A B. Here (A B) is just the set made up of paired elements (a, b) where a A and b B.
More basic ideas of combinations and permutations A permutation of a collection of objects is a reordering of them. For example, there are 6 different permutations of the letters ABC, namely ABC, BAC, BCA, CAB, CBA.
More basic ideas of combinations and permutations A permutation of a collection of objects is a reordering of them. For example, there are 6 different permutations of the letters ABC, namely ABC, BAC, BCA, CAB, CBA. How many ways are there to permute the letters in Hardy?
More basic ideas of combinations and permutations A permutation of a collection of objects is a reordering of them. For example, there are 6 different permutations of the letters ABC, namely ABC, BAC, BCA, CAB, CBA. How many ways are there to permute the letters in Hardy? Permutations of n things taken r at a time. The number of different 3-letter words we could make using the 9 letters of CHERNOBYL is 9 8 7 = 504. In general, the number of distinct ways of permuting r things chosen from n things is n (n 1) (n r + 1).
More basic ideas of combinations and permutations A permutation of a collection of objects is a reordering of them. For example, there are 6 different permutations of the letters ABC, namely ABC, BAC, BCA, CAB, CBA. How many ways are there to permute the letters in Hardy? Permutations of n things taken r at a time. The number of different 3-letter words we could make using the 9 letters of CHERNOBYL is 9 8 7 = 504. In general, the number of distinct ways of permuting r things chosen from n things is This equal to n! (n r)! n (n 1) (n r + 1). and is called P(n, r).
More basic ideas of combinations and permutations A permutation of a collection of objects is a reordering of them. For example, there are 6 different permutations of the letters ABC, namely ABC, BAC, BCA, CAB, CBA. How many ways are there to permute the letters in Hardy? Permutations of n things taken r at a time. The number of different 3-letter words we could make using the 9 letters of CHERNOBYL is 9 8 7 = 504. In general, the number of distinct ways of permuting r things chosen from n things is n (n 1) (n r + 1). n! This equal to (n r)! and is called P(n, r). How many ways are there to permute the letter in GAUSS?
More basic ideas of combinations and permutations A permutation of a collection of objects is a reordering of them. For example, there are 6 different permutations of the letters ABC, namely ABC, BAC, BCA, CAB, CBA. How many ways are there to permute the letters in Hardy? Permutations of n things taken r at a time. The number of different 3-letter words we could make using the 9 letters of CHERNOBYL is 9 8 7 = 504. In general, the number of distinct ways of permuting r things chosen from n things is n (n 1) (n r + 1). n! This equal to (n r)! and is called P(n, r). How many ways are there to permute the letter in GAUSS? The answer isn t 6! but why isn t it?
Mississippi Formula The formula for the amount of permutations of a word is if there are a 1 of the first letter, a 2 of the second letter,..., a n of the n th letter then there are (a 1 + a 2 + a 3 + + a n )! a 1!a 2! a n! different permutations of the word. This is often referred to as the Mississippi formula. (How many permutations are there of the word Mississippi?) If this is new to you I recommend reading the section titled Combinatorial Arguments on page 208.
Example 1 Imagine a piece of graph paper. Starting at the origin draw a path to the point (10, 10) that stays on the grid lines (which are one unit apart) and has a total length of 20. For example, one path is to go from (0, 0) to (0, 7) to (4, 7) to (4, 10) to (10, 10). Another path goes from (0, 0) to (10, 0) to (10, 10). How many possible different paths are there?
Solution Each path can be completely described by a 20-letter sequence of U s and R s, where U means move one unit up and R means move one unit to the right. For example, the path that goes from (0, 0) to (0, 7) to (4, 7) to (4, 10) to (10, 10) would be described by the string UUUUUUURRRRUUURRRRRR. So each path is just a permutation of that word.
Solution Each path can be completely described by a 20-letter sequence of U s and R s, where U means move one unit up and R means move one unit to the right. For example, the path that goes from (0, 0) to (0, 7) to (4, 7) to (4, 10) to (10, 10) would be described by the string UUUUUUURRRRUUURRRRRR. So each path is just a permutation of that word. Using the Mississippi formula we have that the answer is 20! 10!10!.
Example 2 How many different ordered triples (a, b, c) of non-negative integers are there such that a + b + c = 50?
Solution 1 First we note that if a + b + c = 50 then 0 c 50. So we have solutions where c = 0 or c = 1 or,..., or c = 50. So we want the sum of solutions for each case. So how many values are there for a and b if c = n. In this case a + b = 50 n so 0 a 50 n and for each a there is exactly 1 b where b = 50 n a.
Solution 1 First we note that if a + b + c = 50 then 0 c 50. So we have solutions where c = 0 or c = 1 or,..., or c = 50. So we want the sum of solutions for each case. So how many values are there for a and b if c = n. In this case a + b = 50 n so 0 a 50 n and for each a there is exactly 1 b where b = 50 n a. So let s say that C i = {(a, b) : a + b + i = 50} and we have found that C i = 50 i + 1 (the amount of values that a can be). So the amount of solutions are C 0 + C 1 +... + C 50 = 51 + 50 +... + 1 = (51 + 1) + (50 + 2) + (49 + 3) +... + (27 + 25) + 26 52 51 = 2
Solution 2 Before we solve for 50 let s try a smaller number like 11. Recall how we first learned about addition in first or second grade. You practiced sums like 3 + 6 + 2 = 11 by drawing sequences of dots: + + =. Each ordered triple (a, b, c) that add to 11 can be thought of as such a string of 11, s and 2 + s. So all we have to count is the amount of strings like this, ie the permutations.
Solution 2 Before we solve for 50 let s try a smaller number like 11. Recall how we first learned about addition in first or second grade. You practiced sums like 3 + 6 + 2 = 11 by drawing sequences of dots: + + =. Each ordered triple (a, b, c) that add to 11 can be thought of as such a string of 11, s and 2 + s. So all we have to count is the amount of strings like this, ie the permutations. By using the Mississippi formula we get the amount of permutations are 13! 11!2! = ( ) 13. 2
Solution 2 Before we solve for 50 let s try a smaller number like 11. Recall how we first learned about addition in first or second grade. You practiced sums like 3 + 6 + 2 = 11 by drawing sequences of dots: + + =. Each ordered triple (a, b, c) that add to 11 can be thought of as such a string of 11, s and 2 + s. So all we have to count is the amount of strings like this, ie the permutations. By using the Mississippi formula we get the amount of permutations are 13! 11!2! = ( ) 13. 2 So returning to the actual problem at hand we have a string that is 52 characters long and has 50 s. So we have the number of permutations is ( 52 2 ).
The Balls in Urns Formula This last way of solving a problem is a useful general tool which we call the balls in urns formula which is simply stated The number of different ways we can place b indistinguishable balls into u distinguishable urns is ( ) b+u 1 b or in other words the amount of permutations of a string with b s and u 1 s.
Example 3 Let us say that 2 ten digit numbers are sisters if they can be obtained from one another by reordering the digits. And further let us call a set that is as large as possible whose elements are all sisters a sorority. For example, 1, 111, 233, 999 and 9, 929, 313, 111 are sisters who belong to a sorority with 10! 4!3!2!, since the membership of the sorority is just the number of ways of permuting the digits. (we will allow in numbers with leading zeroes into these sororities). The sororities have vastly different sizes. The most exclusive sororities have 1 member (for example, the sorority consisting entirely of 6, 666, 666, 666) yet one sorority has 10! members (the one containing 1, 234, 567, 890). How many sororities are there among the 10 digit numbers?
Solution We note that each sorority is uniquely determined by its collection of 10 digits repetition is allowed. For example one sorority can be named ten 6 s, while another is called three 4 s, one 7, two 8 s, and four 9 s.
Solution We note that each sorority is uniquely determined by its collection of 10 digits repetition is allowed. For example one sorority can be named ten 6 s, while another is called three 4 s, one 7, two 8 s, and four 9 s. So really each sorority is determined by how many of each digit is has; or in other how many balls it has in each digit urn.
Solution We note that each sorority is uniquely determined by its collection of 10 digits repetition is allowed. For example one sorority can be named ten 6 s, while another is called three 4 s, one 7, two 8 s, and four 9 s. So really each sorority is determined by how many of each digit is has; or in other how many balls it has in each digit urn. So since there are 10 urns (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) and 10 balls we have that the amount of sororities is ( 19 9 ).
Sample problem 1 A (female) contestant on a game show is shown three doors by the (male) host. Behind one of these there is a car, and behind each of the other two there is a goat. She chooses one of the doors, hoping of course to get the car. Before the door is opened, the host (who knows what s behind each door) opens one of the remaining two doors to reveal a goat. At this point he offers her the chance to switch her choice if she so wishes. Should she switch, and if so, how does that change the probability of winning a car?
Sample problem 2 You are given one true or false question on a test. If you get the question right, your final mark is increased by one percent, but if you get the question wrong, your final mark is decreased by one percent. You can guess, or you can copy from your neighbour, but you know from experience that you you neighbour answers incorrectly 20% of the time. Besides, there is a 10% chance that the teacher would catch you cheating, and in this case, she would deduct five percent from your final mark. Moral issues aside what should you do: copy from your neighbour, or randomly guess on the question?
Sample problem 3 Two players alternately shoot themselves with a six-shooter, only one chamber of which contains a bullet. (This is called Russian Roulette). You have the first shot, so you decide the rules. Either both players take turns shooting the next chamber, or both players randomly spin the chamber before shooting. Naturally, you want to maximize your chances of surviving. Should you spin first and shoot, or shoot without spinning?
Sample problem 4 Alison, Bernon, and Chantel play the following game. They take turns (in the order A, B, C, A, B, C,...) rolling one die. Alison wins if she rolls 1, 2, or 3 on her turn. Bernon wins if he rolls 4 or 5 on his turn. Chantel wins if she rolls 6 on her turn. They keep repeating this until there is a winner. What is the probability that Alison wins the game? For example if Alison rolls a 5 then it is Bernon s turn. Say her rolls a 6. Then it s Chantel s turn. Say she rolls 6. Then Chantel wins.