The binomial coefficient is equal to , and since , we get that
. By
the well-known formula for the sum of the first positive integers, we have ,
and so the answer to (i) is .
For (ii), we will consider the possible values of . Note that since and are non-negative and , we must have that .
If and , then can be any of the integers from through , for a total of possibilities.
If , then can be any of the integers from through , for a total of possibilities.
In general, if for some
with , then can be any integer from through . There are a total of integers from through .
Therefore, the number of pairs is , but this
is just the sum
written in reverse order. By part (i), we conclude that the number of
non-negative integer pairs
with is .
For (iii), observe that if and
are non-negative integers with
, then there is exactly
one non-negative integer for
which . Therefore, the
number of non-negative integer pairs with is exactly the same as the
number of non-negative integer triples with .
Suppose the
distinguishable cups are labelled Cup , Cup , and so on, up to Cup . Suppose the balls are placed in the cups. If we let be equal to the number of balls in
Cup (noting that it is possible
that ), then we will have
since there
are balls in total. On the other
hand, if we have a non-negative integer solution to , then if we place
balls in Cup for each , then we will have distributed exactly
balls among the cups. Finally, observe that every
placement of balls in the cups leads to a distinct solution to
, and every
solution to this equation gives a distinct way to distribute balls in the cups. This shows that the answers to
questions (i) and (ii) are equal.
We will now show that the answer to question (ii) (and hence, the
answer to question (i)), is .
We need to reimagine the distribution of balls into cups as a choice
of some objects. To give an idea of how it works, we will focus on the
special case with and . That is, we want to count the number
of ways to place
indistinguishable balls in
distinguishable cups.
Imagine laying the balls in a
row. We want to place them among
cups, which means we want to break the balls into groups (where some of the groups could
be empty). We can do this by placing three partitions somewhere among
the balls that are now lined up
in a row. For example, the partitions could be placed as shown in the
diagram below. Each ball is represented by a circle and each partition
is represented by a vertical line.
This corresponds to placing
balls in Cup , balls in Cup ,
ball in Cup , and balls in Cup . For a different example, we could
place the partitions as follows.
In this case, two of the partitions are next to each other with no
balls between them. This corresponds to the third cup having no balls in
it.
In fact, any arrangement of
indistinguishable balls and
indistinguishable partitions will corresponds to an ordered list of four
non-negative integers that have a sum of .
Therefore, there are
positions where the balls and partitions are to be placed, and every way
to choose three of the positions to be partitions corresponds to an
ordering of balls and partitions. Therefore, the answer to
the question in this case is .
In general, if there are cups,
then we need to use partitions.
This means there will be balls
and partitions for a total of
objects. There are positions that need to be chosen for
the partitions. Thus, the number of ways to place indistinguishable balls in distinguishable cups is .
First note that non-negative integers satisfy if and only if
for some
integer with . From part (b), the number
of non-negative integer solutions to is . Therefore, the number
of non-negative integer solutions to is
Now suppose are non-negative integers with . Then we must have
. On the
other hand, if , then there is a unique non-negative integer that satisfies . This shows that the
number of non-negative integer solutions to is the same as the
number of non-negative integer solutions to . From part (b),
the number of solutions to is .
We have shown that the number of non-negative integer solutions to
the inequality is equal to
and it is also equal to , so we conclude that This identity is sometime called the
Hockey Stick Identity. If you are interested in knowing why it
has such a strange name, read about Pascal’s Triangle and try to
interpret this result using Pascal’s Triangle.
The integer having the
property that is the
same as the integer having at
most decimal digits. Thus, every
such integer takes the form where are the
digits of , but some of the
leading digits may be equal to .
For example, to represent the integer in this way, we would write it as
so , , , , , and .
Thus, to count the non-negative integers with a digit sum of , we need to find all non-negative
integer solutions to the equation , but we
have an additional restriction that each is a digit, meaning .
Switching notation slightly to match earlier parts, the answer to the
question is equal to the number of solutions to the equation where each is an integer with .
By part (b), the number of non-negative integer solutions to without the
restriction that is . The
total of includes
some solutions that violate for some . We will now
count the solutions that have for at least one so
we can remove these from the count.
First, we claim that for each index , there are solutions that have . For example, consider a
non-negative integer solution to with . If we set then is a non-negative integer and Also, if we consider a non-negative integer solution to
, and set
, then we obtain
a non-negative integer solution to with . By part (b), there are
non-negative integer
solutions to and so,
equivalently, there are non-negative integer
solutions to with .
Similarly, for each from through , there are exactly non-negative integer
solutions to with . Note that these different
sets of solutions have overlap and so the number of solutions with for at least one is not . This is an overcount. For example, the
solution , , , and is
double counted: once as a solution with and once as a solution with
.
Next, we claim that for every pair of indices and , with , there are
non-negative integer solutions to with and . Each of these solutions
will be double counted in the count . Note that there are no
solutions to that have more
than two variables exceeding and
so this will give us the complete picture.
For an example, consider a non-negative integer solution to with and . If we set and then and are non-negative integers satisfying
.
Similarly, a non-negative integer solution to corresponds
to a non-negative integer solution to with . By part (b), there are
non-negative integer
solutions to and so,
equivalently, there are non-negative integer
solutions to with and .
Similarly, for each pair and
, with , there are solutions to with and .
Since there are
ways to choose two indices, there are
non-negative integer solutions that have two variables exceeding . From this, we conclude that the number
of non-negative integer solutions to that have for at least one is given by .
Therefore, the total number of non-negative integer solutions to
with for all can be calculated as follows:
By the same convention as the previous problem, we can recognize
every positive integer less than as an -digit integer by possibly prepending
some s. Thus, we wish to count the
number of solutions to the equation where
the are integers with for each .
We can rearrange this equation to get , so we
want to count non-negative integer solutions to this equation where
for each .
Since , we must
have that and . For each from through , let denote the number of non-negative
integer solutions to with for , , , and . Since all of the variables have the exact same
restrictions, we also have that
is the number of non-negative integer solutions to where each variable is
no larger than . Thus, for each
from through , there are solutions to the equation with each
side equal to . Therefore, the
answer to the question is Now
suppose that and
with for each . Then ,
and observe that since , we have , and since , we have . You should convince yourself that this shows that the
number of solutions to is the same as the
number of solutions to . In other words,
. When , we also have , so the total we seek is equal to
For through , if , then for each since the total is at most . This means the number of solutions to
where for each is equal to the number of non-negative
integer solutions without restriction. By part (b), if , then .
For through , there are still unrestricted solutions,
but here the count includes solutions with for some . Note that since the total is at most
, it is not possible for more
than one variable to exceed .
Similar to the argument in part (d), for each , there are solutions with . Since there are variables, there are solutions with a
variable exceeding . Therefore,
.
We now just need to compute the total.
Let , , and . Then we have and . Thus, we want to count the
number of non-negative integer solutions to where . Suppose is the number of non-negative integer
solutions to where , , and are all distinct and no larger than
. Since there are six different
arrangements of three distinct integers, exactly one of which puts them
in increasing order, the answer to the given question is .
To compute , we note that by
part (b) there are
non-negative integer solutions to where there are no
restrictions on the non-negative integers , , and . We now need to remove those solutions
where at least two of the variables are equal as well as any that have
at least one of , , and greater than .
If one of , , and is greater than , then the condition and the fact that , , and are non-negative integers implies that
one of the variables equals
and the other two equal . Thus,
any solutions with a variable greater than will also have two variables equal
to each other, so we can just count the number of solutions with at
least two variables equal and subtract this total from . Since is not a multiple of , it is impossible that . Therefore, we need to count the
number of solutions where exactly two variables are equal.
To count the number of solutions to with two variables equal, we
will count the number of solutions with and triple the result since there are
three choices of which two variables are equal.
We now want to count non-negative integer solutions to . This equation implies that
is even, or that for some non-negative integer . Thus, we need to count the number of
solutions to or . By part (b), this is , so the number of solutions to the
equation with two variables equal to each other is .
We now have that , and so the
number of solutions is