Lecture 25 Enoch Cheung December 3 2013 1 How many ways are there to pick two cards from a standard 52 card deck such that the first card is a spade and the second card is not an ace Case 1 First card ace of spades 52 4 48 non aces left 1 48 ways this case can happen Case 2 First card spades but not ace 13 1 12 choices of first card 52 4 1 47 non aces left 12 47 ways this case can happen Therefore there are 48 12 47 ways 2 How many ways are there to deal a 5 card hand such that at least 3 cards have the same rank There are rank is repeated exactly 3 times then 13 choices for the repeated rank Case 1 The there are 43 4 choices of the repeating cards and 48 choices for the remaining cards Case 2 2 The rank is exactly 4 times then there is 1 choice for the four cards and 48 1 48 choices for the remaining card Therefore there are 13 4 48 2 48 ways 3 a How many 5 letter words can be made that have at least 1 vowel It is easier to count the number of 5 letter words that has no vowels There are 26 5 21 consonants Therefore there are 215 words with no vowels There are 265 total words Therefore there are 265 215 words with at least one vowel b How many 6 letter words only have 1 distinct vowel and that vowel occurs exactly twice There are 5 choices of the vowel that appears There are 62 possible positions the vowel can appear in the 6 letter string There are 214 possible choices for the remaining letters Therefore 6 4 the total number is 5 2 21 4 How many anagrams are there of the word BANANA are there First there are 61 possible choices of locations of B Next there are 53 possible choices of locations 6 of the A s The rest are N s Therefore the total number is 61 53 1 3 2 5 Let n N How many lattice paths to 2n 2n go through the point n n Starting from 0 0 there are 2n n lattice paths to n n because each path is a distinct sequence of n UPs and n RIGHTs From n n there are 2n n lattice paths to 2n 2n Therefore there are 2n 2 total paths n 6 How many lattice paths to 11 7 a Pass though both 2 3 and 7 4 Starting from 0 0 there are 2 3 52 lattice paths to 2 3 because each path is a distinct 2 sequence of 2 RIGHTs and 3 UPs Now from 2 3 there are 7 2 4 3 65 lattice paths to 7 2 7 4 Finally from 7 4 to 11 7 there are 11 7 7 4 74 paths 11 7 Therefore total there are 52 65 74 paths b Pass through neither 2 3 nor 7 4 The total number of paths from 0 0 to 11 7 are 11 7 18 11 11 The number of paths that 11 2 7 3 pass through 2 3 are 2 3 52 13 2 11 2 9 The number of paths that pass through 7 4 7 4 11 7 7 4 11 7 are 7 7 4 Note however that if a path goes through both 2 3 and 7 4 11 7 then it will be counted when we count paths that goes through 2 3 and also when we count paths that goes through 7 4 By a there are 52 65 74 such paths Therefore the total number of paths that goes through neither 2 3 nor 7 4 are 18 5 13 11 7 5 6 7 11 2 9 7 4 2 5 4 1 7 How many ways are there to distribute 12 indistinguishable balls among 5 bins Stars and Bars method To divide up 12 balls Stars in a row into 5 bins we need 4 dividers Bars There are 12 4 4 possible positions of the Bars in a sequence of 12 4 objects Therefore there are 16 4 ways If we require at least one ball per bin it is the same as asking how many ways to distribute 12 5 balls among 5 bins where they could be empty then adding 1 ball to each of the 5 bins Therefore 11 there are 12 5 4 4 4 ways 2