Here is a better definition of the problem:
Consider a set of 8 unique items. 3 of these items need to be chosen. Additionally there are 2 items in this set that can not both be chosen simultaneously. The question is: how many ways combinations of these items can be made which satisfy these rules.
My approach was as follows:
I can either choose 3 items which exclude the 2 that have the special requirement. Therefore there are 6 possible items for the first choice, 5 for the 2nd choice, and 4 for the 3rd choice. so 6x5x4=120 permutations. However, since the order of these choices do not matter, there are 3! ways to order 3 objects. I divide the number of permutations by the number of ways to order my items to get 20 possible ways to choose 3 items that do not include the 2 items with special requirements.
I then consider the combinations which include either one of the special items, but not both. Therefore 2 of the choices must be from the set of normal items and 1 of the choices must be from the set of special items. Thus there are 6 possible choices for the 1st choice, 5 possible choices for the 2nd choice, and only 2 possible choices for the 3rd choice. So, 6*5*2=60 permutations. Again, since order does not matter there are 3! ways to arrange three items, and we are left with 10 possible ways to choose 2 items such that only one of these items has the special requirement.
Obviously the total number of combinations would be the sum of these two cases, so 10+20=30 combinations.
However, this answer does not agree with the one provided and its compelling explanation. If I consider all the ways 3 items can be chosen from a set of 8 items without any special conditions. This number is (3 C 8) = 56 combinations. Now if I consider how many of these combinations would not be allowed and subtract that from the 56 combinations I just calculated I would receive the total number of combinations with the special requirement.
The only combinations which would not be allowed are those in which both the special requirement items are present. Hence, there is 1 possibility for the first choice (the first special requirement item), 1 possibility for the second choice (the second special requirement irem), and 6 possibilities for the 3rd choice. So 1x1x6=6 unallowed combinations. This would yield 50 combinations.
Could somebody point out why these two results are not in agreement?
Best Answer
Tip: If you are counting selections of items from sets, go straight to binomial coefficients and do not pass confusion about "order not being important".
${}^n\mathrm C_r$ counts selections of $r$ items from $n$. Then you just need to puzzle out whatfore you are selecting from wherefore.
To count selections of three items from eight when two of the items are incompatible, you:
Count the selections of any three from all eight, minus selections of both incompatible and one from the other six.$$\require{cancel}{}^8\mathrm C_3-\bcancel{{}^2\mathrm C_2}{}^6\mathrm C_1 = \dfrac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}-\frac{6}{1}$$
Count the selection of any three from the other six, plus the selections of one from the incompatibles and two from the other six. $${}^6\mathrm C_3+{}^2\mathrm C_1{}^6\mathrm C_2~=~\frac{6\cdot 5\cdot 4}{3\cdot 2\cdot 1}+\frac{2}{1}\cdot\frac{6\cdot 5}{2\cdot 1}$$