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What is the volume of the set $$S=\{x=(x_1,x_2,x_3,\ldots,x_n) \in \mathbb{R}^n \mid 0\leq x_1 \leq x_2\leq x_3\leq\cdots\leq x_n\leq1\}\text{?}$$

I think this is related to the volume of the unit ball in $\mathbb{R}^n$. Any ideas. Thanks beforehand.

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    I think this question has appeared here before. It's simpler than the volume of the unit ball in $\mathbb R^n. \qquad$ – 2017-01-01
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    I wonder if you meant $0 \le x_1 \leq x_2\leq x_3\leq\cdots\leq x_n\leq1$ rather than $x_1 \leq x_2\leq x_3\leq\cdots\leq x_n\leq1 \text{?} \qquad$ – 2017-01-01
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    @MichaelHardy thanks, modified the post. – 2017-01-01

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The volume of the $n$-dimensional cube $\{(x_1,\ldots,x_n) : 0 \le x_i \le 1 \text{ for } i = 1,\ldots,n\}$ is $1^n=1.$

The volume of $\{(x_1,\ldots,x_n) : 0\le x_1\le x_2\le\cdots\le x_n \le 1\}$ is the same as the volume that you get if any of the other permutations of the set $\{1,\ldots,n\}$ occurs rather than $1,2,3,\ldots,n$. There are $n!$ such permutations. Therefore, the volume of each such set is $1/n!$.

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    Can we use a modification of this argument for the volume: the volume equals$\int_0^1\int_0^1\ldots\int_0^1xdx_1dx_2\ldots dx_n=\frac{1}{n!}$? – 2017-01-01
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    @vidyarthi : The volume of the cube is $$ \int_0^1 \cdots \int_0^1 1\,dx_1\cdots dx_n = 1.$$ But where I wrote $1$ you have $x$, and I don't know what $x$ is. Is it one of $x_1,\ldots,x_n$? Which one? $\qquad$ – 2017-01-01
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    are these n tuples coordinates? – 2017-01-01
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    @MichaelHardy yes, $x=(x_1,x_2\ldots x_n)$ – 2017-01-01
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    @vidyarthi : If that's what $x$ is, then the value of the integral would also be an $n$-tuple rather than a number. $\qquad$ – 2017-01-01
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    @MichaelHardy I mean, how about a line integral? – 2017-01-01
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    @vidyarthi : You have to integrate over an $n$-dimensional region, so a line integral won't do it. – 2017-01-01