1 Combinations & Counting II Samuel Marateck © 2009

2 How many 4-letter words can we form from the letters ABCD w/o replacements?

3 How many 4-letter words can we form from the letters ABCD? This is an example of an ordered list without replacements.

4 How many 4-letter words can we form from the letters ABCD w/o replacements? 4*3*2*1 or 4! or 24

5 How many 2-letter words can we form from the letters ABCD w/o replacements? This is another example of an ordered list

6 How many 2-letter words can we form from the letters ABCD w/o replacements? This is another example of an ordered list 4*3 or 12

7 How many 3-letter words can we form from the letters AAB w/o replacements? This is yet another ordered list; but there are now repetitions of one of the letters!

8 How many 3-letter words can we form from the letters AAB w/o replacements? Lets distinguish between the As by giving them subscripts, A 1, A 2.

9 Lets distinguish between the As by giving them subscripts, A 1, A 2. So the # of words that can be formed is 3*2*1 or 6. They are: B A 1 A 2 B A 2 A 1 A 2 B A 1 A 1 B A 2 A 2 B A 1 A 2 A 1 B A 1 A 2 B

10 If we dont distinguish between the As, the words that can be formed are: B A A A B A A A B We get the number of different words by dividing by the # of ways the As can be arranged or 2!

11 How many 3-letter words can we form from the letters AAB w/o replacements? Its 3!/2! or 3.

12 How many distinct 11-letter words (ordered lists) can we form from the letters in mississippi w/o replacements?

13 How many distinct 11-letter words can we form from the letters in mississippi w/o replacements? There are 4 ss, 4 is, 2 ps and 1 m. The # of ways we can form words with the given repetitions of these letters is 11! How do we account for the repetitions?

14 There are 4 ss, 4 is, 2 ps and 1 m. The # of ways we can form words with repetitions is 11! How do we account for the repetitions. We divide by 4! * 4! * 2! The answer is 11!/(4!*4!*2!)

15 How many paths can we take from point A to B using the lines. B A

16 The 2 directions we can go are North and East. B A

17 This is an example of ordered lists since a path depends upon the order of Ns and Es B A

18 One of the paths is: NENENE B A

19 Another path is: NENEEN B A

20 How many Es and Ns must each path contain? B A

21 There must be 3 Ns and 3 Es in any order in each path. B A

22 This is an example of an ordered list w/o replacements but with repetitions (of N and E). B A

23 The # of paths is therefore 6!/(3! 3!) B A

24 6!/(3! 3!) = 6*5*4*3*2*1/(6*6) = 5*4 or 20

25 Review Ordered list with replacement. Example: How many different lists of length k can we get from tossing a coin. We can get 2 results, a head or tail.

26 How many different lists of length k can we get from tossing a coin. We can get 2 results, a head or tail. Answer: 2 k

27 Ordered list without replacement. Example: How many k different cards can we choose from a deck of n cards?

28 How many k different cards can we choose from a deck of n cards? Answer: n*(n-1)*(n-2)*(n-3)..(n – k +1). 3 cards from a 52 deck: 52*51*50

29 Ordered list without replacement but with repetition. Example: How many 11-letter words can we make from mississippi.

30 Ordered list without replacement but with repetition. Example: How many 11-letter words can we make from mississippi. Answer: 11!/(4!*4!*2!)

31 Unordered list without replacement. Example: How many committees of k people can we choose from a pool of n?

32 Unordered list without replacement. Example: How many committees of k people can we choose from a pool of n? ( n k )

33 Whats the probability of getting 2 kings, 1 queen, 3 jacks and another non-kqj in a hand of 7 cards from a 52 card deck?

34 Whats the probability of getting 2 kings, 1 queen, 3 jacks and another non-kqj in a hand of 7 cards from a 52 card deck? ( 4 2 ) ( 4 1 ) ( 4 3 ) ( 40 1 )/ ( 52 7 ) Note that the sum of the bottom numbers, = 7, the number of cards in a hand. Whereas the top numbers indicate the number of a given type in a deck.

35 Unordered list with replacement (or repetitions). Example: Given an urn with red, green and blue marbles, how many different combinations of marbles can we get if we choose 5 marbles?

36 Given an urn with red, green and blue marbles, how many different combinations of marbles can we get if we choose 5 marbles? Some of the combination are RRRRR, GGGGG, GRBBB. The last one is the same as BBBGR, GBBBR or RGBBB, since order does not count.

37 Lets look at RGBBB. Well let a 0 separate the letters and a 1 for each letter. So we get For RRRBB we get since we need a 0 to separate the last B from the non-existent G. There will always be five 1s and two 0s.

38 There will always be five 1s and two 0s. The # of ways we can permute 7 objects with five repetitions of one and two of the other is 7!/(5!*2!) = 7*6/2 = 21.

39 In general if we have we are choosing n objects and k of them are different, the number of combinations is: (n + k -1)!/(n!*(k-1)!)