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Combinationsby danago
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#1
Apr3007, 06:59 AM

PF Gold
P: 1,131

A subgroup must be formed, with 4 people being chosen from 3 larger groups.
The subgroup of 4 must contain atleast 1 person from each group (A,B,C). How many possible groups are there? Well there are 6 possible choices for the first place, 4 for the next and 3 for the third place. The fourth place can be taken by any of the remaining 10people. The calculation i came up with was: [tex] {}^6C_1 {}^4C_1 {}^3C_1 {}^{10}C_1=720 [/tex] However, that is wrong. What have i done wrong? 


#2
Apr3007, 07:32 AM

P: 3,220

your first choice is 13C1, cause it doesnt matter from which one you choose first. now you are left with 12 people and you should choose from the other two groups, if you first chose from A then you now have 7C1 from either B or C, if you first chose B then now you have 9C1, if first C then now you have 10C1 in the second option you have [10C1+9C1+7C1]/2 cause we don't count repetitions.
now you are left with 11, and you should choose, if the first and second were A,B then now you choose 3C1, if it's A,C then now you have 4C1, and if it were B,C then now 6C1, for the fourth option you have: 10C1. all in all you have: 13C1*([10C1+9C1+7C1]/2)*(6C1+4C1+3C1)*(10C1) but don't take my word, im not that good in counting. (: 


#3
Apr3007, 12:10 PM

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P: 39,339

The problem does not say anything about a "first" person, etc.: this is not a 'permutations' problem.
Here's how I would have done this. Since there must be at least one person from each subgroup, choose one person from A: there are 6 ways to do that. Then choose one from B: there are 4 ways to do that. Choose one person from C: there are 3 ways to do that. Now put the remaining 5+ 3+ 2= 10 people in a single group and choose 1 person from them: there are _{10}C_{1}= 10 ways of doing that. Altogether, there are 6*4*3*10= 720 ways of choosing 4 such people. Except that I used slightly different language, that is exactly what you did. Now, tell me why you say that is wrong! 


#4
Apr3007, 08:13 PM

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P: 1,131

Combinations
Well there are 2 reasons i thought it was wrong.
1) If there are no restrictions placed on the selection process, there are ^{13}C_{4}=715 possible groups. I would have thought that when resctrictions are placed, the number of groups should be less than when there are no restrictions. 2) The answer page says 360. 


#5
Apr3007, 08:25 PM

P: 1,123

The answer is 360, you need to divide by 2 since you decided that ordered mattered when you chose the last person.
You could also think of it like this. You have to choose 2 people from one of the groups, and 1 person from each of the other 2 groups, so you have 3 cases. Case 1 (choose 2 people from group A, 1 person from B, 1 person from C). [tex]\binom{6}{2}\binom{4}{1}\binom{3}{1}[/tex] Case 2 (choose 2 from group B, 1 from A, 1 from C). [tex]\binom{6}{1}\binom{4}{2}\binom{3}{1}[/tex] Case 3 (choose 2 from group C, 1 from A, 1 from B). [tex]\binom{6}{1}\binom{4}{1}\binom{3}{2}[/tex] Add these up and you get 360. 


#6
Apr3007, 08:41 PM

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P: 1,131

Thats for that matt, that makes sense But when you said:



#7
Apr3007, 09:00 PM

P: 1,123

Let me give you a concrete example to help clear what I probably cannot explain too well.
Let Group A be the people a,b,c,d,e,f Group B consist of the people g,h,i,j And Group C consist of the people k,l,m The way you counted above, you counted everything twice. For example: You chose a from Group A, g from Group B, k from Group C, and then say b from Group A. You then counted b from Group A, g from Group B, k from Group C, and then a from Group A. But these are the same groups, so you are double counting, and you can use the same reasoning to show that you double counted every group. That is why you need to divide by 2. 


#8
Apr3007, 09:07 PM

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P: 1,131

Ahh ok thanks very much for that



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