Multiplication Principle and Addition Principle

1. Multiplication Principle and Addition Principle Multiplication Principle: Suppose a task is accomplished by n steps and each step requires a choice from a number of available choices. Let these numbers be A1, A2, . . . An Then, the total number of ways to accomplish this task is A1 x A2 x A3 x . . . . An. • Addition Principle:Suppose a task is • accomplished by choosing an object from the union of disjoint sets with cardinalities B1, B2, B3, . .. Bn. Then, the total number of ways to accomplish this task is • B1+ B2+ B3 + . . . . . Bn.

4. Set Theoretic Descriptionsof the two principles • The cardinality of the Cartesian product of n sets A1, A2, . . An is given by n (A1 x A2 x . .. An) = n(A1). n(A2) . . .n(An). • The cardinality of the disjoint union of n sets B1, B2, . . . . Bn is given by n (B1 U B2 U . . . Bn)= n(B1) + n(B2)+ . . . + n(Bn)

5. Examples • On a table, there is a pile of 10 apples and another pile of 15 pears. 1. How many choices do you have if you are instructed to pick an apple and a pear? 2. How many choices do you have if you are instructed to pick an apple or a pear?

6. Mistakes you must avoid • Suppose there are 4 doctors and 5 chess players in a room with 7 people. If you are instructed to pick a doctor or a chess player in this room (and slap the face) how many choices do you have? Is your answer 4+5 = 9 ?Then, review your knowledge of addition principle.

7. Read the problems carefully. In a room, there are 7 people. Everyone in that room is either a chess player or a doctor. 4 of them are doctors and 5 of them are chess players. In how many ways can one accomplish each of the following tasks? 1. Task is to get into the room, pick a chess player and play a chess with. Then, pick a doctor and ask for an opinion about your asthma symptom. 2. Task is to pick a team of two people, consisting of one chess player and one doctor.

8. Addition Principle is frequently applied in a form of `subtraction’ principle. n(A \ B) = n(A) – n(B) Example: In a basket of 30 Easter eggs, there are 7 green eggs. Your task is to pick an egg that is not green. How many choices do you have? The answer: 30-7= 23. 23 choices. Another Example: (See the Venn diagram of chess players and doctors.) We will approach the second question in the previous slide in the following way. Initially, we write a name of a chess player and (next to it) write a name of a doctor. Consider this writing as a tentative list of 2-team members. There are 5x4=20 possible ways of writing a pair of names in this way. But, exactly 2 of them are a pair of names of a same person (which is not permitted). So, the answer to the second question ( of forming a team of chess_player-doctor pair) is 20-2=18.

9. Multiplication Principle is frequently applied in the form of Division Principle. • Division Principle: If A=B x C (cartesion product of B and C), then n(B) = n(A) / n(C). EXAMPLE 1: Gregor Samsagoes to walk every morning wearing a hat and carrying a cane. He has 2 hats, which are identical except that one is grey and the other is brown. He has 3 canes. He is color blind. In how many ways, can he choose a hat and a cane to go out for a walk based on his discerning ability? Answer: If he were able to tell brown from grey, there would be 2 x 3= 6 ways. But, now that all the hats are identical to him from his view, there are 6 / 2 = 3 ways.

10. Division Principle re-stated: Suppose a task A consists of accomplishing a task B followed by a task C. Then, the number of ways to accomplish task B is given by the number of ways to accomplish the task A divided by the number of ways to accomplish task C. EXAMPLE 2. How many different words can be formed by rearranging the letters in the word “ELEMENT”? Answer: 7! / 3! ( Explain. . . . .)

11. Permutation • Informally, a permutation on a set of n elements is an ordering of these elements. • Formally, a permutation on a set of n elements is a one-to-one correspondence between the set and itself. • Number of permutations on a set of n elements • Generalized Permutation: Let A be a set of n elements and r< n. Any one-to-one function from the set {1,2,3, . .. . r} into the set A is called a generalized permutation of the k elements on the set A. • Number of generalized permutation of r elements on the set of n elements is denoted by nPr. • nPr x (n-r)! = n! by multiplication principle.

12. Combination • Let r<n. An r-combination on the set of n elements is a subset with cardinality r. • nCr denotes the number of all the r-combinations on the set of n elements. • nPr and nCr are related by the following equation nPr = nCr x r! (illustrated below)

15. Picking a generalized permutation, a one-to-one function from {1,2, . . r} into The set A of n elements is equivalent to STEP1: Picking a subset of A that has r elements. STEP 2: Picking a permutation on these r elements. Therefore, n P r = n C r xr !

16. Summary • n! = number of permutations on n objects. • nPr = number of r-permutations on n objects = • nCr = number of r-sets that can be formed from n objects = nPr / r! =

17. Partition • Let A be a set of n elements. If, for some subsets A1, A2, . . . , Ak that are pairwise disjoint, A= A1 U A2 U . . . U Ak then we say {A1, A2, . . . Ak} is a partition of A.

18. Ordered Partition • Let a1+a2+ . . . ak= n and a1< a2< . . ak ( < here means less than or equal to) The sequence of subsets A1, A2, . . . . Ak with cardinalities a1, a2, . . . . . ak is called an ordered partition if these sets are mutually disjoint and their union is the set A.

19. Number of ordered partitions • How many ordered partitions of a given type does a set of n element have? • Experiment: Consider a set of 7 elements. How many ordered partitions of type (2,2,3) are there? The construction of a partition of a given type can be considered as a multi-step task. The number of these steps is 1 less than the number of subsets that form the partition. In this particular experiment, 2 steps. STEP 1: Pick a subset of 2 elements (from 7 elements) STEP 2: Pick a subset of 2 elements (from the remaining 5 elements) Now, we employ the multiplication principle: 7 C 2 x 5 C 2 =

20. Counting Ordered Partitions of a Given Type Let a1+a2+ . . . ak = n. The number of type (a1, a2, . . . . ak) partitions of a set of n elements is given by

21. Application (of partition counting) • How many different words can be formed by rearranging the letters in the word “ MISSISSIPI” ? Discussion: Compare this problem with the following problem: `` How many ordered partitions of the type (1,1,4,4) are there for a set of 10 elements?”. How are these two problems related?

22. Given a 10 element set, imagine these elements are doors along a hall way of a hotel. Imagine your duty is to write letter M on 1 door letter P on 1 door letter S on 4 doors letter I on 4 doors This job is equivalent to forming a word by an arrangement of letters in “MISSISSIPI”. This job is also equivalent to forming a type (1,1,4,4) ordered partition of the 10 element set.