A committee of 4 is to be selected from a group of 30 people. How many different committees are possible given the following conditions?

Question

1. There is no distinction between the responsibilities of the members 
2. One person is the chair and the rest are general members 
3. Each person is assigned to a particular duty based on what order they picked

Answer 1

1. With no distinction between the committee members, we're given 30 people and want to choose 4: the answer is ₃₀C₄ = 30!/[(30-4)!*4!] = 27405

2. If instead we first want to select one to be chair and 3 to be general members, I can think of two nice approaches; while we only need one, it's always fun to see how the different methods work out to the same value.

1. First choose 1 of the 30 people to be the chair; there are ₃₀C₁ ways to do this.

Then choose 3 of the remaining 29 people to be the general members; there are ₂₉C₃ = 29!/[(29-3)!*3!] ways to do this.

Then we have a total of 30 * 29!/[(29-3)!*3!] possible committees with a leader.

2. But what if instead, we first choose all 4 people for the committee as in the first prompt, which we already established there are ₃₀C₄ = 30!/[(30-4)!*4!] ways to do.

Then, let's select one of the four to be chair: there are ₄C₁ = 4 ways to do this.

We get a total of 30!/[(30-4)!*4!] * 4 possible committees with a leader.

Hm... our answers look different. Did we do something wrong?

Nope! We just need to do a little arithmetic manipulation of the expressions.

For the first expression, note that 30*29! = 30! - our numerators now match up.

For the second, note that while there's a 4! in the denominator, we then multiply the fraction by 4... cancel these 4/s out and now we have 3!, matching the righthand part of the other expression.

But what about the (29-3)! vs. (30-4)!, huh? Well, notice that 29-3 = 30-4 = 26 either way :)

So these processes are arithmetically equivalent, and both evaluate to 109620 different ways to select this type of committee.

3. For our final process, let's pick each member one by one, as we're told order matters.

The classical way to do this is just say "ok then ₃₀P₄ is our answer" but I'd like to give a little more insight into what's going on with that.

I think the easiest way to understand selecting these 4 committee members out of the 30, in order, is to:

1. pick a first one; there's 30 ways to do this
2. now, pick a second; with 1 person already picked, there's 29 left we could choose here
3. 28 options for our third member
4. and 27 for our fourth

... gives us a total of 30*29*28*27 ways of selecting these committee members with specific tasks assigned based on the order in which they're selected.

What does this have to do with ₃₀P₄? Well, the formula you use to compute that is 30!/(30-4)! - notice that this multiplies everything from 30 down to 1 together, and then immediately cancels everything from 26 down back out. This is just a nicer way of writing "multiply all the integers from 30 down to 27 together", or to put it even more clearly: "multiply 4 descending consecutive integers together, starting at 30".

Another way I sometimes like to think of it - and this will give us insight toward where the formula for Choose comes as well - is that if we wanted to be really inefficient, we could organize ALL THIRTY people in a row: 30*29*28*27*26*25*...*4*3*2*1, and then immediately undo the organization of all but the first four of those by dividing out by how many ways we could have organized the last 30-4=26 of them: 26*25*...*4*3*2*1. This exactly gets us 30!/(30-4)! with some insight into *why* that's what we're doing: we're choosing one person at a time 'til we've organized all of them, and then we're disorganizing all the ones we didn't actually want organized.

This sounds silly and inefficient at first, but the reason I explain it this way is because now we can see how we get from the Permutation formula to the Choose formula: if I wanted those same 4 out of 30 people but I don't want them in order either, at this point we've seen the way to forget the order on n people is to just divide by the number of ways to order those things: n! is exactly the additional term in the denominator when you compare the formulas for _nP_r = n!/(n-r)! vs. _nC_r = n!/[(n-r)!*r!]. In each case, we can see the formula as coming from just putting everything in order, and then disordering the parts that we don't want ordered anymore. Disordering the first 4 and the last 26 still leaves us with those two distinct groups: we haven't disordered all 30, but split it into a first group of 4 and a second group of 26.

... this lets us get formulas without too much more thought, for things like "how many ways are there to split a class of 12 students into four groups of 3?" Now we've got more than 2 categories, but the same organize-than-disorganize lets us deal with that just fine: the answer in this case would be 12!/[3!*3!*3!*3!].

The really classic question of this "multichoose" family is "how many ways can you rearrange the letters in MISSISSIPPI?" We just look at how many copies of each letter there are: 1 M, 4 I's, 4 S's, and 2 P's for a total of 1+4+4+2 = 11 letters. If we arranged them all without recognizing some of them were the same, we'd have 11! total rearrangements; if we then divide through by the 4! ways you can switch the 4 I's around while keeping everything else still, then by the 4! ways you can do likewise with the 4 S's, and the 2! ways we can do so with the 2 P's, we get a total of 11!/[4!*4!*2!] ways to rearrange the letters. Sometimes we write it as 11!/[1!*4!*4!*2!] so the one M doesn't feel left out, even though multiplying by 1 doesn't change the value... it gives us further insight into where the computation came from.