### Video Transcript

In this video, we will learn how to
use properties of permutations to simplify expressions and solve equations involving
permutations. First, letβs consider what we know
about permutations. A permutation is an arrangement of
a collection of items where order matters and repetition is not allowed. So we say permutations represent
counting without replacement in which order matters.

We calculate the number of possible
permutations for a collection of items with the formula πPπ equals π factorial
over π minus π factorial. This is the number of ways we
select π elements from a collection of π elements. For example, if we have four P two,
we have a collection of four objects, and we want to know how many different ways we
can select just two of the objects. If in our box we have four letters
and weβre going to choose two randomly, there would be 12 different sets of two, 12
different permutations.

Notice here since weβre dealing
with permutation and order does matter, π΄π΅ is not the same ordering as π΅π΄, which
means we get 12 different ways of choosing two out of four. If we wanna see this with our
formula, that would be four factorial over four minus two factorial, four factorial
over two factorial, and we expand that to be four times three times two times one
over two times one. The two times one in the numerator
and the denominator cancel out, and we get four times three, which is 12.

Before we move on, we should say
that there are quite a few different notations for permutations. The one weβll primarily use looks
like this, πPπ. But you might also see it written
like this where π is a superscript and π is a subscript. Pππ can also mean this where the
π and the π are on the same side or Pπ,π or even P open parentheses π, π
closed parentheses.

Now weβll move on and look at
properties that all permutations share. We use these properties to solve
different types of equations. The first property weβll consider
looks like this. For a permutation πPπ, it will be
equal to π times the permutation of π minus one π minus one. Letβs consider this for six P
three. According to this property, six P
three will be equal to six times the permutation of five choose two. Six P three will be equal to six
factorial over six minus three factorial. And the other side of the equation
will be six times five factorial over five minus two factorial.

If we do a bit of simplifying and
then expand the factorials, on both sides of the equation, three factorial in the
numerator and the denominator cancel out, leaving us with six times five times four
on the left and six times five times four on the right, which will be equal to one
another, which shows us that πPπ is equal to π times π minus one Pπ minus
one.

Another property we want to
consider has to do with π factorial. π factorial plays a big role in
permutations, and π factorial is equal to π times π minus one factorial. We use this property very often to
simplify expressions that have factorials in the numerator and the denominator. We could say four factorial is
equal to four times four minus one factorial, which is four times three
factorial. By expanding both sides of the
equation, we get four times three times two times one equals four times three times
two times one.

And there might be times when we
want to even further extend this property. π factorial is equal to π times
π minus one times π minus two factorial. Using our example of four, thatβs
four factorial is equal to four times three times two factorial. Using this property of factorials,
we can show that zero factorial equals one. Since we know that π factorial
equals π times π minus one factorial, one factorial will be equal to one times one
minus one factorial. One factorial equals one times zero
factorial. Then zero factorial must be equal
to one, which means weβll need to remember that both zero factorial and one
factorial are equal to one.

We wanna develop fluency and using
the definition of permutation and its properties along with these factorials, using
all of that weβll be able to solve equations involving permutations. So letβs consider some
examples.

Solve the following equation for
π: five Pπ is equal to 120.

We know that we calculate πPπ by
taking π factorial over π minus π factorial. In this case, we donβt know the
π-value. We know that our set has five
elements, but we donβt know how many weβre trying to choose. To find π, we need to find out how
many decreasing consecutive integers, starting with five, we should multiply
together to equal 120. We know that π factorial is equal
to π times π minus one factorial. Five Pπ is then equal to five
factorial over five minus π factorial. If we expand the factorial in the
numerator, we get five times four times three times two times one. And we know that five Pπ must be
equal to 120, but five factorial equals 120. So we end up with the equation 120
equals 120 over five minus π factorial.

If we multiply both sides of the
equation by five minus π factorial, we get 120 times five minus π factorial equals
120. And then if we divide both sides by
120, on the left, we have five minus π factorial, and on the right, 120 divided by
120 equals one. This means we need an π-value that
will make five minus π factorial equal to one. Based on the properties of
factorials, we know that there are two places where a factorial equals one, zero
factorial and one factorial, which means five minus π must be zero or five minus π
must be one. If five minus π is zero, then π
equals five. And if five minus π is one, then
π equals four.

Remember, at the beginning, we said
that π would be equal to the number of decreasing consecutive integers beginning
with five we multiply together to get 120. And so there is one other strategy
we can use when we know the π-value of a permutation but we donβt know the
π-value. We know the number of permutations
we can have is 120. And we know that we are beginning
with a set of five. If we start by dividing 120 by
five, we get 24. Now we take 24 and we divide by the
consecutive integer below five. So 24 divided by four equals
six.

Again, weβll take that value and
divide it by the integer that is below four. So six divided by three equals
two. Two divided by the integer below
three, which is two, equals one, which shows us that π could equal four, that four
consecutive decreasing integers beginning with five multiply together to equal
120. Two times three times four times
five does equal 120. However, we could follow the
pattern one final time because one divided by one equals one and one times two times
three times four times five also equals 120, which gives us π equals four or π
equals five.

In our next example, weβll look at
a case when we arenβt given the number of elements we started with.

Find the value of π given that πP
four equals 24.

We know that πPπ equals π
factorial over π minus π factorial and that πP four equals 24, which means we
donβt know how many items our original set held, but we do know weβre selecting four
of them. In cases where we arenβt given the
π-value, it might not seem obvious where we should start, so letβs start by
plugging in what we know into our formula, which will give us πP four equals π
factorial over π minus four factorial. We know that π factorial equals π
times π minus one factorial. We can use this property to rewrite
our numerator so that π factorial is equal to π times π minus one factorial.

But that doesnβt get us any closer
to simplifying the fraction. But we can expand π minus one
factorial, which would be π minus one times π minus one minus one factorial, which
would be π minus two factorial. And we could expand π minus two
factorial to be π minus two times π minus three factorial. And π minus three factorial would
be equal to π minus three times π minus four factorial. This allows us to cancel out the π
minus four factorial in the numerator and the denominator. And since we know that πP four
equals 24, we can say that 24 will be equal to π times π minus one times π minus
two times π minus three. This expression tells us we need
four consecutive integers that multiply together to equal 24. And the π-value will be the
starting point, the largest of those four values.

For smaller numbers like 24, we can
try to solve this with a factor tree. 24 equals two times 12, 12 equals
two times six, and six is two times three. So we can say 24 equals two times
three times four. But remember, for this problem, we
need four consecutive integers that multiply together to equal 24 and not three. One, two, three, and four are four
consecutive integers, and when multiplied together, they equal 24. Remember that our π-value is the
largest value. If we let π equal four, 24 does
equal four times three times two times one and confirms that π equals four.

This strategy worked when we were
dealing with a set of a small size. In our next example, weβll again
look at a case when we donβt know the number in our set, the π-value, but weβre
dealing with a much larger set of permutations.

Find the value of π such that πP
three equals 32,736.

We know the πPπ equals π
factorial over π minus π factorial. And weβve been given that πP three
equals 32,736. We donβt know how many items are in
our original set, but we do know that weβll be choosing three of them. So we can set up an equation that
says π factorial over π minus three factorial. And because we know the property π
factorial equals π times π minus one factorial, we can do some expanding and
simplifying. We can expand π factorial to be π
times π minus one factorial, which is further expanded to π times π minus one
times π minus two factorial, and finally to π minus three factorial. From there, weβre able to cancel
out the terms π minus three factorial in the numerator and the denominator.

We know that πP three equals
32,736. This π times π minus one times π
minus two tells us we need three consecutive integers that multiply together to
equal 32,736. Sometimes when weβre looking for
this π-value, we might use a factor tree. However, with a number this large,
this would be pretty difficult. But there is another strategy we
can use. Since we know weβre looking for
three consecutive integers, we can try to estimate one of those by taking the cube
root of 32,736. When we do that, we get 31.9895
continuing.

And so we can try to divide 32,736
by the integers on either side of this value, 31 or 32. If we start with 32, 32,736 divided
by 32 equals 1,023. And weβll take that 1,023 and
divide it by the integer below 32. 1,023 divided by 31, when we do
that, we get 33. And so, if weβre looking carefully,
we end up with 31, 32, and 33 as factors of 32,736. Weβre saying 32,736 is equal to 33
times 32 times 31. And these are three consecutive
integers that multiply together to equal ~~32,733~~ [32,736]. π will be the largest of these
three values. Weβre saying that if we have a set
of size 33 and weβre choosing three of them, we will end up with 32,736
permutations.

In our final example, weβll look at
how we would simplify a ratio between two different permutations.

Evaluate 123P10 over 122P nine.

We know to calculate πPπ, we have
π factorial over π minus π factorial, which means the numerator is 123 factorial
over 123 minus 10 factorial. And our denominator is 122
factorial over 122 minus nine factorial, which we simplify to be 123 factorial over
113 factorial all over 122 factorial over 113 factorial. If we rewrite this with division,
it looks like this. And we know dividing by a fraction
is multiplying by the reciprocal, which will be 123 factorial over 113 factorial
times 113 factorial over 122 factorial. And then we have 113 factorial in
the numerator and the denominator. So we have 123 factorial over 122
factorial. And we know that π factorial
equals π times π minus one factorial, which means we could rewrite the numerator
as 123 times 122 factorial. The 122 factorial in the numerator
and the denominator cancel out, and this ratio simplifies to 123.

However, we could have saved
ourselves some work by remembering an additional property of permutations. And that is that πPπ is equal to
π times π minus one Pπ minus one. If we notice that 122 equals 123
minus one and nine equals 10 minus one, we could rewrite our numerator as 123 times
122P nine, which would mean we would have a permutation of 122P nine in the
numerator and in the denominator. And so the fraction would reduce to
123. When it comes to evaluating and
simplifying permutations, we always want to look for patterns that can lead us back
to properties which will help us simplify.

Before we finish, letβs review our
key points. The number of permutations of size
π taken from a set of size π is given by πPπ equals π factorial over π minus
π factorial. Some properties we use for solving
problems involving permutations are as follows: π factorial equals π times π
minus one factorial and πPπ equals π times π minus one Pπ minus one.