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Today, we're going to do a fun
problem called rooks on a
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chessboard.
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And rooks on a chessboard is a
problem that's going to test
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your ability on counting.
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So hopefully by now in class,
you've learned a few tricks to
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approach counting problems.
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You've learned about
permutations, you've learned
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about k-permutations, you've
learned about combinations,
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and you've learned
about partitions.
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And historically for students
that we've taught in the past
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and many people, counting
can be a tricky topic.
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So this is just one drill
problem to help you get those
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skills under your belt.
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So what does the rooks on a
chessboard problem ask you?
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Well, you're given an 8-by-8
chessboard, which I've tried
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to draw here.
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It's not very symmetrical.
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Sorry about that.
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And you're told that you
have eight rooks.
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I'm sure most of you guys
are familiar with chess.
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But if any of you aren't,
chess is a
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sophisticated board game.
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And one of the types of pieces
you have in this game is
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called a rook.
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And in this particular problem,
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there are eight rooks.
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And your job is to place all
eight rooks onto this 8-by-8
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chessboard.
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Now, you're told in the problem
statement that all
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placements of rooks are
equally likely.
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And you are tasked with finding
the probability that
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you get a safe arrangement.
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So that is to say, you place
your eight rooks on the board.
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What is the probability
that the way you
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placed them is safe?
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So what do I mean by "safe"?
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Well, if you're familiar with
the way chess works, so if you
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place a rook here, it can move
vertically or it can move
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horizontally.
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Those are the only two
legal positions.
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So if you place a rook here
and you have another piece
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here, then this is not a safe
arrangement, because the rook
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can move this way
and kill you.
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Similarly, if you have a rook
here and another piece here,
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the rook can move horizontally
and kill you that way.
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So two rooks on this board are
only safe from each other if
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they are neither in the same
column nor in the same row.
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And that's going to be key for
us to solve this problem.
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So let's see-- where
did my marker go?
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I've been talking a lot,
and I haven't really
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been writing anything.
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So our job is again, to find the
probability that you get a
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safe arrangement.
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So I'm just going to do
"arrange" for short.
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Now, I talked about this
previously, and you guys have
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heard it in lecture.
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Hopefully you remember something
called the discrete
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uniform law.
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So the discrete uniform law is
applicable when your sample
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space is discrete and all
outcomes are equally likely.
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So let's do a quick
check here.
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What is our sample space
for this problem?
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Well, a logical choice would
be that the set of all
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possible outcomes is the set
of all possible spatial
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arrangements of rooks.
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And hopefully it's clear to
you that that is discrete.
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And the problem statement
furthermore gives us that
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they're equally likely.
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So the discrete uniform
law is in fact
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applicable in our setting.
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So I'm going to go ahead and
write what this means.
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So when your sample space is
discrete and all outcomes are
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equally likely, then you can
compute the probability of any
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event, A, simply by counting
the number of outcomes in A
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and then dividing it by the
total number of outcomes in
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your sample space.
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So here we just have to find
the number of total safe
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arrangements and then divide
it by the total number of
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arrangements.
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So again, as you've seen in
other problems, the discrete
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uniform law is really nice,
because you reduce the problem
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of computing probabilities to
the problem of counting.
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And so here's where we're
going to exercise those
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counting skills, as I
promised earlier.
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Now, I would like to start
with computing the
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denominator, or the total
number of arrangements,
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because I think it's a slightly
easier computation.
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So we don't care about the
arrangements being safe.
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We just care about
how many possible
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arrangements are there.
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Now, again, we have eight rooks,
and we need to place
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all of them.
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And we have this 8-by-8 board.
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So pretty quickly, you guys
could probably tell me that
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the total number of square
is 64, because this is
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just 8 times 8.
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Now, I like to approach
problems sequentially.
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That sort of really helps me
think clearly about them.
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So I want you to imagine a
sequential process during
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which we place each rook
one at a time.
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So pick a rook.
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The chessboard is
currently empty.
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So how many squares can you
place that rook in?
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Well, nobody's on the board.
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You can place it in 64 spots.
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So for the first rook that you
pick, there are 64 spots.
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Now, once you place this rook,
you need to place the second
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rook, because again, we're
not done until
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all eight are placed.
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So how many possible
spots are left.
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Well, I claim that there are 63,
because one rule of chess
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is that if you put a piece in
a particular square, you can
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no longer put anything
else on that square.
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You can't put two
or more things.
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So the first rook is occupying
one spot, so there's only 63
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spots left.
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So the second rook has 63 spots
that it could go in.
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Similarly, the third
rook has 62 spots.
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Hopefully you see the pattern.
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You can continue this down.
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And remember, we have to
place all eight rooks.
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So you could do it out
yourself or just
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do the simple math.
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You'll figure out that the
eighth rook only has 57 spots
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that it could be in.
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So this is a good start.
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We've sort of figured out if
we sequentially place each
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rook, how many options
do we have.
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But we haven't combined these
numbers in any useful way yet.
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We haven't counted the number
of total arrangements.
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And this may already be obvious
to some, but it wasn't
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obvious to me when I was first
learning this material, so I
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want to go through
this slowly.
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You have probably heard in
lecture by now about the
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counting principle.
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And what the counting principle
tells you is that
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whenever you have a process that
is done in stages and in
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each stage, you have a
particular number of choices,
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to get the total number of
choices available at the end
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of the process, you simply
multiply the number of choices
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at each stage.
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This might be clear to you,
again, simply from the
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statement, for some of you.
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But for others, it might
still not be clear.
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So let's just take
a simple example.
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Forget about the rook problem
for a second.
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Let's say you're at
a deli, and you
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want to make a sandwich.
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And to make a sandwich, you need
a choice of bread and you
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need a choice of meat.
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So we have a sandwich-building
process,
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and there's two stages.
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First, you have to pick the
bread, and then you have to
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pick the meat.
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So let's say for the choice
of bread, you can
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choose wheat or rye.
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So again, you can always use
a little decision tree--
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wheat or rye.
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And then let's say that
for the meats,
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you have three options.
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You have ham, turkey,
and salami.
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So you can have ham,
turkey, or salami--
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ham, turkey, or salami.
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How many total possible
sandwiches can you make?
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Well, six.
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And I got to that
by 2 times 3.
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And hopefully this makes sense
for you, because there's two
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options in the first stage.
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Freeze an option.
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Given this choice, there's
three options
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at the second stage.
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But you have to also realize
that for every other option
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you have at the first stage, you
have to add an additional
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three options for the
second stage.
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And this is the definition
of multiplication.
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If you add three two times,
you know that's 3 times 2.
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So if you extrapolate this
example to a larger, more
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general picture, you will have
derived for yourself the
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counting principle.
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And we're going to use the
counting principle here to
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determine what the total number
of arrangements are.
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So we have a sequential process,
because we're placing
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the first rook and then the
second rook, et cetera.
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So at the first stage,
we have 64 choices.
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At the second stage,
we have 63 choices.
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At the third stage, we have
62 choices, et cetera.
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And so I'm just multiplying
these numbers together,
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because the counting principle
says I can do this.
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So my claim is that this product
is equal to the total
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number of arrangements.
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And we could stop here, but I'm
going to actually write
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this in a more useful way.
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You guys should have
been introduced to
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the factorial function.
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So you can express this
equivalently as 64 factorial
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divided by 56 factorial.
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And this is not necessary for
your problem solution, but
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sometimes it's helpful to
express these types of
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products in factorials,
because you can see
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cancellations more easily.
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So if it's OK with everybody,
I'm going to erase this work
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to give myself more room.
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So we'll just put our answer for
the denominator up here,
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and then we're going to get
started on the numerator.
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So for the numerator, thanks to
the discrete uniform law,
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we only need to count the number
of safe arrangements.
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But this is a little bit more
tricky, because now, we have
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to apply our definition
of what "safe" means.
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But we're going to use the same
higher-level strategy,
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which is realizing that we can
place rooks sequentially.
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So we can think of it as
a sequential process.
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And then if we figure out how
many choices you have in each
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stage that sort of maintain the
"safeness" of the setup,
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then you can use the counting
principle to multiply all
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those numbers together
and get your answer.
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So we have to place
eight rooks.
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Starting the same way we did
last time, how many spots are
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there for the first rook
that are safe?
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Nobody is on the board yet, so
nobody can harm the first rook
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we put down.
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So I claim that it's just
our total of 64.
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Now, let's see what happens.
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Let's pick a random
square in here.
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Let's say we put our
first rook here.
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Now, I claim a bunch of spots
get invalidated because of the
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rules of chess.
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So before, I told you a rook can
kill anything in the same
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column or in the same row.
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So you can't put a rook here,
because they'll kill each
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other, and you can't
put a rook here.
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So by extension, you can see
that everything in the column
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and the row that I'm
highlighting in blue, it's no
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longer an option.
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You can't place a
rook in there.
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Otherwise, we will
have violated
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our "safety" principle.
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So where can our
second rook go?
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Well, our second rook can go in
any of the blank spots, any
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of the spots that are not
highlighted by blue.
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And let's stare at this
a little bit.
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Imagine that you were to take
scissors to your chessboard
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00:13:06,560 --> 00:13:09,370
and cut along this line
and this line and this
250
00:13:09,370 --> 00:13:10,200
line and this line.
251
00:13:10,200 --> 00:13:14,190
So you essentially sawed off
this cross that we created.
252
00:13:14,190 --> 00:13:18,470
Then you would have four
free-floating chessboard
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00:13:18,470 --> 00:13:23,070
pieces-- this one, this one,
this one, and this one.
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00:13:23,070 --> 00:13:27,650
So this is a 3-by-4 piece,
this is 3-by-3, this is
255
00:13:27,650 --> 00:13:30,020
4-by-3, and this is 4-by-4.
256
00:13:30,020 --> 00:13:33,600
Well, because you cut this part
out, you can now slide
257
00:13:33,600 --> 00:13:36,000
those pieces back together.
258
00:13:36,000 --> 00:13:39,950
And hopefully you can convince
yourself that that would leave
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00:13:39,950 --> 00:13:43,390
you with a 7-by-7 chessboard.
260
00:13:43,390 --> 00:13:48,650
And you can see that the
dimensions match up here.
261
00:13:48,650 --> 00:13:53,000
So essentially, the second rook
can be placed anywhere in
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00:13:53,000 --> 00:13:55,970
the remaining 7-by-7
chessboard.
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00:13:55,970 --> 00:14:00,280
And of course, there are 49
spots in a 7-by-7 chessboard.
264
00:14:00,280 --> 00:14:03,430
So you get 49.
265
00:14:03,430 --> 00:14:07,150
So let's do this experiment
again.
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00:14:07,150 --> 00:14:11,860
Let me rewrite the reduced
7-by-7 chessboard.
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00:14:11,860 --> 00:14:14,540
You're going to have to forgive
me if the lines are
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00:14:14,540 --> 00:14:16,680
not perfect--
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00:14:16,680 --> 00:14:22,760
one, two, three, four, five,
six, seven; one, two, three,
270
00:14:22,760 --> 00:14:23,210
four, five, six, seven.
271
00:14:23,210 --> 00:14:24,630
Yep, I did that right.
272
00:14:24,630 --> 00:14:32,030
And then we have one, two,
three, four, five, six, seven.
273
00:14:32,030 --> 00:14:36,800
That's not too bad for
my first attempt.
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00:14:36,800 --> 00:14:39,980
So again, how did I get this
chessboard from this one?
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00:14:39,980 --> 00:14:43,210
Well, I took scissors and I cut
off of the blue strips,
276
00:14:43,210 --> 00:14:46,690
and then I just merged the
remaining four pieces.
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00:14:46,690 --> 00:14:50,160
So now, I'm placing
my second rook.
278
00:14:50,160 --> 00:14:54,350
So I know that I can place my
second rook in any of these
279
00:14:54,350 --> 00:14:58,770
squares, and it'll be
safe from this rook.
280
00:14:58,770 --> 00:15:00,830
Of course, in reality, you
wouldn't really cut up your
281
00:15:00,830 --> 00:15:01,390
chessboard.
282
00:15:01,390 --> 00:15:05,470
I'm just using this as a visual
aid to help you guys
283
00:15:05,470 --> 00:15:08,220
see why there are 49 spots.
284
00:15:08,220 --> 00:15:11,020
Another way you could see 49
spots is literally just by
285
00:15:11,020 --> 00:15:15,890
counting all the white squares,
but I think it takes
286
00:15:15,890 --> 00:15:17,720
time to count 49 squares.
287
00:15:17,720 --> 00:15:20,501
And this is a faster
way of seeing it.
288
00:15:20,501 --> 00:15:23,930
So you can put your second
rook anywhere here.
289
00:15:23,930 --> 00:15:26,340
Let's actually put in the
corner, because the corner is
290
00:15:26,340 --> 00:15:27,590
a nice case.
291
00:15:27,590 --> 00:15:30,700
If you put your rook in the
corner, immediately, all the
292
00:15:30,700 --> 00:15:35,360
spots in here and all the spots
in here become invalid
293
00:15:35,360 --> 00:15:38,230
for the third rook, because
otherwise, the rooks can hurt
294
00:15:38,230 --> 00:15:40,070
each other.
295
00:15:40,070 --> 00:15:45,130
So again, you'll see that if you
take scissors and cut off
296
00:15:45,130 --> 00:15:47,940
the blue part, you will have
reduced the dimension of the
297
00:15:47,940 --> 00:15:49,730
chessboard again.
298
00:15:49,730 --> 00:15:52,820
And you can see pretty quickly
that what you're left with is
299
00:15:52,820 --> 00:15:55,100
a 6-by-6 chessboard.
300
00:15:55,100 --> 00:16:02,820
So for the third rook, you get a
6-by-6 chessboard, which has
301
00:16:02,820 --> 00:16:06,020
36 free spots.
302
00:16:06,020 --> 00:16:08,880
And I'm not going to insult
your intelligence.
303
00:16:08,880 --> 00:16:11,280
You guys can see the pattern--
304
00:16:11,280 --> 00:16:13,130
64, 49, 36.
305
00:16:13,130 --> 00:16:16,310
These are just perfect
squares decreasing.
306
00:16:16,310 --> 00:16:21,740
So you know that the fourth
rook will have 25 spots.
307
00:16:21,740 --> 00:16:24,840
I'm going to come over here
because I'm out of room.
308
00:16:24,840 --> 00:16:27,970
The fifth rook will
have 16 spots.
309
00:16:27,970 --> 00:16:31,230
The sixth rook will
have nine spots.
310
00:16:31,230 --> 00:16:33,900
The seventh rook will
have four spots.
311
00:16:33,900 --> 00:16:37,780
And the eighth rook will
just have one spot.
312
00:16:37,780 --> 00:16:39,360
And now, here we're going
to invoke the
313
00:16:39,360 --> 00:16:40,880
counting principle again.
314
00:16:40,880 --> 00:16:43,930
Remember the thing that I just
defined to you by talking
315
00:16:43,930 --> 00:16:46,680
about sandwiches.
316
00:16:46,680 --> 00:16:49,240
And we'll see that to get
the total number of safe
317
00:16:49,240 --> 00:16:50,860
arrangements, we can
just multiply
318
00:16:50,860 --> 00:16:52,910
these numbers together.
319
00:16:52,910 --> 00:16:54,750
So I'm going to go ahead
and put that up here.
320
00:16:54,750 --> 00:17:02,360
You get 64 times 49 times
36 times 25 times 16
321
00:17:02,360 --> 00:17:05,079
times 9 times 4.
322
00:17:05,079 --> 00:17:07,810
And in fact, this
is our answer.
323
00:17:07,810 --> 00:17:10,859
So we're all done.
324
00:17:10,859 --> 00:17:15,630
So I really like this problem,
because we don't normally ask
325
00:17:15,630 --> 00:17:18,690
you to think about different
spatial arrangements.
326
00:17:18,690 --> 00:17:22,339
So it's a nice exercise, because
it lets you practice
327
00:17:22,339 --> 00:17:27,069
your counting skills in a
new and creative way.
328
00:17:27,069 --> 00:17:31,220
And in particular, the thing
that we've been using for a
329
00:17:31,220 --> 00:17:33,730
while now is the discrete
uniform law.
330
00:17:33,730 --> 00:17:36,730
But now, I also introduced
the counting principle.
331
00:17:36,730 --> 00:17:39,310
And we used the counting
principle twice--
332
00:17:39,310 --> 00:17:41,910
once to compute the numerator
and once to compute the
333
00:17:41,910 --> 00:17:44,210
denominator.
334
00:17:44,210 --> 00:17:49,480
Counting can take a long time
for you to absorb it.
335
00:17:49,480 --> 00:17:52,850
So if you still don't totally
buy the counting
336
00:17:52,850 --> 00:17:54,450
principle, that's OK.
337
00:17:54,450 --> 00:17:59,440
I just recommend you do some
more examples and try to
338
00:17:59,440 --> 00:18:02,280
convince yourself that it's
really counting the right
339
00:18:02,280 --> 00:18:04,020
number of things.
340
00:18:04,020 --> 00:18:07,510
So counting principle is
the second takeaway.
341
00:18:07,510 --> 00:18:10,520
And then the other thing that
is just worth mentioning is,
342
00:18:10,520 --> 00:18:12,980
you guys should get really
comfortable with these
343
00:18:12,980 --> 00:18:19,960
factorials, because they will
just show up again and again.
344
00:18:19,960 --> 00:18:21,960
So that's the end of the
problem, and I'll
345
00:18:21,960 --> 00:18:23,210
see you next time.
346
00:18:23,210 --> 00:18:27,134