Distinct Variants

Question

I'm really having some trouble on this problem:

Find the number of distinct variants that the functions have.

I'm working with the following function (technically an expression) $$x_1x_2x_3^2$$

From what I read, a variant is basically another way of re-expressing a function by permuting their variables. Two variants are said to be distinct variants if they differ as functions over $\mathbb{C}$. I'm looking at this problem and it turns out that there are $3$ distinct variants here (odd problem so I looked in the back of the book). I thought it was only $1$ but why is it $3$? Can someone explain it to me? My textbook didn't provide any concrete examples as of how to approach these kinds of problems.

$x_1x_2x_3^2 \ne x_1x_2^2x_3 \ne x_1^2x_2x_3 \ne x_1x_2x_3^2$. The $2$ invariants are in the middle. — woogie, Mar 29 '17 at 01:44
$x_1x_2x_3^2 \ne x_1x_2^2x_3$, $x_1x_2^2x_3 \ne x_1^2x_2x_3$ and $x_1^2x_2x_3 \ne x_1x_2x_3^2$. Note that $\ne$ is not transitive. — woogie, Mar 29 '17 at 01:51

benguin · Accepted Answer · 2017-03-29T03:39:57.250

2

Notice there are six (possibly redundant) variants if you look at all the permutations of the variables:

$$x_1x_2x_3^2$$ $$x_1x_3x_2^2$$ $$x_2x_1x_3^2$$ $$x_2x_3x_1^2$$ $$x_3x_1x_2^2$$ $$x_3x_2x_1^2.$$

Notice that some of these are redundant though, i.e.: $$x_3x_2x_1^2 = x_2x_3x_1^2$$ $$x_1x_2x_3^2 = x_2x_1x_3^2$$ $$x_1x_3x_2^2 = x_3x_1x_2^2$$

Thus, there are no more than three distinct variants. It is easy to check that there are no more "redundancies" between the remaining three variants and thus those three are distinct.

edited Mar 29 '17 at 03:39

answered Mar 29 '17 at 01:59

benguin

3,846

Ahh okay that makes sense. I believe that the suggested solution in my book then is incorrect. – John W. Smith Mar 29 '17 at 02:01
Small follow-up question: If I were to do this for $x_1^3x_2^3x_3^3$, as pointed out by woogie in the comment section, would there only be $1$ distinct variant? – John W. Smith Mar 29 '17 at 02:02
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Yes there would only be one distinct variant because in that case, because all 6 variants that you get from permuting the variables are all redundant (equal) to one another. – benguin Mar 29 '17 at 02:05
Ahh now that makes sense. – John W. Smith Mar 29 '17 at 02:06
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My appologies benguin. The solution in the back of my book does say $3$, so we do count the original permutation. – John W. Smith Mar 29 '17 at 02:15
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I've edited the original question as well. – John W. Smith Mar 29 '17 at 02:15

Distinct Variants

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