Taken from finals on discrete mathematics;
How many permutations of the set $\{a,b,c,d,e,f,g,h,i,j\}$ are there such that:
a) Each of the patterns ab, de, gh and ij appears
b) None of the patterns ab, de, gh and ij appears
c) At least one of the patterns ab, de and gh appears.
d) Exactly one of the patterns ab, de and gh appears.
e) Atleast two of the patterns ab, de and gh appear.
f) The patterns ab and de appear, but the patterns gh and ij do not.
Expain your answers.
a) Was sticking the letters together and thus yielding $6!$ total permutations.
b) None is total minus the size of intersection of all sets with patterns {ab} {de} etc. and was calculated using I-E principle.
c) Intersection of the sets, again found using I-E principle.
Got stuck at figuring out d, e and f. I managed to find a formula for exact and atleast conditions on https://www.youtube.com/watch?v=D1T3xy_vtxU but haven't derived it.
Furthermore, any idea how to do f?
Part (f) is just more inclusion-exclusion. There are $8!$ permutations that include both $ab$ and $de$. There are $7!$ that contain $ab$, $de$, and $gh$, and there are $7!$ that contain $ab$, $de$, and $ij$. Finally, there are $6!$ that contain $ab$, $de$, $gh$, and $ij$. Thus, there are
$$8!-2\cdot7!+6!=40320-2\cdot5040+720=30960$$
permutation that include $ab$ and $de$ but not $gh$ or $ij$.
This should help you quite a bit with (d). As for (e), suppose that you know how many permutations contain both $ab$ and $de$; then you also know how many contain both $ab$ and $gh$ and how many contain $de$ and $gh$. If you add those up, how many times have you counted the permutations that contain all three of these patterns?