Counting Review R 1. R 3 receiver. sender R 4 R 2

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1 Counting Review Purpose: It is the foundation for many simple yet interesting examples and applications of probability theory. For good counting we need good crutches (fingers?). Hence good images for the things we wish to count. Good notation for certain types of counts. Want to be able to count large amounts. Example 1 (Defective Relays or Craced Rivets): Suppose we have a sequence of relay stations. A signal can be passed through this sequence as long as not two consecutive stations are defective. Suppose there are m + n stations, m operating and defective. In how many ways can we arrange these m + stations in sequence such that a signal may pass through? Similarly, one could have m good rivet holes and craced rivet holes. How many ways are there of having no two neighboring rivets are craced? Tough! sender R 1 R receiver R 2 R 4 Special case: m 2, 2; symbol for defective and 0 symbol for operating. Represent all possible arrangements 1

2 0 0 y 0 0 y 0 0 n 0 0 y 0 0 n 0 0 n out of 6 will pass the signal. How about general m and? Product Rule(Basic Principle of Counting or Ordered Pairs): n 1 items or choices for the first position n 2 items or choices for the second position how many paired choices for first and second position Counting in n 1, n 2 times, efficient addition. Give rectangular image and tree branch count. Example 2 (Dice): red and green dice: pairs (first choice, second choice)? n 1 n 2 paired (ordered) choices Example (Dice without doubles): We throw a pair of red and green dice until the faces don t match: possible pairs Generalized Product Rule (Ordered tuples): n 1 choices for the first position n 2 choices for the second position n choices for the th position how many ordered tuples can be chosen? (first choice, second choice,..., th choice) Example 4 (Bytes): 2 n binary sequences of length n bytes n 1 n 2... n possible choices 2

3 Example 5 (Birthdays): If a birthday could be any one of 65 possible days in a year how many possible birthday arrangements are there for 2 persons? Example 6 (License Plates): If a licence plate consist of three letters followed by three digits, how many different licence plates are possible? , 576, 000. If the six positions could be any letter or any digit, then there are 6 6 2, 176, 782, 6 possibilities. Example 7 (Anagrams): How many ways are there to arrange the three letters a, b, c in sequence? Answer: 2 1 6, ways to fill the first position, two remaining ways to fill the second position and one way to fill the last position with the remaining letter. Each such arrangement of the three letters is called a permutation of the three letters. Permutations: Suppose we have n distinct objects then there are n(n 1)(n 2) 2 1 n! permutations of these objects, i.e. arrangements of these distinct objects in order. n! is read n factorial. Example 8 (Horse Race): In a horse race with 8 horses there are 8! 40, 20 possible finishes if we disregard the possibility of ties. Permutations With Repetitions: If some of the objects to be permuted are indistinguishable from each other then some of the permutations will also be indistinguishable, namely those where the repeated objects are permuted among each other. For example, of the six letters in the word PEPPER only, namely P,E,R, are distinct. If we distinguish among the P s and E s by labelling them, then there are! 2! distinct ways to write the word PEPPER. For example, P 1 E 1 P 2 P E 2 R is distinct from P 2 E 1 P 1 P E 2 R as far as the labelling is concerned. There are 6! permutations of the symbols P 1 P 2 P E 1 E 2 R but only 6!/(! 2!) 60 are distinguishable once the lables are dropped. In general, if there are n objects of which n 1 are alie, n 2 are alie,..., n r are alie then there are n! n 1! n 2! n r!

4 permutations (distinguishable arrangements) of these objects. Example 9 (Treatment/Placebo Assignments): Suppose ten patients in an experimental drug treatment program are to be divided into two groups of size five each, the first five to receive the new treatment and the other five to receive the placebo drug. How many possible ways are there to mae the drug assignments? Answer: Line the patients up in order and create an arrangement of five T s and five P s. Thus there are 10!/(5! 5!) 252 possibilities. Suppose there are two drugs (A and B) and a placebo and 15 patients are to be equally divided into three groups each receiving the respective drug then there are 15!/(5! 5! 5!) 756, 756 possible assignments. Generalized Permutation: Suppose we have n distinct objects, however we only select ( n) of them to arrange them in order. How many possible ways are there? Answer: n(n 1) (n ( 1)) n(n 1) (n + 1) (n) If n then (n) n n!, reducing to the full permutation. Example 10 (Baset Ball Finals): At the end of the season the first four teams in a baset ball league of 20 teams advance to the finals. How many possible finishes for the first four positions are possible? Answer: (20) 4 116, 280. Combinations: Suppose we have n objects and we are interested in finding the number of possible groups that can be formed by taing r (r n) objects from these n, the order of objects in the group being immaterial. Any such group is called a combination of n taen r at a time. The number of such combinations is denoted by r n r n! r! (n r)! (n) r r! where the computational formula follows as in example 9. Convention: 0 1 and n n(n 1) (n r + 1) r! 0 if r < 0 or r > n. r Approximation: For large numbers n it becomes tedious (and sometimes not very manageable on calculators) to calculate n! because n! gets too large. The right order of magnitude may often be quite sufficient, especially if ratios of factorials are required. Stirling s formula states: ln(n!) ( n + 1 ) ln n n ln(2π) + θ 12n 4

5 where 0 < θ < 1 and θ/(12n) is usually neglected but indicates the size of the possible error. In direct form this is n n! 2πn with e e Example 11 (Washington Lottery): How many possible choices are there in the Washington Lottery for picing 6 numbers out of 49? Answer: ( ) 49 49! 1, 98, ! 6! Using Stirling s formula we get: 14, 182, with about 1.4% relative error. The error is large because Stirling s formula is not accurate when applied to 6!. If we use Stirling s formula only on 49! and 4! we get 1, 987, with about.024% relative error! Example 1 (Survey Sample): A sample of 1, 000 (without replacement) is taen from a population of size 100, 000. If you are a member of the population what fraction of all possible such samples will contain you? Answer: ( ) 100, 000 possible samples 1, 000 ( ) 99, samples with you in it 999 Fraction with you in it 1, 000! 99, 000! 99, 999! 100, 000! 99, 000! 999! 1, , Example 14 (Rivets and Relays Revisited): m good rivets, craced rivets, how many arrangements without neighboring rivets? Answer: Line up the good rivets in sequence. There are m m + 1 slots between and at the end of these operating relays. If m + 1 then it is possible to put at most one craced rivet into each one of these slots and there are ( ) m+1 ways of doing this. If we have 75 good rivets and 20 craced ones, then there are ( ) sequences without neighboring craced rivets. Compare this to the ( ) sequences of 95 rivets with 20 craced ones, i.e., the fraction of dangerous sequences is r Combinatorial Identity (Pascal s Triangle): ( ) ( ) ( ) n n 1 n 1 + r r 1 r 5 for 1 r n

6 proof: choose r from n in one of two possible ways: 1. Choose r outright: ( ) n r possibilities 2. Designate one of the n items as special. Then we may either choose the special item plus r 1 others in 1 (n 1 ) ( ) ways or we may choose r without the special in n 1 ways. r 1 Since these are mutually exclusive possibilities the claim follows Pascal ( ), Tartaglia ( ), Omar Khayyam ( 112) and earlier in Persia and China, cf. Edwards (1987) Pascal s Arithmetical Triangle Binomial Theorem: (x + y) n n 0 Proof by multiplying out and counting similar terms. Example 15: x y n (x + y) 4 x 4 + 4x y + 6x 2 y 2 + 4xy + y 4 r Example 16 (Possible Subsets): A set of n elements has how many subsets? Thus 2 n (1 + 1) n subsets of size n n n 0 is the total number of subsets counting the empty subset. Otherwise we would have 2 n 1 non empty subsets. Note also: 0 ( 1 + 1) n n 0 ( 1) 1 n ± 2 n

7 i.e. there are as many subsets of odd size as there are of even size. Multinomial Coefficients: Divide n distinct items into r distinct groups of respective sizes n 1, n 2,..., n r, n n 1 + n n r. How many possible ways? Answer: ( )( ) ( ) n n n1 n n1 n 2 n r 1 n 1 n! (n n 1 )!n 1! n 2 (n n 1 )! (n n 1 n 2 )!n 2! (n n 1 n 2 n r 1 )! 0!n r! ( ) n! n 1!n 2! n r! : n n 1, n 2,..., n r n r Example 17 (Sat): In the German game of Sat a dec of 2 cards is dealt to three players with 10 cards per player and two in the middle (the sat). How many possible deals? Answer: ( ) 2 10, 10, 10, 2 2! !10!10!2! or using Stirling s approximation on 2!, resulting in.26% relative error. Multinomial Theorem: (x 1 + x x r ) n (n 1, n 2,..., n r ) : n 1 + n n r n ( n n 1, n 2,..., n r where the summation is over all non negative integer valued r vectors (n 1, n 2,, n r ) such that n 1 + n n r n. Example 18: (x + y + z) ) x n 1 1 x n 2 2 x nr r ( ) ( ) ( ) x y 0 z 0 + x 0 y z 0 + x 0 y 0 z, 0, 0 0,, 0 0, 0, ( ) ( ) ( ) + x 2 y 1 z 0 + x 2 y 0 z 1 + x 1 y 2 z 0 2, 1, 0 2, 0, 1 1, 2, 0 ( ) ( ) ( ) + x 0 y 2 z 1 + x 0 y 1 z 2 + x 1 y 0 z 2 0, 2, 1 0, 1, 2 1, 0, 2 7

8 ( ) + x 1 y 1 z 1 1, 1, 1 x + y + z + x 2 y + x 2 z + xy 2 + y 2 z + yz 2 + xz 2 + 6xyz More complicated counts: 1 r 1 ways to write x 1 + x x r n, integer x i 1 Proof: choose r 1 gaps of the n 1 gaps between n 1 s. For n 10 and r an example choice is: x 1 + x 2 + x r 1 r 1 ways to write x 1 + x x r n, integer x i 0 Proof: choose r 1 positions for r 1 dividers among the n + (r 1) positions for the n 1 s and r 1 divider places. For n 10 and r 5 an example choice is: x 1 + x 2 + x + x 4 + x