CSC2110 Tutorial 11

1. CSC2110 Tutorial 11 More Counting Tom Chan (Room: SHB 115) 22/11/2007

2. T/F Question 4 (splitting into m sets) • There are 132! / (4!33 x 33!) ways to arrange a class of 132 students into groups of 4, if the groups are unnumbered • If the groups are numbered or forms a sequence(order MATTERS), # ways = 132! / (4!33) ((ABCD) = (BCAD) = … — 4! combinations each) • But now groups are unnumbered (order DOES NOT matter), (ABCD) (EFGH) … is the same as (EFGH) (ABCD) … — 33! orderings of groups • True CSC2110 Tutorial 11: More Counting

3. T/F Question 5 (splitting into m sets) • A sequence of 100 balls with 5 colours will have the greatest number of permutations if we have the same number of balls for each colour. • Recall # combinations for 2 coloursis the largest when • Similar for multinomial. True CSC2110 Tutorial 11: More Counting

4. MC Question 5 (splitting into m sets) • How many ways are there to go from (0, 0, 0) to (6, 6, 6) by x-step, y-step and z-step only?(a) (18 6) (b) 18! / (6!3) (c) 63 (d) 6!3 • Map set of paths to set of sequences of 18 chars xyzzxyxyz… with exactly 6 x’s, 6 y’s and 6 z’s. Similar to the “MISSISSIPPI” problem seen before • (b) 18! / (6!3) CSC2110 Tutorial 11: More Counting

5. Short Question 8 (binomial theorem) • Evaluate the coefficient of the term x3y7 in the expansion of (x− 3y)10. • By the binomial formula, • now we haveand set • Coefficient • Remember the +/- sign CSC2110 Tutorial 11: More Counting

6. Short Question 10 (Combinatorial proof) • Show the Vandermonde’s Identityusing combinatorial proof. (Hint: consider a pool of m red cards and n black cards.) • # ways to choose r cards from mred cards and n black cards: • Alternatively, choose k black cards and r − kred cardsalso gives us r cards: CSC2110 Tutorial 11: More Counting

7. Short Question 10 (con’t) • Alternatively, choose k black cards and r − k red cards also gives us r cards: • Vary k from 0 to r, we have total # ways to pick r cardsequals • So CSC2110 Tutorial 11: More Counting

8. Short Question 12 (Inclusion-Exclusion) • Using Inclusion-Exclusion Principle, calculate the number of primes less than 100. A listing of primes will score no mark. • Define • # primes CSC2110 Tutorial 11: More Counting

9. Short Question 12 (con’t) • By Inclusion-Exclusion principle, Sum of size of individual setsminus Sum of all two-way intersectionsplus Sum of all three-way intersectionsminus Sum of all four-way intersections. • So • Number of primes = 98 − 73 = 25 CSC2110 Tutorial 11: More Counting

10. T/F Question 6 (Pigeonhole Principle) • It is always possible to find two subsets with equal sum from a set of 150 numbers, each with less than or equal to 43 digits. • Pigeons = # subsets: 2150≈ 1.417 × 1045 • Pigeonholes = sum of subsets: ≤ 150 x 1043 = 1.5x 1045 • # pigeons not necessarily larger than # pigeonholes • False • What if all numbers are exactly 43 digits? CSC2110 Tutorial 11: More Counting

11. T/F Question 6 (con’t) • It is always possible to find two subsets with equal sum from a set of 150 numbers, each with exactly 43 digits. • Pigeons = # subsets: 2150≈ 1.417 x 1045 (unchanged) • Pigeonholes = sum of subsets:≤ 150 x 1043 x 0.9 = 1.35x 1045 (why 0.9?) • # pigeons larger than # pigeonholes • True CSC2110 Tutorial 11: More Counting

12. Count the number of 8-bit strings that do not contain three consecutive ones.(a) 129 (b) 139 (c) 149 (d) 159 Count the # strings that contain three consecutive ones 111????? (A) ?111???? (B) ??111??? (C) ???111?? (D) ????111? (E) ?????111 (F) 01111110 is in (B), (C), (D) and (E) Too many repetitions! A better choice? MC Question 7 (Miscellaneous) CSC2110 Tutorial 11: More Counting

13. Maybe this one? 111????? (A) 0111???? (B) ?0111??? (C) ??0111?? (D) ???0111? (E) ????0111 (F) Almost disjoint, except (A) and (E): 1110111? (A) and (F): 111?0111 (B) and (F): 01110111 # strings with 3 consecutive ones = 32 + 16 x 5 − 2 − 2 − 1 = 107 (c) (256 − 107 = 149) Another way of counting? Next week MC Question 7 (con’t) CSC2110 Tutorial 11: More Counting

14. Long Question 2 (Miscellaneous) • A password used on a particular system consists of 4 characters chosen from the set {0, 1, …, 9, X, Y} • (a) How many combinations are there if no characters are allowed to repeat? • 12! / (12 − 4)! = 11880 • (b) How many combinations are there if the password must consists of exactly 3 of the characters (e.g. 0112, X1X2, etc.)? • 4! / (2! x 1! x 1!) ways to arrange “AABC” • C(3, 1) ways to choose which of the three to appear twice • C(12, 3) ways to choose 3 characters from the set • C(12, 3) x C(3, 1) x 4! / 2 = 7920 CSC2110 Tutorial 11: More Counting

15. Long Question 2 (con’t) • (c) How many combinations are there if the password must consists of exactly 2 of the characters, each appear twice (e.g. 0011, 1010, etc.)? • 4! / (2! x 2!) ways to arrange “AABB” • C(12, 2) ways to choose 2 characters from the set • C(12, 2) x 4! / 2! / 2! = 396 • (d) With reference to answers in the previous parts, calculate the total number of passwords that no character appears more than twice. • The possible cases are those in (a), (b) and (c) and they are disjoint • 11880 + 7920 + 396 = 20196 CSC2110 Tutorial 11: More Counting

16. Just for fun (optional) a b c balls • n balls into n bins • # combinations? • Any rules? Partition Numbers A000041 Bell / Exponential Numbers A000110 bins 1 2 3 A001700 A000312 See http://www.research.att.com/~njas/sequences for details CSC2110 Tutorial 11: More Counting