### Video Transcript

Find π given that 105 πΆπ is
greater than one and ππΆ 103 is greater than one.

To answer this question, we recall
the definition of a combination. The combination ππΆπ represents
the number of different ways to select π objects from a total of π distinct
objects, where the order does not matter.

Given that π and π are positive
integers and π is greater than or equal to π, the formula for a combination is
ππΆπ equals π factorial divided by π minus π factorial multiplied by π
factorial. We are asked to find the value of
π, satisfying both 105 πΆπ is greater than one and ππΆ 103 is greater than
one.

Letβs consider what we know about
π from the combination 105 πΆπ. In this case, π equals 105 and π
equals π. We know in general that π is
greater than zero and π is greater than or equal to π. Therefore, in this case, π is
greater than zero and 105 is greater than or equal to π.

Now we will consider the
combination ππΆ 103. Here, we have π equals π and π
equals 103. According to the definition of a
combination, in this case, π is greater than zero and π is greater than or equal
to 103. We already know that π is a
positive integer. However, we now have more specific
information about the value of π. We have 105 is greater than or
equal to π, and we have π is greater than or equal to 103. Pulling these statements together,
we have π is greater than or equal to 103 and less than or equal to 105.

Letβs consider what positive
integers are in this range. The possible values for π include
103, 104, and 105. Letβs see what happens with ππΆ
103 if π equals 103. If π equals 103, then we have 103
πΆ 103, which equals 103 factorial divided by 103 minus 103 factorial multiplied by
103 factorial. 103 minus 103 factorial simplifies
to zero factorial which equals one. After simplifying the denominator,
we have 103 factorial divided by 103 factorial. Thus, 103 πΆ 103 equals one.

But this is a problem because we
are told that ππΆ 103 must be greater than one. Therefore, π cannot equal 103. Something similar happens with 105
πΆπ, when we let π equal 105. We find that 105 πΆ 105 equals one
as well. Therefore, π cannot equal 105. In fact, any combination of the
form πCπ will always equal one, because there is only one way to select all the
objects.

It seems that 104 is the only
possibility that remains. To be sure 104 satisfies both given
inequalities, we will substitute 104 for π. If π equals 104 and we use the
first combination formula 105 πΆπ, we find that 105 πΆ 104 equals 105, and 105 is
certainly greater than one.

Now we move on, to check π equals
104 in the second combination ππΆ 103. We find that 104 πΆ 103 equals
104. This satisfies the second given
inequality, because 104 is greater than one.

In conclusion, we have shown that
π equals 104, which satisfies both of the conditions given.