# Probability of a subset occuring within a section of a permutation

I have a list of permutations of a vector of length 24. In each position of the vector is a number between 1 and 24 and repeats are not allowed.

An online tool tells me that this will give 6.20448e+23 possible permutations.

I was wondering how many of these possible permutations have the numbers 1, 2 and 3 in the first three positions, in any order, and how to calculate this?

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Welcome to Math.SE. Thank you for your question. We will better be able to answer it if you let us know what you've tried and any related tools you're familiar with. –  vadim123 Apr 27 '13 at 17:52
–  vadim123 Apr 27 '13 at 17:54

There are $3!=6$ ways to arrange the numbers $1,2$, and $3$ in the first three positions. There are then $21!$ ways to arrange the remaining $21$ numbers in the remaining $21$ positions. The total number of acceptable arrangements is therefore
$$3!\cdot21!\approx 3.065456530303\times 10^{20}\;,$$
since any arrangement of $1$-$3$ can be combined with any arrangement of $4$-$24$.