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Sec 4.7
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#1
26*26*26*10*10*10*10 =
(26^3)*(10^4) =175,760,000
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#29! = 362,880
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#310! = 10*(9!) = 10*(362,880) =3,628,800(check using =fact:3,628,800
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#9C(8,4) = (8*7*6*5)/(4*3*2*1) =1680/24 =70(check using =combin:70
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C(9,2) = (9*8)/(2*1) =72/2 =36(check using =combin:36
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C(7,6) = (7*6*5*4*3*2)/(6*5*4*3*2*1) =7(check using =combin:7
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C(10,3) = (10*9*8)/(3*2*1) =72/2 =36(check using =combin:36
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#10
We need to count how many different pairs of people there are among the 20 people, i.e., how many different ways to choose 2 people from 20:
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C(20,2) = (20*19)/(2*1) =380/2 =190(check using =combin:190
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#11
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(a)
We need to count how many ways there are choose 4 classes from 5:
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C(5,4) = (5*4*3*2)/(4*3*2*1) =5(check using =combin:5
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Note that C(n,n-1) = C(n,1) = n for any integer n. (Do the calcuations to see why!) One way to understand C(n, n-1) = n is that you get the n different combinations by excluding one of the objects. Thus, the student's 5 different choices in this example consists of (1) excluding French, and choosing the other 4 classes; (2) excluding Spanish; (3) excluding History; (4) excluding Physics; and (5) excluding English Lit. This understanding lets us answer part (b):
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(b)From (a), the sample size has 5 outcomes in it. To compute the probability, we need to count how many of these outcomes include both French and Spanish. Clearly there are 3 such outcomes: the combinations that exclude History, exclude Physics, and exclude English Lit. Hence:
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P(French & Spanish chosen) = 3/5 =0.6
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