Combinations

Question

In how many ways can one divide twelve books
1) into four equal bundles
2) equally among four boys.

elementbrdr · Answer

Just curious, was the language you used in your post the language from the actual question? I understand that multiplying 12c3*9c3*6c3*3c3 by 1/(4!) removes the 4! ways that the 4 combinations of 3 books could be allocated. But it took me a while to see that the question was actually asking for that modification to the calculation. Was the original question clearer? Obviously fluke understood what was going on, and it makes sense in retrospect, but it took me a little while to understand what the questions was calling for.

KarishmaB · Answer

When the question says that you need to make n groups/bundles/teams that are not distinct, you need to divide by n!. If the groups/bundles/teams are distinct then you do not divide by n!
e.g. 
In part 2 of the question above, there are 4 boys. Obviously they are distinct. So you have to divide the books into 4 distinct bundles.
In part 1, say you have 4 identical folders. You need to put 3 books in each folder. Since the bundles are identical, you will divide by 4! to reduce the number of combinations (compared with the case above).
Since the question says 'four bundles', we assume that they are not distinct. On the other hand, if they were numbered or had positions e.g. 'Bundle 1', 'Bundle 2' etc, then they would be distinct and we would be back to case 2.

gijoedude · Answer

Why did you raise 3! to the power of 4?

elementbrdr · Answer

The expression for counting the number of ways the books can be allocated equally among 4 boys is 12c3*9c3*6c3*3c3. Each of the four combinations in the expression reflects the number of ways that the remaining books can be distributed to one of the boys who has not yet received books. After a boy receives a bundle of 3 books, there are 3 fewer books remaining to be given to the next boy, so each successive combination expression reflects 3 fewer elements. When you expand the combination expression (by applying the definition of a combination, which is n!/k!(n-k)!), you get the following:  12!/3!(12-3)! * 9!/3!(9-3)! * 6!/3!(6-3)! * 3!/3!(3-3)! 
The (n-k) in the denominator of each combination expression simplifies and cancels the numerator of the next combination expression, except for (3-3)!, which equals 1 (0! = 1 by definition). So you're left with the following:
 12!/3! * 1/3! * 1/3! *1/3! 
This simplifies to 12!/3!^4

maheshsrini · Answer

In a similar way can you explain Why is the answer to the 1st question divided by 4!?

gladvijay · Answer

after dividing them into 4 bundles.. you can arrange them in 4! ways.. by dividing you eliminate repetitions.. the boys are unique so you dont have to divide there..

Combinations

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