Let the probability of the event that the permutations  $a_1{a}_2{a}_3{a}_4{a}_5{a}_6$ of the six integers from 1 to 6 is such that for all  i from 1 to 5,  $a_{i+1}$ does not exceed  $a_i$  by 1 is  $\frac{p}{q}\left(p,q\in N,H.C.F(p,q)=1\right).$  Find  $q-p$.

Let S be the set of permutations of the six integers from 1 to 6. Then  $\left|S\right|=6!=720$. Define  $P\left(i\right)$ to be the subset of S such that the digit i+1 follows immediately  $i,i=1,2,3,4,5.$ Then  $\sum _i\left|P\left(i\right)\right|=\left(\begin{array}{c}5\\ 1\end{array}\right)5!$,

$\sum _{i_1<i_2}\left|P\left(i_1\right)\cap P_{i_2}\right|=\left(\begin{array}{c}5\\ 2\end{array}\right)4!$

$\sum _{i_1<i_2<i_3}\left|P\left(i_1\right)\cap P\left(i_2\right)\cap P\left(i_3\right)\right|=\left(\begin{array}{c}5\\ 3\end{array}\right)3!$

$\sum _{i_1<i_2<i_3<i_4}\left|P\left(i_1\right)\cap P\left(i_2\right)\cap P\left(i_3\right)\cap P\left(i_4\right)\right|=\left(\begin{array}{c}5\\ 4\end{array}\right)2!$

$\left|P\left(1\right)\cap P\left(2\right)\cap P\left(3\right)\cap P\left(4\right)\cap P\left(5\right)\right|=\left(\begin{array}{c}5\\ 5\end{array}\right)1!$
By the principle of Inclusion and Exclusion, the required number is 

$$\begin{gathered} 6!-\left(\begin{array}{c}5\\ 1\end{array}\right)5!+\left(\begin{array}{c}5\\ 2\end{array}\right)4!-\left(\begin{array}{c}5\\ 3\end{array}\right)3!+\left(\begin{array}{c}5\\ 4\end{array}\right)2!-\left(\begin{array}{c}5\\ 5\end{array}\right)1!=309 \\ \frac{p}{q}=\frac{309}{720}=\frac{103}{240} \end{gathered}$$