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Welcome back.
Or maybe you didn't even watch the last video, but anyway, we
are on problem number 15.
The Acme Plumbing Company will send a team of three plumbers
to work on a certain job.
The company has four experienced
plumbers and four trainees.
If a team consists of one experienced plumber and two
trainees, how many such teams are possible?
So we're going to send three plumbers, right?
Three plumbers.
So it's one, two, three.
And one of them has to be experienced, right?
And then these are
inexperienced, these are trainees.
So this is the team, one experienced and two trainees.
And now we're going to have to pick from a set.
Let's say my four experienced plumbers are
plumbers A, B, C, D.
These are experienced.
And let's say my trainees are E, F, G, H.
So these four are trainees.
So how many different experienced plumbers can I
have heading up the team?
Well, there's four possibilities, right?
So there's four different people I can stick in here.
Four choices, or four possibilities.
Now, independent of who I put into the experienced slot, how
many trainees-- let's say this is trainee slot one and
trainee slot two, not that there's a difference-- so now
if I'm picking who can go into trainee slot one, how many
choices do I have?
Well, I have four choices because none of them have been
staffed yet.
So I can put four people here.
Not four people, I'm putting one person, but I can pick one
of four, so there's four possibilities here.
Once I've already filled trainee slot one, one of these
people are going to disappear.
Then the set is going to be one less, right?
There's only going to be three trainees left once
I've filled slot one.
So how many trainees can I put in slot two?
Well, there's only three possibilities I could put
because I already stuck one in slot one.
So there's four possible experienced people, four
possible choices for trainee in slot one.
And then since one trainee will already have been staffed
in slot one, there's only three choices for slot two.
So it's 4 times 4 times 3.
So what is that?
4 times 12, which is 48 possibilities.
Next problem.
And the trick here is just realizing that once you staff
one of the trainees, you only have 3 left that you can put
in the other one, in the second trainee slot.
Next problem.
Problem 16.
All right, we've got some circles going on here.
Let me draw a circle.
So they have one circle like that.
And they have a bigger circle that looks something like--
oh, I don't know if I-- no, let me undo that.
Edit, undo.
OK, let me see if I can make it more even.
That's pretty good.
OK, let me fill it in with a pleasant color.
All right.
The figure above consists of two circles that
have the same center.
If the shaded area is 64 pi square inches and the smaller
circle has a radius of 6 inches, so that distance right
there is 6, what is the radius in inches
of the larger circle?
So, we want to figure out radius of the larger circle.
Radius of the larger circle I'll do in this green color.
Let's call that x.
So what's going to be the area of the shaded circle?
How would we figure it out if we knew the radius of the
larger circle?
What would be the area of the large circle?
Area of the large circle minus the area of the small circle
is going to equal 64 pi, right?
Because if the whole thing was filled in in mauve or whatever
color this is, then it would just be the area of the large
circle, but we're subtracting out the area of the small
circle to get the mauve region.
So what's the formula for the area of the large circle?
It's pi times its radius squared.
So we already set that radius to be equal to x, so it's pi x
squared-- and that's what we're trying to figure out in
this problem-- minus the area of the small circle.
What's the area of the small circle?
Well, its radius is 6 so it's pi r squared, so
it's pi times 6 squared.
And that is going to be equal to 64 pi.
And just to simplify things, we can just divide both sides
by pi just to get them out of the way.
But you don't have to do that, that just simplifies things
for my simple mind.
So you get x squared minus 36 is equal to 64.
Add 36 to both sides, you get x squared is equal to 100.
So x could be equal-- if you just solved it, it would be
plus or minus 10, and of course, a radius can't be
negative, so it's plus 10.
So x is 10, or the radius of the larger circle is 10.
Next problem.
Problem 17.
If p, r, and s are three different prime numbers
greater than two-- let me just write them down, p r, and s--
three different prime numbers greater than two and n is
equal to p times r times s, how many factors including one
and n does n have?
How many factors?
This is a good problem.
This is essentially the prime factorization of n, right?
And we want to know how many positive factors including 1
and n, right?
So what are all of the factors here?
So if we just made a factor tree-- let's make a
combination of all the numbers that when I multiply the pair
can equal n.
So one times n is equal to n.
Another number that would equal n is pr times s would
also equal n, right?
Because obviously p times r times s is n, so p times r
times s is n, but I'm saying p times r is a number, right?
Hopefully that makes some sense.
Similarly, we could say that p times rs is equal to n.
Same logic.
I'm saying r times s is a number, right?
Some number.
And then finally we could say that sp times r, with sp as a
number, is also equal to n, right?
So all of these numbers that I said-- 1, 2, 3, 4, 5, 6, 7,
8-- all of these numbers are factors of n.
So that's how many factors there are, there are eight
factors including 1 and n.
If this seemed a little bit abstract to you, try it out
with some numbers.
Actually, that's the best way you could do it if this was
too abstract.
You could say, what if p is 3, r is 5, and s is 7.
Then what is n?
Well, n is going to be 15 times 7.
So n is 15 times 7, what is that?
70 plus 35, n is 105.
So if n is 105, this is 1 times 105.
pr, this is 15 times 7.
This would be 3 times 35.
Is that right?
Oh no, sorry, 85.
No, sorry.
If 5 times 7 is 35 times 3, oh right, it's 105, right?
So we would write 3 times 35, and sp is 21 times 5.
So those would be the actual numbers.
So hopefully that makes a little bit more sense to you.
But this is the way you could do it in a general way.
It looks like I'm running out of time, so I will do number
18 in the next video.
But this is a little bit of a non-intuitive problem.
Just remember, this is the prime factorization, and then
you can construct its-- what I call the normal
factorization-- you can construct that by just taking
combinations of the prime numbers.
And hopefully the actual numbers make a little bit more
sense to you.
See you in the next video.