I have been trying this problem for a while. But, couldn't find any solution. How do I solve this?
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1Because the chance of any of them being equal is $0$, we can ignore that. What do you get from all of the possible orderings of the variables being equally likely? – Joshua Wang Jan 27 '21 at 05:09

1$3!$ total cases and only one happens so ? – Daman Jan 27 '21 at 05:25
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$\displaystyle \int_0^1\int_0^z\int_0^ydxdydz=\int_0^1\int_0^zydydz=\int_0^1\frac{z^2}{2}=\frac 1 6$
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You can also choose another of the six possible integration orders:
$\displaystyle \int_0^1\int_x^1\int_x^zdydzdx=\int_0^x\int_x^1(zx)dzdx=\int_0^1\left[\frac{z^2}2xz\right]_{z=x}^1dx=\int_0^1\left(\left(\frac 1 2 x\right)\left(\frac{x^2}2x^2\right)\right)dx=\int_0^1\left(\frac 1 2 x+\frac {x^2}2\right)dx=\left[\frac x 2  \frac {x^2} 2+\frac{x^3}6\right]_0^1=\frac 1 6$
Jellyfish
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Could you please write this integral in other forms as well? like $dydzdx,dzdydx,$ – Daman Jan 27 '21 at 05:26