### Video Transcript

In this video, we will learn how to
find the factorial of any number π, which is the product of all integers less than
or equal to π and greater than or equal to one. We will also learn how to find
factorials to solve problems and solve problems containing permutations and
factorials. Letβs begin by looking at a written
and algebraic definition of a factorial.

The factorial of a positive integer
π is the product of all the positive integers less than or equal to π. We use the notation π followed by
an exclamation mark, which is read as π factorial. π factorial is equal to π
multiplied by π minus one multiplied by π minus two and so on multiplied by two
multiplied by one. We also define the factorial of
zero to be equal to one; that is, zero factorial equals one. We also know that for any integer
π greater than or equal to one, π factorial is equal to π multiplied by π minus
one factorial. We can see that from the general
rule for π factorial above. This property will be really useful
when solving more complicated problems in this video. However, we will begin by solving a
straightforward problem.

Evaluate four factorial.

We recall that the factorial of any
positive integer π is the product of all the positive integers less than or equal
to π. This means that π factorial is
equal to π multiplied by π minus one multiplied by π minus two and so on all the
way down to one. Four factorial is therefore equal
to four multiplied by three multiplied by two multiplied by one. Four multiplied by three is equal
to 12. Multiplying this by two gives us
24, and multiplying 24 by one is also 24. We can multiply the integers four,
three, two, and one in any order to give us an answer of 24. Therefore, four factorial equals
24.

In our next question, we will solve
a more complicated problem.

Simplify the expression six
factorial over four factorial minus 27 factorial over 28 factorial. Give you answer as a fraction.

We recall that π factorial is
equal to π multiplied by π minus one multiplied by π minus two and so on all the
way down to one. This means that we could calculate
six factorial by multiplying six by five, by four, by three, by two, and by one. Whilst this wouldnβt be too
difficult for the first fraction, working out 27 factorial and 28 factorial in this
way would be very time-consuming. We can therefore recall another
rule for calculating π factorial. It is equal to π multiplied by π
minus one factorial. We could also see that π factorial
is equal to π multiplied by π minus one multiplied by π minus two factorial. This allows us to rewrite six
factorial as six multiplied by five multiplied by four factorial.

The first term in our question
simplifies to six multiplied by five multiplied by four factorial all divided by
four factorial. As the four factorials cancel, we
are left with six multiplied by five. This is equal to 30. We can use this method again for
the second fraction as 28 factorial is equal to 28 multiplied by 27 factorial. This time, the 27 factorials
cancel, leaving us with one over 28. We need to subtract one over 28 or
one twenty-eighth from 30. This is equal to the mixed number
29 and twenty-seven twenty-eighths.

In order to write our answer just
as a fraction, we will need to convert this into an improper or top-heavy
fraction. We do that by first multiplying the
whole number 29 by the denominator 28. This is equal to 812. We then add the numerator of 27,
giving us 839. The expression six factorial over
four factorial minus 27 factorial over 28 factorial is equal to the fraction 839
over 28.

In our next question, weβll use our
knowledge of factorials to solve an algebraic equation.

Find the solution set of one over
π plus seven factorial plus one over π plus eight factorial is equal to 256 over
π plus nine factorial.

There are lots of ways of starting
this question. One way would be to multiply both
sides by π plus nine factorial. Multiplying the first term by π
plus nine factorial gives us π plus nine factorial over π plus seven
factorial. The second term on the left-hand
side becomes π plus nine factorial over π plus eight factorial. As π plus nine factorial divided
by π plus nine factorial is equal to one, the right-hand side becomes 256.

We recall that π factorial is
equal to π multiplied by π minus one factorial. This means that π plus nine
factorial can be rewritten as π plus nine multiplied by π plus eight factorial or
π plus nine multiplied by π plus eight multiplied by π plus seven factorial. The first term therefore simplifies
to π plus nine multiplied by π plus eight. The second term simplifies to π
plus nine. π plus nine multiplied by π plus
eight plus π plus nine is equal to 256.

We can distribute the parentheses
or expand the brackets using the FOIL method. Multiplying the first terms gives
us π squared, the outer terms eight π, the inner terms nine π, and the last terms
72. We now have an equation π squared
plus eight π plus nine π plus 72 plus π plus nine is equal to 256. By collecting like terms, the
left-hand side simplifies to π squared plus 18π plus 81. We can then subtract 256 from both
sides of the equation such that π squared plus 18π minus 175 is equal to zero.

We can now factor this quadratic
expression into two sets of parentheses. The first term in each of them is
π, as π multiplied by π is π squared. The second terms will have a sum of
18 and a product of negative 175. 25 multiplied by seven is 175. This means that positive 25
multiplied by negative seven is negative 175. The numbers positive 25 and
negative seven also have a sum of 18. As this expression equals zero, one
of our parentheses must be equal to zero. This means that either π is equal
to negative 25 or π equals seven. Factorials are only defined for
nonnegative integers. This means we can discard the
solution π equals negative 25. The value of π that satisfies the
equation is π equals seven. The solution set of the equation
just contains the number seven.

Our last question involves
permutations and factorials. Before moving on to this, we will
recall the definition of a permutation. A permutation is a rearrangement of
a collection of items. It is defined as the number of ways
we can order π elements from a set of π elements with no repetition. We write this as subscript π
capital P subscript π. It is just read as πPπ. It is defined by π factorial
divided by π minus π factorial. For example, nine P five is equal
to nine factorial divided by nine minus five factorial. This simplifies to nine factorial
divided by four factorial.

Using the property that π
factorial is equal to π multiplied by π minus one factorial, nine factorial is
equal to nine multiplied by eight multiplied by seven multiplied by six multiplied
by five multiplied by four factorial. As the four factorial cancels, we
can then multiply the five integers nine, eight, seven, six, and five, giving us
15,120.

We will now answer a question
involving permutations and factorials.

Given that πPπ is equal to 504
and π factorial equals six, find the values of π and π.

We recall that when dealing with
permutations, πPπ is equal to π factorial over π minus π factorial. We are also told in the question
that π factorial is equal to six. This is a factorial we can work out
quite easily. We know that three multiplied by
two multiplied by one is equal to six. This means that three factorial is
equal to six. Our value of π is therefore equal
to three.

We were told that πPπ is equal to
504. Therefore, πP three equals
504. Substituting π equals three into
our general formula for permutations, we have π factorial divided by π minus three
factorial is equal to 504. π factorial can be rewritten as π
multiplied by π minus one multiplied by π minus two multiplied by π minus three
factorial. Dividing this by π minus three
factorial, we are left with π multiplied by π minus one multiplied by π minus
two. This is equal to 504. We are therefore looking for three
consecutive integers that multiply to give us 504.

We could try to work out these
values using trial and improvement. However, there is a trick we can
use to find three consecutive integers that multiply to give a number. We begin by taking the cube root of
that number. The cube root of 504 is 7.958 and
so on. Well, how does this help us? This is not an integer. What we can do is take the integers
on either side of this number. In this case, these are seven and
eight. We can now divide our number, in
this case 504, by the two integers. 504 divided by seven is equal to
72. Therefore, seven multiplied by 72
is 504. We now divide 72 by the second
integer eight. 72 divided by eight is equal to
nine. Therefore, eight multiplied by nine
is 72.

We have now written 504 as the
product of three consecutive integers. These integers are seven, eight,
and nine, which correspond to π minus two, π minus one, and π, respectively. Our value of π is nine. If πPπ equals 504 and π
factorial is six, π equals three and π equals nine. There are slightly different
methods we couldβve used to calculate π and then calculate π. Letβs firstly go back to the fact
that we know that π factorial is equal to six.

When trying to find an unknown
integer given its factorial, we can divide by consecutive positive integers. This means that we begin by
dividing our number six by one. This is equal to six. We then divide by the next positive
integer two. Six divided by two is equal to
three. We then divide by the next positive
integer three, and three divided by three is equal to one. As six divided by one divided by
two divided by three is equal to one, then six is also equal to three multiplied by
two multiplied by one. We have once again proved that
three factorial is equal to six. Therefore, π equals three.

When we got to the stage that π
multiplied by π minus one multiplied by π minus two was equal to 504, we couldβve
used our knowledge of prime factors to try and rearrange them into consecutive
integers. 504 is equal to two multiplied by
252. 252 is equal to two multiplied by
126. Repeating this process by dividing
by prime numbers, we can write 504 as a product of its prime factors. 504 is equal to two multiplied by
two multiplied by two multiplied by seven multiplied by three multiplied by
three. This can be rewritten as two cubed
multiplied by seven multiplied by three squared. Two cubed is equal to eight, and
three squared is equal to nine. Once again, we have three
consecutive integers seven, eight, and nine such that seven is equal to π minus
two, eight is equal to π minus one, and nine is equal to π.

We will now summarize the key
points from this video. The factorial of a positive integer
π is defined as the product of all positive integers less than or equal to π. The key property of the factorial
is that π factorial is equal to π multiplied by π minus one factorial. We can use this to simplify
expressions using factorials and also solve factorial equations. When trying to find an unknown
integer given its factorial, we divide by consecutive positive integers until we
reach an answer of one. Finally, we saw that the number of
permutations of size π taken from a set of size π is given by πPπ is equal to π
factorial divided by π minus π factorial.