Determine the number of permutations of $ \ \{1,2,3,4,5,6,7,8,9,10\} \ $ that have exactly 3 numbers in their natural position Determine the number of permutations of $ \  \{1,2,3,4,5,6,7,8,9,10\} \  $  that have exactly  3 numbers in their natural position. $$ $$ Is it $ \ \ \begin{pmatrix} 10 \\ 3 \end{pmatrix} \times 7 ! \  $ ?

You need choose $3$ elements that stay on their positions in $\binom{10}{3} = 120$ and multiply it by the number $!7$ of permutations of special kind called derangement on remaining $7$ elements. There are many ways to compute number $!k$ (see link), particularly it is known to be equal to $\frac{k!}{e}$ rounded to the nearest integer. Therefore $\frac{5040}{e} \approx 1854 =\,!7$ and the desired number of permutations on $10!$ elements keeping exactly $3$ of them in place is $\binom{10}{3} \cdot\,!7 = 120 \cdot 1854 = 222\,480$.

Another way to compute number of such permutation is inclusion-exclusion principle. There are $\binom{10}{3}\cdot 7!$ ways to select $3$ elements to stay in place and permute all remaining. But for each of $\binom{10}{4}$ ways to select $4$ elements each permutation that keeps them in place was computed $4$ times, but should not be computed at all. Continuing these thoughts we get that the right number of desired permutations is
$$\binom{10}{3}\cdot 7! - \binom{10}{4} \cdot 6! \cdot \binom{4}{3} + \binom{10}{5} \cdot 5! \cdot \binom{5}{3} - \binom{10}{6} \cdot 4! \cdot \binom{6}{3}\\
 + \binom{10}{7} \cdot 3! \cdot \binom{7}{3} - \binom{10}{8} \cdot 2! \cdot \binom{8}{3} + \binom{10}{9} \cdot 1! \cdot \binom{9}{3} - \binom{10}{10} \cdot 0! \cdot \binom{10}{3}\\
 = 604800 - 604800 + 302400 - 100800 + 25200 - 5040 + 840 - 120 = 222480.$$
