Eulers candy division problem

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Member of group 4:
• Trương Khánh Huyền
• Vũ Thị Khánh Linh
• Nguyễn Thị Lan
• Trần Đình Vũ


Introduction
Euler's formula is considered a special formula in the high school math
curriculum. In this book, we will introduce you to the definition as well as
the applications of this formula in problem solving.
The Euler candy division problem is a well-known counting problem with
many applications in other counting problems. In this lesson, I present the

basic original problem and some applied counting problems that would be
very difficult to count in the normal way, but when you understand the
counting of Euler's problem, the problem becomes simple. Hope they
help you to get high score!


Problem: repeating combination and
Euler's candy division problem
I. Repeating combination
II. Around Euler's candy division problem


I.Repeating combination
Problem: A person goes to a store to buy school supplies as a gift including pens, books,
and notebooks, in total he only buys 5 items. Given that there are 5 identical pens, 6
books, and 10 notebooks in a store, how many ways are there to choose pens, books, and
notebooks for a gift?
We see that the number of pén, books and notebooks is greater than the quantity to be
purchased, so the only problem returned to the counter is how many sets of lists are there
for a total of 5, where each set has or has none.
There are three objects: pens, books, and notebooks, which we denote by A=P, B, N. A gift
consists of 5 pieces, so the gift can be X= P, P, P, B, N consisting of 3 pens and 1 book, 1
notebook, or a set of Y=P, P, B, N, N . We can see that the objects P, N is repeated. Then we
say the combination of X, and Y is r-combination.


Definition: Let the set A = a1, a2, …, ak. A
mapping from p: A↦N, then P is called a
multiset of A.
Proposition : Given the set A = a1, a2, …, ak the

number of mappings p : A ↦ N such that p(a1) +
p(a2) +⋯+ p(ak) is


Prove

Each mapping, we give corresponds to a binary sequence of length
n+k−1, where p(a1) the first digit is 0, followed by 1, then p(a2) digit 0 …
and finally p(a) digit 0. For example, the set of PPBNN corresponds to
the sequence 0010100.
This is a 1 - 1 correspondence, so the number of mappings pp is equal
to the number of binary sequences, so we only need to count the
number of binary sequences.
We see that the sequence has n+k−1 digits in which there are k−1
digits, so the number of binary sequences is just the number of ways to
choose positions for k−1 digits, so the number of binary sequences is
So the mapping number p is


Going back to the above problem, we see that
the number of gifts with 5 is a convolutional
combination of 5 of books, writing, and books,
so the number of possible gifts is =
Note: In the above problem, make sure the
number of each product type is not less than 5
pieces.


II. Around Euler's candy division problem.
Starting from a problem “There are n identical candies divided by m babies. How

many ways are there to divide the candy in all?” The math seems to be very simple
but it is a difficult problem for many students. There is a combinatorial problem in
the National Examination for Good Students, the solution of which can be
prevented by applying the results of Euler's candy division.


From the actual problem, deduce the results of Euler's
candy division problem.
We all know that, in a binary sequence, the elements
take the value 0 or 1. The number of valid binary
sequences has length n and in each sequence, there are
exactly k (0 ≤ k ≤ n) elements that take the value equal
to 1.
The opening problem
Given a grid of squares. The nodes are numbered from
0 to m from left to right and from 0 to n from bottom to
top. How many different paths are there from the node
(0,0) to node (m;n) if only the edges of the squares are
allowed to go from left to right or from bottom to top?


Solution :

Such a path is considered to consist of (m+n) segments (each
segment is a square edge). At each segment, only one of two
values can be selected to go up (we encode 1) or to the right (we
encode 0). The number of segments going up is exactly equal to
n and the number of segments to the right is exactly m. The
problem leads to finding how many binary sequences of length
( m + n ) in which there are exactly n elements with the value 1.

The desired result is.


Euler's candy division problem

There are n identical candies divided by m babies. How many ways are there
to divide the candy in all?
Or the problem itself : Find the number of non-negative integer
solutions of the equation x1 + x2 +…+ xm = n (m,nN).

According to the opening problem, the number
of tasks to find is .