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Chapter 1: Problem 32

a) Provide a combinatorial argument to show that if \(n\) and \(k\) are positive integers with \(n=3 k\), then \(n ! /(3 !)^{k}\) is an integer. b) Generalize the result of part (a).

Short Answer

Expert verified
The expression \(n ! /(3 !)^{k}\) can be interpreted combinatorially as the number of permutations of \(k\) groups of 3 objects, considering that the order of the groups does not matter. Hence the result will always be an integer if \(n=3k\). The generalization follows as there will always be a whole number of permutations when dividing the total number of arrangements by the arrangements of each group, for any \(n\) and \(k\), where \(n=m*k\).

Step by step solution

01

Evaluate \(n ! /(3 !)^{k}\)

Firstly, when \(n=3k\), replace \(n\) in the formula with \(3k\). This changes the equation to \((3k)! / (3!)^{k}\). Expand out both the numerator and denominator.
02

Simplify the Expansion

The result is a series of numbers multiplied together that can also be organized into \(k\) groups of size 3. Each group in the numerator forms a multiple of 3!. And upon simplification, each of these groups will be cancelled out by one of the 3! in the denominator.
03

Derive Combinatorial Reasoning

The simpler version of the equation as described above describes a permutation scenario. The question is basically asking whether all permutations of \(k\) groups of 3 distinct items are possible, where the order of the groups does not matter. Since each group of 3 items can be arranged in 3! ways, all \(k\) groups would be arranged in \((3!)^k\) ways. In the case when the order of the groups is not important, the number of permutations of these \(k\) groups will be a multiple of \((3!)^k\), hence the division yields an integer.
04

Generalize The Finding

In general, for any positive integers \(n\) and \(m\), if \(n = m*k\), then \((mk)! / ((k)!)^{m}\) is an integer. This is because the number of ways to order \(mk\) objects can always be divided exactly by the number of ways to order \(m\) groups of \(k\) objects. In other words, each of the \(k\) groups can be arranged in \((k!)^{m}\) ways, hence making \((mk)! / ((k)!)^{m}\) an integer.

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