Math question, counting principles (permutation, combination).?

Question

A club of 40 members wants to elect a president, a vice-president, and treasurer from the other members, and an advisory committee of for people is to be selected. In how many ways can this be done?
Help I don't know how to do it but I do have a sense.

Answer 1

Sorry, the problem is a bit unclear as written.
If I'm reading it correctly, you have: one pres, one VP, one treasurer, and 4 committee members, selected from a population of 40. Is this correct?

Assuming so, let's just go and pick them in that order. So we have 40 choose 1, then 39 choose 1, then 38 choose 1, then 37 choose 4. Thus the answer would be
40 * 39 * 38 * (37 4).

[EDIT: 12/10/07]

The reason that the answer is NOT 40 choose 7 [which I'll denote henceforth as (40 7)] is that not all of the slots are interchangeable. (40 7) would be correct if you were just picking 7 committee members, but here you want to treat as distinct cases the one where Joe is President and Sue is VP from the one where Sue is Prez and Joe is VP.

Answer 2

Well I did it in Excel... 40 things taken 7 at a time.

Combinations

Excel says the answer is 18,643,560

Answer 3

the quantity of attainable alternatives partly A is back to a determination upon 2 of three so the form of attainable combos is "3 want 2" or 3^2. even if this consists of each choosing one million & one million (and a pair of&2,3&3) and replica pairings like one million&2 or 2&one million. So the quantity is (3^2 - 3)/2 (subtract 3 doubles (one million&one million,2&2,3&3), and divide by making use of skill of two to account for combos (one million&2 vs 2&one million)) This leaves 3 for section A For area B its the equivalent element yet "4 come to a determination upon 2" (4^2 - 4)/2 this leaves 6 potentialities for area B. the total will must be the made of those 2, you will ought to get 18 diverse checks I evaluate. ***for sure, no there are 18 unique combos of questions even if there are greater with the aid of fact the you're asking with reference to the ORDER. So redo what i've got above, do merely no longer divide by making use of way of two. area A (3^2 - 3) = 6 area B (4^2 - 4) = 12 form of diverse orderings is an element A * area B = seventy 2!