You have won first place in a contest and are allowed to choose 2 prizes from a table that has 6 prizes numbered 1 through 6:
Prizes 1&2 are mobile phones,
Prizes 3&4 are laptops,
Prizes 5&6 are Tv's.
How many different combinations of 2 prizes could you possibly choose without getting a prize from the same category twice?
In this example, we are taking a subset of 2 prizes (r) from a larger set of 6 prizes (n). Looking at the formula, we must calculate “6 choose 2.”
C (6,2)= 6!/(2! * (6-2)!) = 6!/(2! * 4!) = 15 Possible Prize Combinations
The 15 potential combinations are {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, {5,6}
But here we got repeated prizes from same categories:
{1,2}, {3,4}, {5,6}
How can I find the correct number of conditioned combination?
And what about a bigger scales (e.g. 3 categories, each category contained of 4 different items).
Thanks.
Best Answer
There is an easier way to do this. Assume there are $m$ categories each consisting of $n$ items. Then we want to choose $2$ differet categories which can be done in $m \choose 2$ ways. Now, there are $n$ choices for the item to pick in each of the categories, this is done in $n^2$ ways. Overall, we get $${m \choose 2} \times n^2$$ ways.