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Permutations and Combinations

Posted on 2004-04-29
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Last Modified: 2008-01-09

Ok, Ithought I was getting the hang of this again and since I remember doing permutations in school (over 15 years ago) like the number of ways you could pick 6 numbers out of 42 to win the lottery 42*41*40*39*38*37/6*5*4*3*2*1, I thought I'd be able to handle this.  

However I've been set a question which involves a group of 10 students doing a postgrad course. 4 are from business backgrounds, 4 from science and 2 from arts.  
They need to select a group of 5 from the 10, so I know the number of combinations is 10*9*8*7*6/5*4*3*2*1...straight forward enough.  
They then start introducing constraints into it like how many groups when each group must have 1 student from arts and 2 from business and science (question is vague...do i assume they mean 2 from each group meaning 4 in total?)  

I'm not sure how to include these constraints. Any suggestions ?
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Question by:Penfold
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7 Comments
 
LVL 13

Expert Comment

by:SteH
ID: 10947311
With constraints you are not picking 5 out of 10 but
1 out of 2 for arts
2 out of 4 for business and science.
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Author Comment

by:Penfold
ID: 10947775
But how do I then combine these 2 choices....Do i just add them or multiply them or what ?
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LVL 12

Accepted Solution

by:
andrewjb earned 275 total points
ID: 10947847
Multiply, providing the sets of choices in each category don't overlap.
(Which they wont, here, 'cos the options are independent, but something like '1 arts, 2 science, 2 girls, 1 boy' overlap, which would make it more complex)
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LVL 13

Expert Comment

by:SteH
ID: 10947934
Depending on the problem one might think of taking possible order of choosing them into account. But from the question I understand it to just find the probability as multiplication of now 3 pickings.
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LVL 5

Expert Comment

by:PointyEars
ID: 10948069
Number of art students: a = 2
Number of business students: b = 4
Number of science students: s = 4

Let's say that we call Cn,k the combinations of k objects out of n (without repetitions and ignoring the ordering).  As you said:
  Cn,k  =  n!/(k! * (n-k)!)  =  (n * (n-1) * ..  * (n-k+1)) / k!

If you need to take i students of arts, j students of business, and k students of science, the total number of combinations is given by:
  Ca,i * Cb,j * Cs,k

If you need 1 art student, 2 business students, and 2 science students:
  C2,1 * C4,2 * C4,2 = 2 * 4*3/2 * 4*3/2 = 72

Indeed, all possible choices for arts are:
  A1, A2
for business:
  B1+B2, B1+B3, B1+B4, B2+B3, B2+B4, B3+B4
for science:
  S1+S2, S1+S3, S1+S4, S2+S3, S2+S4, S3+S4
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LVL 13

Expert Comment

by:SteH
ID: 10950734
Hi Penfold

I feel that the accepted comment is not obviuosly answering your question. That comment only explains my first comment. In that cases it would be nice to give some feedback why you choose that specific answer.
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LVL 5

Expert Comment

by:PointyEars
ID: 10950873
I gave you a detailed answer but conceptually didn't actually add to what SteH and andrewjb said.  Therefore, I accept that you didn't give me any points.  Still, SteH gave you an esential hint on which andrewjb built up the solution.  I feel that you should have split the points.
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