Lecture 27 Enoch Cheung December 4 2013 1 a n 2 X n k 0 k 2n n Right hand side counts the number of combinations of n objects from 2n objects Left hand side counts the same thing Paint n of the objects red and n of the objects blue To pick n objects there are cases for each picking i red objects 0 k n where Case k represents n and n k blue objects There are nk ways to pick red objects and n k nk ways to pick blue 2 objects Therefore for case k there are nk choices Therefore the total number of choices are Pn n 2 k 0 k 2 n X n 2n 1 b k n k n 1 k 1 Given 2n people n male and n female We are counting the numbers of ways to pick a male team leader and n 1 more team leaders to make a team of n people Left hand side we are considering cases for 1 k n where Case k is the case where there are k male and n k females where one of the k males is the leader so for Case k there are k ways to pick the leader from the k male chosen nk ways to choose the k male team members 2 n and nk n k ways to pick the n k female team members Therefore k nk ways in each 2 Pn Case k so k 1 k nk total Right hand side pick one of the n male leaders then from the 2n 1 remaining people pick n 1 more team members n X n n 2n 1 c k k k 1 Given n people and we have to pick a team with a leader can be any size 1 Left hand side cases for each 1 k n where Case k is the case where the team has k people so there are nk ways to pick the team and in the team of k there are k choices of leaders Right hand side we first pick a leader of which there are n choices And for each of the remaining n 1 people we decide if they are in or out of the team 2 choices per person so there are 2n 1 choices Thus total there are n 2n 1 choices n X 2n d 22n 1 2k k 0 Given 2n people we have to pick a team of even size Left hand side there are case cases for 0 k 2n where Case kPisn the2n where the team has 2k people so there are 2n choices for Case k Thus there are k 0 2k total choices 2k For the right hand side order the 2n people in some way For the first 2n 1 people we decide whether they are in or out of the team 2 choices per person For the last person if our team is already even the the last person is out otherwise the team is odd and the last person is in Thus there are 22n 1 choices m m 2 m N m3 3 m 3 3 2 1 Consider arbitrary m N We are counting 3 lettered words from an alphabet of size m Left hand side we pick 3 letters one by one to form the word so m3 ways Right hand side there are 3 cases Case 1 is the case where all 3 letters of the word are distinct so we pick a subset of 3 letters from m so there are m 3 ways to do this then we count permutations of the 3 chosen letters to make words 1 and there are 3 permutations Thus Case 1 has 3 0 which makes sense m 3 ways Note that if m 3 then this number is Case 2 is the case where the word has 2 distinct letters and one of which is repeated There are m 2 ways to pick the subset of letters that are used There are 2 ways to pick which letter is unique and which is repeated and 3 choices of positions for the unique letter Therefore case 2 has 2 3 n2 3 n2 ways Case 3 is the case where all 3 letters in the word are the same There are m 1 choices of the repeated letter m m Therefore the total number of ways are 3 m 3 3 2 1 2