pbanavara wrote:I guess it's 80 - 4c1*5p2
The first entree can be chosen in 4c1 ways
The two side dishes can be chosen in 5p2 ways because they have to be different - you cannot choose the same dish twice. If this limitation was not there then it would be 5c2 I guess ..
I'll let others chime in - P&C is not my cup of T anyways
- pradeep
We use permutations when order matters and combinations when it doesn't.
Let's say our 5 side dishes are salad, rice, fries, corn and jelly beans. Do we care if we pick salad then corn or corn then salad? No, we end up with the same set of dishes.
Since order doesn't matter, we use combinations.
The combinations formula is:
nCk = n!/k!(n-k)!
in which n = total # of objects and k = # chosen.
In this question, since we're choosing entrees AND side dishes, we apply the forumla twice and then multiply the results. If we were choosing entrees OR side dishes, we'd apply the formula twice and then add the results.
So, as noted above, we have:
4C1 * 5C2
4!/1!3! * 5!/2!3!
4*3*2*1/3*2*1 * 5*4*3*2*1/2*1*3*2*1
4 * 10
40
(Note that I wrote out the entire factorial, but there are lots of short cuts for cancelling out and, on test day, we certainly don't need to write down all those 1s, since they don't have any impact on our calculations.)
