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  1. Default Combinations/Perumutations problem.

    Twenty-one telephones have just been received at an authorized service center. Seven of these telephones are cellular, seven are cordless, and the other seven are corded phones. Suppose that these components are randomly allocated the numbers 1, 2, . . . , 21 to establish the order in which they will be serviced.
    (a) What is the probability that all the cordless phones are among the first fourteen to be serviced? (Enter your answer to three decimal places.)
    (b) What is the probability that after servicing fourteen of these phones, phones of only two of the three types remain to be serviced? (Enter your answer to three decimal places.)
    (c) What is the probability that two phones of each type are among the first six serviced? (Enter your answer to three decimal places.)

    No idea how to approach the problems. for a, I thought it would be:
    (7! * 14P7)/21P14, but that rounded to 0, which makes me question myself on how to approach the problem.

  2. Lazy Mathematician Female
    IGN: MsJudith
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    Default Re: Combinations/Perumutations problem.

    nvm~ i'll go think about this for a sec

    [(14 choose 7) * 14!/(7!7!)]/(21!/(7!7!7!)) =0.0295149....

    1st term: choose the 7 spots out of the first 14 spots that the cordless phones go into. Since they are all the same there's no need to permute them.

    2nd term: permute the other 14 phones in the open spots. There are 14 open spots... and two sets of 7 are identical phones.

    division term: permuting 21 items with three sets of 7 identical items.

    b)[3*(14 choose 7) * 14!/(7!7!)]/(21!/(7!7!7!))=0.08854....

    multiply the top part by 3, so that any of the three phone types is entirely in the first 14 phones. This will leave the last 7 phones to be two of the three types.

    c)[ (6!/(2!2!2!)) * (15!/(5!5!5!))]/ (21!/(7!7!7!))=0.1706656....

    1st term: permute two of each phone type into the first six slots
    2nd term: permute the rest of the phones
    division term: all possible permutations.

    hope these are right.... definitely sound right in my head.... but combinatorics is a very tricky (and fun) subject :D

  3. Default Re: Combinations/Perumutations problem.

    In (a) you can look at it as picking the 'last 7' which aren't cordless first: 14!/7! ways to take those from the other two groups, and the remaining 14 can be sorted 14! ways. Then divide that by 21! which is the total number of permutations. That 7 of them are identical doesn't matter here. 14!^2 / (21!*7!) gives 0.029514 agreeing with shouri.

    My tendency is to not use npr/ncr and just do straight up factorials cause it's easier for me to not screw up; 14!/7! is obviously the same thing as 14p7. When you have (7!*14p7)/(21p14) it expands to (7!*14!/7!)/(21!/14!) = 14!^2/21!, which is off by that one extra factor of 7!. 14p7/21p14 = 0.0295 so I'm not sure what issue you had that made it come out 0. Some calculators don't handle factorials well; I used Wolfram Alpha.



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