Penfold
asked on
Permutations and Combinations
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*
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...str
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 ?
ASKER
But how do I then combine these 2 choices....Do i just add them or multiply them or what ?
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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.
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
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
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.
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.
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.
1 out of 2 for arts
2 out of 4 for business and science.