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I am asked to find the number of elements in the quotient set of $A\times A \quad$ $A=\{100000,...,999999\}$ considering that the defined relationship n~m $\Leftrightarrow$ n and m they have exactly the same digits except for the order in which they appear, for example 347222~272432 and 110001~100011.

What is the correct way to do this exercise? I have tried several ways without success.

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  • $\begingroup$ What exactly have you tried so far? $\endgroup$
    – Leo Sera
    Commented Jul 17, 2020 at 0:37
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    $\begingroup$ In the English-speaking world, $999.999$ means $999+\frac{999}{1000}$. I suspect you mean $999999$. $\endgroup$
    – TonyK
    Commented Jul 17, 2020 at 0:38
  • $\begingroup$ Rephrase the question... "How many 6 digit numbers with non-strictly decreasing numbers are there?" To answer this, stars-and-bars is ideal, and take note that $000000$ is not a six-digit number $\endgroup$
    – JMoravitz
    Commented Jul 17, 2020 at 0:42
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    $\begingroup$ @JMoravitz: Not strictly decreasing digits, but rather non-increasing digits. (At least to me strictly decreasing excludes repeated digits.) $\endgroup$
    – Brian M. Scott
    Commented Jul 17, 2020 at 0:44
  • $\begingroup$ Right, I misspoke, a bit distracted., but the suggestion still stands, and the ability to recognize that this is a stars-and-bars problem in disguise is arguably the most important step $\endgroup$
    – JMoravitz
    Commented Jul 17, 2020 at 0:44

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