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How many ways are there to travel in \(x y z\) space from the origin \((0,0,0)\) to the point \((4,3,5)\) by taking steps one unit in the positive \(x\) direction, one unit in the positive \(y\) direction, or one unit in the positive \(z\) direction? (Moving in the negative \(x, y,\) or \(z\) direction is prohibited, so that no backtracking is allowed.)

Short Answer

Expert verified
There are 27720 ways.

Step by step solution

01

- Understand the problem

The goal is to count the number of different paths from the origin \(0, 0, 0\) to the point \(4, 3, 5\) by moving only in positive directions of x, y, and z coordinates.
02

- Determine the total number of steps

To go from \(0, 0, 0\) to \(4, 3, 5\), you need to take a total of \(4 + 3 + 5 = 12\) steps.
03

- Consider steps distribution

Of the 12 steps, 4 steps must be in the x direction, 3 steps must be in the y direction, and 5 steps must be in the z direction.
04

- Calculate the number of combinations

Use the multinomial coefficient to determine how to arrange these steps. The formula for the number of ways to arrange \(n\) items is: \[\frac{n!}{n_{1}! n_{2}! \dots n_{k}!}\]
05

- Apply the formula

Inserting our values, the number of ways to arrange 4 x-steps, 3 y-steps, and 5 z-steps is: \[\frac{12!}{4!3!5!}\].
06

- Compute the factorials

Calculate 12!, 4!, 3!, and 5! separately: \(12! = 479001600\), \(4! = 24\), \(3! = 6\), and \(5! = 120\).
07

- Divide to find the number of paths

Substitute the calculated factorials into the expression: \[\frac{479001600}{24 \times 6 \times 120} \].
08

- Simplify the expression

Solve the division step-by-step: \[\frac{479001600}{24 \times 6 \times 120} = \frac{479001600}{17280} = 27720\].

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Key Concepts

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

Multinomial Coefficient
In this exercise, the multinomial coefficient plays a central role. The multinomial coefficient generalizes the binomial coefficient to more than two groups. It tells us how many ways we can arrange a set of items divided into several distinct groups.

The formula for the multinomial coefficient is:
\[ \frac{n!}{n_1! n_2! \dots n_k!} \]
Here, \(n!\) is the factorial of the total number of items we are arranging. The \(n_1!, n_2!, \dots, n_k!\) in the denominator are the factorials of the counts of each distinct group of items.

For our example, to count the paths from \(0,0,0\) to \(4,3,5\), we compute:
\[ \frac{12!}{4!3!5!} \]
This formula helps us manage the different paths by accounting for all unique permutations of the 4 steps in the x-direction, 3 steps in the y-direction, and 5 steps in the z-direction.
Factorials
Factorials are an essential concept in combinatorics. The factorial of a positive integer \(n\), denoted \(n!\), is the product of all positive integers less than or equal to \(n\).

For example:
  • \(4! = 4 \times 3 \times 2 \times 1 = 24\)
  • \(3! = 3 \times 2 \times 1 = 6\)
  • \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
  • \(12! = 479001600\)

When computing combinatorial objects, factorials are used to handle large numbers of permutations. In our problem, multiplying and dividing these factorials gives us the multinomial coefficient, the count of unique paths to the destination. We use the values: \[ \frac{479001600}{24 \times 6 \times 120} = 27720 \]
Combinatorial Mathematics
Combinatorial mathematics, or combinatorics, is the study of counting, arrangement, and combination of elements in sets. It is a critical branch of discrete mathematics and is widely used in fields such as computer science, statistics, and optimization.

Our problem involves counting the number of distinct paths from the origin in a 3D space to a specific point with given constraints. Combinatorial approaches often include techniques like permutations, combinations, and the use of factorials, as seen in our multinomial coefficient calculation.

Combinatorics helps us solve problems like determining the number of ways to arrange a set of items, ensuring we account for all possibilities without omission or repetition. This problem-solving method uses structured formulae and systematic approaches to handle complex counting tasks efficiently.
Discrete Mathematics
Discrete mathematics studies mathematical structures that are fundamentally discrete rather than continuous. This includes topics like combinatorics, graph theory, and finite state machines. Discrete math forms the mathematical basis for many computer algorithms and data structures.

In our exercise, we dealt with a problem within the realm of discrete mathematics: counting paths on a discrete grid. Each step in our path can be thought of as moving from one discrete point to another (no fractions or decimals). We use combinatorial techniques to count these distinct paths.

Discrete mathematics is crucial because it allows us to model and solve problems where variables and options are countable and distinct, much like board games, scheduling problems, and network flows. By understanding and applying discrete structures and combinatorial principles, we can efficiently solve problems in technology and logistics, amongst many other fields.

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Most popular questions from this chapter

Data are transmitted over the Internet in datagrams, which are structured blocks of bits. Each datagram contains header information organized into a maximum of 14 different fields (specifying many things, including the source and destination addresses) and a data area that contains the actual data that are transmitted. One of the 14 header fields is the header length field (denoted by HLEN), which is specified by the protocol to be 4 bits long and that specifies the header length in terms of 32-bit blocks of bits. For example, if HLEN = 0110, the header is made up of six 32-bit blocks. Another of the 14 header fields is the 16-bit-long total length field (denoted by TOTAL LENGTH), which specifies the length in bits of the entire datagram, including both the header fields and the data area. The length of the data area is the total length of the datagram minus the length of the header. a) The largest possible value of TOTAL LENGTH (which is 16 bits long) determines the maximum total length in octets (blocks of 8 bits) of an Internet datagram. What is this value? b) The largest possible value of HLEN (which is 4 bits long) determines the maximum total header length in 32-bit blocks. What is this value? What is the maximum total header length in octets? c) The minimum (and most common) header length is 20 octets. What is the maximum total length in octets of the data area of an Internet datagram? d) How many different strings of octets in the data area can be transmitted if the header length is 20 octets and the total length is as long as possible?

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