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

i would say 40

4C1 * 5C2 = 4 * 10 = 40

Its a combination prob since same 1 entree same 2 dish is the same meal no matter u pick the entree first or the 2 dishes first

40

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.)

What's a good example of when order does matter for use of the Permutation formula N!/(n-r)!

Say you have to pick pairs of letters from the three letters A,B,C.

- If order matters, then AB and BA are two different pairs (ie it matters whether A occupies the first slot or the second). So we have 3p2 = 3!/(3-1!) = 6 possibilities {AB, BA, AC, CA, BC, CB}

- If order doesn't matter, then AB and BA are the same pair. So we have 3c2=3!/(2!*(3-2)!) = 3 possibilities {AB/BA, AC/CA, BC/CB}.