# Number of equivalence classes A = $\{100000,...,999999\}$

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.

• What exactly have you tried so far? Commented Jul 17, 2020 at 0:37
• In the English-speaking world, $999.999$ means $999+\frac{999}{1000}$. I suspect you mean $999999$. Commented Jul 17, 2020 at 0:38
• 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 Commented Jul 17, 2020 at 0:42
• @JMoravitz: Not strictly decreasing digits, but rather non-increasing digits. (At least to me strictly decreasing excludes repeated digits.) Commented Jul 17, 2020 at 0:44
• 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 Commented Jul 17, 2020 at 0:44