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Counting problemo

  • Thread starter sapiental
  • Start date
  • #1
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Homework Statement



a) if jack has 10 friends, in how many ways can he invite 5 of them to dinner.
b) suppose 2 friends dont like each other, and if one is invited, the other cant come.
c) what if 2 of the friends are married and if they invite that friend, the spouse must come.

Homework Equations



Rule of products

The Attempt at a Solution



a) I get 10*9*8*7*6 = 30240 ways.
b) I simply subtract one friend initially to get 9*8*7*6*4 = 15120 ways
c) I subtract to 8 friends because their spouses fill the spots of the other 2.

8*7*6*5*4 = 6720 ways

I'm unsure if i'm misinterpreting the question. My method goes by the order in which they are invited..any help is much appreciated
 

Answers and Replies

  • #2
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a) I'm sure the order they are invited is not important here.
b) there are two cases: 1. you invite one of the two enemies and 4 others. 2. you invite 5 out of the 8 other friends. I don't think order matters here either.
c) you don't know that there will be 2 friends with spouses, there could be 0, 1 or 2 of them. Do those cases seperately.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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In c, you were told specifically that there "2 of the friends are married". You seem to be interpreting that to mean that two of the friends are married to someone not included in the original 10 friends. I would interpret it to mean that two of the friends are married to each other.
 
  • #4
118
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Hi,

thanks for all the input it really helped me gain more insight into the problem

My new approach does not depend on the order and I employed the permutation and combination method.

For a) I did C(10,5) = 10!/[(5!)(5!)] and got 252 possible combinations

for b) i did the following C(2,1) deals with the enemies for 1 spot, and C(8,4) deals with the friends who smoke the peace pipe. C(2,1)*C(8,4) I get 140 different ways. Then for the case of 5 out of 8 friends. C(8,5) = 56

so case1 + case2 = 196 different ways.
c) still working on this one
am i on the right track? Thanks!
 
Last edited:
  • #5
118
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for c I did out all the cases

case 1. where no married couples are invited, we have 6 friends for 5 spots = 6 different ways to invite

case 2 one of the married couples for 2 spots, 6 singles for (6,3) = 20 different ways.

since there are 2 married couples I multiply by 2 for a total of 40 different ways.

case 3. 2 married couples fill 4 spots, and 1 spot for the lucky 5th wheel. = 6 different ways

I add all together and get 52 possibilities. Could someone confirm this pretty please. Thanks!
 
  • #6
454
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a and b are correct.

I do now think that 2 friends are married means "to each other" so there is only one
married couple. another option is that they are married to non-friends who must come
along and use up a spot, but they won't reduce the number of non-married friends to choose from

you gave the answer for 2 married couples amongst the friends. I don't think that interpretation is justified.

case 1. If you leave out the 2 married ones, there are 8 friends left for 5 spots.

case 2. there is only 1 married couple. for the other 3 spots there are also 8 friends left.
this is the same number as case one. instead of chosing 3 friends to invite you can choose
5 friends to not invite.
 
  • #7
118
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the problem states "assuming 10 friends consist of 6 single people and 2 married couples" sorry for not clarifying. would it still be off?
 
  • #8
454
0
the problem states "assuming 10 friends consist of 6 single people and 2 married couples" sorry for not clarifying. would it still be off?
Then it is OK.
 

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