Combinations and Probabilities

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1 Combinations and Probabilities Tutor: Zhang Qi Systems Engineering and Engineering Management November 2014 Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

2 Combination Review Theorem Let x 1, x 2,...x m be distinct elements, where x i occurs k i times. The number of different ( sequences that ) can be obtained by permuting these k1 + k k m elements is k 1 k 2... k m Theorem (Multinomial Theorem) For any x 1, x 2,...x m and some integer n 1 ( ) (x 1 + x x m ) n n = x k 1 k 1 k 2... k m k k m=n 1 x k 2 2 and particularly, it turns into Binomial Theorem when m = 2....x km m Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

3 Combination Application Examples: Coefficients of expansion 1. What is the( coefficient ) of x 12 y 13 in the expansion of (2x 3y) 25? 25 Solution: 2 12 ( 3) Hint: take j = 13 for (2x + ( 3y)) 25 = ( ) j=0 (2x) 25 j ( 3y) j j 2. What is the coefficient of x1 3x 2x3 2 in the expansion of (2x 1 3x 2 ( + 5x 3 ) 6? ) 6 Solution: 2 3 ( 3) Hint: take k 1 = 3, k 2 = 1, k 3 = 2 for (2x 1 3x 2 + 5x 3 ) 6 = ( ) 6 k 1 +k 2 +k 3 =6 x k 1 1 k 1 k 2 k x k 2 2 x k Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

4 Combination Application Examples: Simplify Sums Method 1. By induction. Method 2. Construct combinatorial models. Method 3. Apply multinomial theorem. Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

5 Constructing combinatorial models. Example Prove ( ) n n k=0 k2 = n(n + 1)2 n 2 k Hint: Count in two ways of selection from a set of n elements. Then pick two not necessarily distinct elements from this subset. Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

6 Example Cont Prove ( ) n n k=0 k2 = n(n + 1)2 n 2 k Solution: The left-hand side ( ) corresponds to for all k, choosing k elements n from the n elements thus possible subsets. And at the same time, k from each subset with k elements, we pick at random 2 elements and these 2 elements needn t to be distinct, followed by k 2 ways. For the right-hand side, we consider picking the 2 elements first. We split the problem into two cases. If the 2 elements are distinct, there are n(n-1) ways. Then the k subset needs k-2 other elements from the whole set, that is 2 n 2. On the other hand, if the 2 elements are identical, that is n ways. Then the k subset needs k-1 other elements from the whole set, which has 2 n 1 ways. Summing up the two cases we get n(n 1)2 n 2 + n2 n 1 = n(n + 1)2 n 2. Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

7 Applying multinomial theorem. 1. ( ) n n k=0 = 2 n Hint: (1 + 1) n k 2. ( ) n n k=0 ( 1)k = 0 Hint: ( 1 + 1) n k Be careful with the index! Example 17 k=1 ( ) ( 1) k 13 k = k k=1 17 = k=0 ( ) 17 ( 13) k 1 17 k k ( ) 17 ( 13) k 1 17 k k = ( ) 17 1 = ( 12) 17 1 ( ) 17 ( 13) 0 (1) 17 0 Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

8 Example Prove n ( n 2 0 ( 1) k) k ( ) if n is odd = 2m ( 1) m if n=2m k=0 m Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

9 Solution: Let S n = n k=0 ( 1)k ( n k) 2. When n is odd, S n = Thus S n = 0 when n is odd. = = k=0 k=0 ) 2 n ( n ( 1) k+1 k n ( n ( 1) n k n k n ( ) n 2 ( 1) i i i=0 = S n ) 2 Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

10 Solution: When n is even, say n = 2m, we observe that S n is actually the coefficient of x n in (1 + x) n (1 x) n, as S n = = k=0 k=0 ) 2 n ( n ( 1) k k n ( )( ) n n ( 1) k 1 n k k n k On the other hand, (1 + x) n (1 x) n = (1 x 2 ) n and the coefficient of x n in (1 x 2 ) n is ( ) 2m ( 1) m m Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

11 Calculation Probability step1. List all possible outcomes to get the sample space S. step2. Calculate the probability of each element in S and make sure they sum up to 1. step3. Define the event set T, which should be a subset of S. step4. Calculate the probability of T according to Pr(T ) = x T Pr(x). Especially, when we have finite n possible outcomes and they are equally likely with probability 1 n, then Pr(T ) = T S Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

12 Example Suppose we choose a sequence i 1, i 2,..., i n of integers between 1 and n at random. (1)What is the probability that the chosen sequence is a permutation of 1,2,...,n? Solution: The sample space S is the set of all possible sequences of length n and each of whose terms is one of the integers 1,2,...n. Hence S = n n and each is assigned with equal probability. The event T is the set of permutation, thus T = n!. So Prob(T ) = T S = n! n n Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

13 Example Cont Suppose we choose a sequence i 1, i 2,..., i n of integers between 1 and n at random. (2)What is the probability that the sequence contains exactly n-1 different integers? Solution: The event T is the set of sequences containing exactly n-1 different integers. So for each element in T, among 1,2,...,n, exactly one integer repeats twice and exactly one is missing. ( ) n There are n choices for the repeated integer, and possible positions 2 to put it twice. Then n-1 choices for the missing integer. After that the remaining n-2 integers can be put in the remaining n-2 positions and will lead to (n 2)! ways. Thus ( ) n n! 2 T = n (n 1)(n 2)! = 2 2!(n 2)!. Pr(T ) = T S = n! 2 2!(n 2)!n n Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

14 Example Suppose in chess, a rook can move horizontally or vertically in any direction any distance on the board. Two rooks are said to be attacking each other if they are both in the same row or column; so one can get, in a single move, to the position occupied by the other. Now we put 5 identical rooks at random in nonattacking positions on an 8-by-8 board. What is the probability that the 5 rooks are both in rows 1,2,3,4,5 and in columns 4,5,6,7,8? Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

15 Example Cont What is the probability that the 5 rooks are both in rows 1,2,3,4,5 and in columns 4,5,6,7,8? Solution The sample space S consists all placement of 5 rooks on the 8-by-8 board. As all rooks are identical, we ( just ) need to firstly pick 5 8 different rows to put them and there are ways. Then we pick 5 5 different columns and match ( ) them with columns one by one to fix 5 8 positions and there are 5! ways. So S = 8!2 5 3! 2 5!. T is the event that the 5 rooks are in the positions prescribed above. Actually, it is equivalent to the situation that we put 5 rooks in nonattacking positions on 5-by-5 board. Similarly, there are 5! ways and thus T = 5! So Pr(T ) = T S = 5!2 3! 2 8! 2 Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

16 Thank you for your time! Tutor: Zhang Qi (SEEM) Tutorial 7 November / 16

Inclusion and Exclusion Principle

Inclusion and Exclusion Principle Tutor: Zhang Qi Systems Engineering and Engineering Management qzhang@se.cuhk.edu.hk November 2014 Tutor: Zhang Qi (SEEM) Tutorial 7 November 2014 1 / 14 Review Theorem