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Chapter 1: Problem 48

Problem 47 has a geometric interpretation in a coordinate plane. A lattice path in the plane is a "curve" made up of line segments that either go from a point \((i, j)\) to the point \((i+1, j)\) or from a point \((i, j)\) to the point \((i, j+1),\) where \(i\) and \(j\) are integers. (Thus lattice paths always move either up or to the right.) The length of the path is the number of such line segments. (a) What is the length of a lattice path from (0,0) to \((m, n) ?\) (b) How many such lattice paths of that length are there? (h) (c) How many lattice paths are there from \((i, j)\) to \((m, n),\) assuming \(i, j,\) \(m,\) and \(n\) are integers? (n)

Short Answer

Expert verified
a) m+n; b) \( \binom{m+n}{m} \); c) \( \binom{(m-i)+(n-j)}{m-i} \)

Step by step solution

01

Understand the movement in lattice paths

A lattice path consists of segments moving either to the right or upwards. Each move increases either the x-coordinate (move right) or the y-coordinate (move up).
02

Calculate the length of the path from (0,0) to (m,n)

To get from \(0,0\) to \(m,n\), you need to move right \(m\) times and up \(n\) times. Thus, the total number of moves (or the path length) is \(m+n\).
03

Find the number of lattice paths from (0,0) to (m,n)

The number of distinct paths to get from \(0,0\) to \(m,n\) is given by the number of ways to arrange \(m\) right moves and \(n\) up moves in a sequence of \(m+n\) moves. This is a combinatorial problem, and the number of such paths can be calculated using the binomial coefficient \(\binom{m+n}{m} \) or equivalently \(\binom{m+n}{n} \).
04

Generalize the lattice paths from (i,j) to (m,n)

For a general starting point \( (i,j) \), the movement remains the same: right or up. To reach \((m,n)\) from \((i,j)\), you need \(m-i\) right moves and \(n-j\) up moves. The total moves will be \((m-i) + (n-j) = (m+n)-(i+j)\). The number of such paths is given by the binomial coefficient \(\binom{(m-i) + (n-j)}{m-i} \) or equivalently \(\binom{(m-i) + (n-j)}{n-j} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorial Mathematics
Combinatorial mathematics deals with counting, arranging, and combining objects. When solving problems about lattice paths, we look at different ways to arrange moves.
To understand combinatorial properties of lattice paths, imagine you need to move from point \((0,0)\) to \((m,n)\). You'll move right (R) and up (U) in sequences. For example, moving to \((2,2)\) requires sequences like \(RRUU\) or \(URRU\).
The length of the lattice path shows the total moves needed, which is \(m+n\).
To choose the arrangement, we use combinations. The binomial coefficient \(\binom{m+n}{m}\) or \(\binom{m+n}{n}\) helps us count these ways. These coefficients come from combinatorial algebra, specifically choosing \(m\) positions from \(m+n\) total moves.
  • Arrangements of different sequences
  • Understanding path lengths
  • Applying combinatorial principles
  • Using binomial coefficients

These combinatorial ideas make it easier to calculate the number of lattice paths for any \((m,n)\) point.
Coordinate Geometry
In coordinate geometry, every point in the plane is defined by two numbers \((i,j)\), representing the x-coordinate and y-coordinate.
Lattice paths navigate this grid-based system. Starting at point \((i,j)\), each move can either increase the x-coordinate by 1 (move right) or the y-coordinate by 1 (move up).
To go from \((0,0)\) to \((m,n)\), we make \(m\) rightward moves and \(n\) upward moves. This graphical interpretation helps visualize these paths.
The coordinate steps involved include measuring distances and visualizing grid paths:
  • Moving right increases the x-value
  • Moving up increases the y-value
  • Calculating total path length as \(m+n\)
  • Using grid-based coordinates to find paths

Coordinate geometry simplifies understanding lattice paths by providing graphical interpretations and visual aids.
Binomial Coefficients
Binomial coefficients are crucial in combinatorial problems, including counting lattice paths.
Denoted as \(\binom{n}{k}\), it represents the number of ways to choose \(k\) elements from a set of \(n\) elements.
For lattice paths, the formula \(\binom{m+n}{m}\) or \(\binom{m+n}{n}\) helps count how many distinct sequences of right and up moves exist.
For example, to go from \((0,0)\) to \((2,2)\), the number of possible paths is \(\binom{4}{2} = 6\).
Key aspects:
  • Definition of binomial coefficients
  • Applications in counting and arranging
  • Using combinations to find path numbers
  • Examples with lattice paths

The binomial coefficients make it easier to calculate the several paths, ensuring accuracy in combinatorial problems.
Mathematics Education
Mathematics education aims to build strong foundational skills in solving problems, understanding theories, and applying concepts.
Teaching lattice paths is excellent for enhancing students' grasp of combinatorial mathematics and coordinate geometry.
By learning to count paths, students practice analysis, creative problem-solving, and logical thinking.
Core educational objectives include:
  • Explaining mathematical concepts clearly
  • Providing real-world applications and visualizations
  • Encouraging problem-solving and critical thinking
  • Building confidence in working with combinations and coordinates

Incorporating visual aids, step-by-step solutions, and interactive exercises can improve students' engagement and understanding.
Utilizing lattice paths and binomial coefficients not only instructs but also dramatically increases comprehension, preparing students for more advanced mathematical challenges.

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Most popular questions from this chapter

From the symmetry of the binomial coefficients, it is not too hard to see that when \(n\) is an odd number, the number of subsets of \(\\{1,2, \ldots, n\\}\) of odd size equals the number of subsets of \(\\{1,2, \ldots, n\\}\) of even size. Is it true that when \(n\) is even the number of subsets of \(\\{1,2, \ldots, n\\}\) of even size equals the number of subsets of odd size? Why or why not? (h)

Another name for a list, in a specific order, of \(k\) distinct things chosen from a set \(S\) is a \(k\) -element permutation of \(S .\) We can also think of a \(k\) -element permutation of \(S\) as a one-to-one function (or, in other words, injection) from \([k]=\\{1,2, \ldots, k\\}\) to \(S .\) How many \(k\) -element permutations does an \(n\) -element set have? (For this problem it is natural to assume \(k \leq n\). However the question makes sense even if \(k>n\). What is the number of \(k\) -element permutations of an \(n\) -element set if \(k>n ?\)

Now suppose we are thinking about a set \(S\) of functions \(f\) from \([m]\) to some set \(X\). (For example, in Problem 6 we were thinking of the set of functions from the three possible places for scoops in an ice-cream cone to 12 flavors of ice cream.) Suppose there are \(k_{1}\) choices for \(f(1)\). (In Problem 6 , \(k_{1}\) was \(12,\) because there were 12 ways to choose the first scoop.) Suppose that for each choice of \(f(1)\) there are \(k_{2}\) choices for \(f(2)\). (For example, in Problem \(6 k_{2}\) was 12 if the second flavor could be the same as the first, but \(k_{2}\) was 11 if the flavors had to be different.) In general, suppose that for each choice of \(f(1), f(2), \ldots f(i-1),\) there are \(k_{i}\) choices for \(f(i) .\) (For example, in Problem 6 , if the flavors have to be different, then for each choice of \(f(1)\) and \(f(2),\) there are 10 choices for \(f(3) .)\) What we have assumed so far about the functions in \(S\) may be summarized as \- There are \(k_{1}\) choices for \(f(1)\) \- For each choice of \(f(1), f(2), \ldots, f(i-1),\) there are \(k_{i}\) choices for \(f(i)\). How many functions are in the set \(S ?\) Is there any practical difference between the result of this problem and the general product principle?

Show that if we have a function from a set of size \(n\) to a set of size less than \(n,\) then \(f\) is not one-to-one. \((\mathrm{h})\)

If we make a sequence of \(m\) choices for which \- there are \(k_{1}\) possible first choices, and \- for each way of making the first \(i-1\) choices, there are \(k_{i}\) ways to make the \(i\) th choice, then in how many ways may we make our sequence of choices? (You need not prove your answer correct at this time.)
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