Answered: How many permutations of…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

How many permutations of {1,2,3,4,5,6,7} have exactly one element fixed?

Expert Solution

Step 1

Using inclusion-exclusion principle to compute the given permutation:

There are $(\begin{matrix} 7 \\ 1 \end{matrix}) \cdot 6!$ ways to select 1 element to stay in place and permute all remaining.

But for each of $(\begin{matrix} 7 \\ 2 \end{matrix})$ ways to select 2 elements, each permutation that keeps them in place was computed 2 times, but should not be computed at all.

Thus, continuing in this manner, the desired permutation can be given by:

$(\begin{matrix} 7 \\ 1 \end{matrix}) \cdot 6! - (\begin{matrix} 7 \\ 2 \end{matrix}) \cdot 5! \cdot (\begin{matrix} 2 \\ 1 \end{matrix}) + (\begin{matrix} 7 \\ 3 \end{matrix}) \cdot 4! \cdot (\begin{matrix} 3 \\ 1 \end{matrix}) - (\begin{matrix} 7 \\ 4 \end{matrix}) \cdot 3! \cdot (\begin{matrix} 4 \\ 1 \end{matrix}) + (\begin{matrix} 7 \\ 5 \end{matrix}) \cdot 2! \cdot (\begin{matrix} 5 \\ 1 \end{matrix}) - (\begin{matrix} 7 \\ 6 \end{matrix}) \cdot 1! \cdot (\begin{matrix} 6 \\ 1 \end{matrix}) + (\begin{matrix} 7 \\ 7 \end{matrix}) \cdot 0! \cdot (\begin{matrix} 7 \\ 1 \end{matrix})$

$= \frac{7!}{1! 6!} \cdot 6! - \frac{7!}{2! 5!} \cdot 5! \cdot \frac{2!}{1! 1!} + \frac{7!}{3! 4!} \cdot 4! \cdot \frac{3!}{1! 2!} - \frac{7!}{4! 3!} \cdot 3! \cdot \frac{4!}{1! 3!} + \frac{7!}{5! 2!} \cdot 2! \cdot \frac{5!}{1! 4!} - \frac{7!}{6! 1!} \cdot 1! \cdot \frac{6!}{1! 5!} + \frac{7!}{7! 0!} \cdot 0! \cdot \frac{7!}{1! 6!}$

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