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1 Combinations & Counting II Samuel Marateck © 2009

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

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

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

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

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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

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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!

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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.

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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

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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!

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

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

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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?

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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!)

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

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

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

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18 One of the paths is: NENENE B A

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19 Another path is: NENEEN B A

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

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

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

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

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

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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.

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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

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

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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

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

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

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

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

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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?

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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, 2+1+3+1 = 7, the number of cards in a hand. Whereas the top numbers indicate the number of a given type in a deck.

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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?

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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.

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37 Lets look at RGBBB. Well let a 0 separate the letters and a 1 for each letter. So we get 1010111. For RRRBB we get 1110110 since we need a 0 to separate the last B from the non-existent G. There will always be five 1s and two 0s.

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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.

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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)!)

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