21. Six numbers are randomly selected from the integers 1 to 45 inclusive. Let $p$ be the probability that at least three of the numbers are consecutive. Find the value of $\lfloor 1000 p\rfloor$. (Note: $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$ ).
21. Answer: 56
The number of selections such that no two of the numbers are consecutive $=$ The number of binary sequences containing $6^{\prime} 1$ 's and $34^{\prime} 0$ 's $=\binom{40}{6}$.
The number of selections such that exactly two of the numbers are consecutive $=$ The number of binary sequences containing a '11', 4 '1's and 35 '0's $=\binom{5}{1}\binom{40}{5}$.
The number of selections with exactly two sets of two consecutive numbers but no three numbers are consecutive $=$ The number of binary sequences containing 2 ' 11 's, 2 '1's and 36 ${ }^{\prime} 0 ' s=\binom{4}{2}\binom{40}{4}$
The number of selections with exactly three sets of two consecutive numbers but no three numbers are consecutive $=$ The number of binary sequences containing $3^{\prime} 11^{\prime} \mathrm{s}$ and $37^{\circ} 0$ 's $=$ $\binom{40}{3}$
Thus, $1000 p=1000\left[\binom{45}{6}-\binom{40}{6}-\binom{5}{1}\binom{40}{5}-\binom{4}{2}\binom{40}{4}-\binom{40}{3}\right] \div\binom{ 45}{6}=56.28$.