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1 Combination Symbols A supplement to Greenleafs QR Text Compiled by Samuel Marateck ©2009

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2 How many 4-card hands consisting of 1 king and 3 queens can be chosen from a deck?

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3 How many 4-card hands consisting of 1 king and 3 queens can be chosen from a deck? Since order does not matter and there are four kings and four queens in the deck, the answer is: ( 4 1 ) ( 4 3 )

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4 What is the meaning of ( 4 1 )? Its the number of ways we can choose one thing from four, independent of the order. It is pronounced four choose one.

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5 Similarly ( 4 3 ) is the number of ways we can choose three things from four independent of the order. It is pronounced four choose three.

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6 In ( 4 1 ) ( 4 3 ), why do we multiply the two?

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7 For each king there are three queen pairings. These are the pairings for the king of spades: k Q Q Q

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8 But there are also k, k and k. So there are 16 different combinations, four for each King.

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9 What is the probability of choosing 4-card hands consisting of 1 king and 3 queens from a deck?

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10 What is the probability of choosing 4-card hands consisting of 1 king and 3 queens from a deck? ( 4 1 ) ( 4 3 ) / ( 52 4 )

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11 We divide by ( 52 4 ) since this is the number of ways we can choose four cards at random from a deck.

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12 Lets evaluate ( 4 1 ) ( 4 3 ) / ( 52 4 )

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13 ( 4 1 ) ( 4 3 ) / ( 52 4 ) is: 16/(52*51*50*49/(4*3*2*1)) =0.00006 or.006%

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14 Out of how many hands would you expect to get this hand?

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15 Out of how many hands would you expect to get this hand? 0.00006 is 6x 10 -5, so in 10 5 hands you would expect to get 6 such hands or in one out of 16,666 hands you would get this hand.

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16 How many 5-card hands can you get that have three aces?

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17 How many 5-card hands can you get that have three aces? The number of ways we can choose three aces is ( 4 3 ). How many cards are left in the deck?

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18 How many non-aces are in the deck? There are 48 non-aces left in the deck and there are two more cards to choose for our hand.

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19 So there are ( 4 3 ) ( 48 2 ) ways we can get three aces: 4*48*47/2 = 4*47*24 = 4512 ways.

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20 What is the probability of getting three aces in a 5-card hand?

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21 What is the probability of getting three aces in a 5-card hand? ( 4 3 ) ( 48 2 ) / ( 52 5 ) = 4512/((52*51*50*49*48)/(5*4*3*2*1)) = 4512/2598960 =.00174

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22 What is the probability of winning the lottery?

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23 What is the probability of winning the lottery? There are 54 numbers that you can choose from; the numbers 1 to 54. You must choose the five correct numbers independent of their order. The answer is:

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24 P(winning) = 1/( 54 5 ) ( 54 5 ) = 54*53*52*51*50/120 1/( 54 5 ) = 3.16 x 10 -7

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25 If there are 6 pegs distributed in a circle and a line is drawn from each peg to each other peg, how many lines are there?

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26 For each peg 5 lines are drawn; but there are 6 pegs. Since, however, each line connects two pegs, we are overcounting by 2, so we must divide by 2. What is the answer?

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27 # of lines is 5*6/2 or 15.

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28 Another way of looking at this is: From the first peg, 5 lines are drawn. From the second peg, 4 lines are drawn since it is already connected to the first peg. From the third peg, 3 lines are drawn, since it is connected to the first two, and so on,

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29 For the six pegs, 5+4+3+2+1 or 15 lines are drawn. For n pegs n-1 + n-2 + n-3 +..+ 1 lines are drawn. We know what the sum from 1 to m is.

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30 The sum is: m(m+1)/2. Substituting n-1 for m, the sum from 1 to n-1 is (n-1)(n-1 +1)/2 =?

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31 (n-1)(n-1 +1)/2 = n(n-1)/2 which is the answer we got before. Can we do this with combination symbols?

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32 If there are 6 pegs distributed in a circle and a line is drawn from each peg to each other peg, how many lines are there?

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33 There are 6 slots:....... 1 2 3 4 5 6 How many ways can we place two item in these slots?

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34 How many ways can we place two item in these slots? The answer is ( 6 2 ). For n pegs its ?

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35 For n pegs its ( n 2 ).

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36 How many ways can we choose a 5-card hand so that no two cards have the same face values?

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37 How many ways can we choose a 5-card hand so that no two cards have the same face values? For the first card we have ( 52 1 ) ways we can choose the first card. How many choices do we have for the second card?

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38 How many choices do we have for the second card? 48, since one face value has been eliminated. So the number of ways we can choose the second card is:

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39 ( 48 1 ). The third card is?

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40 ( 44 1 ). So the final answer is: ( 52 1 ) ( 48 1 )( 44 1 ) ( 40 1 ) ( 36 1 ). What is the probability?

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41 P(each card has a different face value) = ( 52 1 ) ( 48 1 )( 44 1 ) ( 40 1 ) ( 36 1 ) ( 52 5 )

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42 In a class of 25, what is the probability that two or more people have the same birthdate?

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43 In a class of 25, what is the probability that two or more people have the same birthdate? We will first calculate the probability that no one has the same birthdate.

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44 Given the first person, the probability that the second one has a different birth date is 364/365. That the first, second and third ones have different birth dates is: 1* 364/365*363/365. For all 25 people?

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45 For all 25 people? P(different birth dates) = 364*363*362*361…341/365 24 = 0.47

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46 P(2 or more have same birth dates) =.53

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47 There are 25 people to be chosen for a Committee or 5. What is my probability of being chosen?

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48 What is the probability of my being chosen? ( 1 1 ) ( 24 4 )/ ( 25 5 ).

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49 An urn contains 10 red balls and 40 black ones. What is the probability you will draw 2 red balls.

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50 ( 10 2 ) ( 40 0 )/ ( 50 2 ) = 10*9/2 /(50*49/2) = 45/1225

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51 An urn contains 17 red balls and 33 black ones. What is the probability you will draw 7 red balls if you choose 10 randomly?

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52 ( 17 7 ) ( 33 3 )/ ( 50 10 )

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53 A jury pool contains 98 men and 75 women. 12 jurors are chosen at random. What is the probability that 6 will be women

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54 ( 98 6 ) ( 75 6 )/ ( 173 12 )

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