Question

How many ways are there to travel in xyz 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

Therefore, there are ways \(27,720\) to travel in xyz space from the origin to \((0,0,0)\;to\;(4,3,5)\)point 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.

Step-by-step solution

Step 1: Assumptions and substitution

To move from\((0,0,0)\;to\;(4,3,5)\) we will have to travel four times in the positive x-direction, three times in the positive y-direction, and five times in the position z-direction

\(\begin{array}{l}n = 4 + 3 + 5 = 12\\k = 3\end{array}\)

\(\begin{array}{l}{n_1} = 4\\{n_2} = 3\\{n_3} = 5\end{array}\)

Step 2: Further simplification

Distributing n distinguishable objects into k distinguishable boxes such that \({n_i}\) objects are placed in box i (\(i = 1,2,3,4,5\)) can be done in \(\frac{{n!}}{{{n_1}!{n_2}!......{n_k}!}} = \frac{{12!}}{{4!3!5!}} = 27,720\)

Therefore, there are\(27,720\)ways to travel in xyz space from the origin to \((0,0,0)\;to\;(4,3,5)\)point 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.