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There is this problem I found in a book on combinations. It goes roughly like this:

There are 4 type A machines and 5 type B machines. Three of the machines are removed from the piles. What is the total number of possibilities that the result contains at least one A machine and at least one B machine?

The answer on the book was given as 1C4 * 2C5 + 2C4 * 1C5 = 70 total possibilities. However, the way I did it was by first enumerating the total methods of choosing one type A and one type B, i.e. 20 different methods, and then multiplying by 7, the number of machines left over after choosing 2. This results in an answer of 140, two times too much.

Why was my answer wrong?

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you differentiate between machines, but actually they are all identical, i.e. indistinguishable –  juampa Dec 29 '12 at 16:48
    
No they're all different as the book states. –  barto Dec 29 '12 at 16:50
    
I mean the ones belonging to the same group. Machines of group A are identical but different from those of group B –  juampa Dec 29 '12 at 16:53
    
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  Davide Giraudo Dec 29 '12 at 17:54

1 Answer 1

up vote 1 down vote accepted

Because if you choose $A_1$ and $B_1$ as the first 2, and later chose $A_3$ to be the last one, it is exactly the same as choosing $A_3$ and $B_1$ as the first 2, and later choosing $A_1$ to be the last one.

This also shows why you counted by twice as much, since every such sequence will be repeated twice.


A simple way to convince yourself, is to consider the scenario where there is 1 type A machine and 2 type B machines, and we want to choose 3 machines, at least 1 of each type. Clearly, there is only 1 way to do so. However, if we were to use your method, for the first 2 machines, there is 1 way to choose a type A machine, 2 ways to choose a type B machine, and finally there is 1 way to choose what's left. This will give $1 \times 2 \times 1$, which is not 1.

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Nice example, second paragraph. –  André Nicolas Dec 29 '12 at 16:34

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