Originally Posted by

**Soroban** Hello, william;!

c) The consonants are together.

We have 5 consonants and 3 vowels.

Duct-tape the consonants together.

. . Then we have 4 "letters" to arrange: .$\displaystyle \boxed{LMPCS},\,O,\,I,\,Y$

And there are $\displaystyle 4!$ arrangements of the four "letters".

But in each arrangement, the 5 consonants can be ordered in $\displaystyle 5!$ ways.

Therefore, there are: .$\displaystyle 4! \times 5! \:=\:24 \times 120 \:=\:2,\!880$ arrangements

. . with the consonants together.

d) O and S are not together.

We know there are $\displaystyle 8!$ possible arrangements.

Let's count the arrangement in which O and S **are** together.

Duct-tape the O and S together.

. . Then we have seven "letters" to arrange: .$\displaystyle \boxed{OS},\,L,\,Y,\,M,\,P,\,I,\,C $

And there are $\displaystyle 7!$ arrangements of the seven "letters."

But in each arrangement, the $\displaystyle OS$ could be ordered $\displaystyle SO.$

Hence, there are: .$\displaystyle 2 \times 7!$ arrangements with O and S together.

Therefore, there are: .$\displaystyle 8! - 2\!\cdot\!7! \:=\:40,320 - 10,080 \:=\:30,240$ arrangements

. . with O and S nonadjacent.