Permutation Questions from iB HL math So I started combinatronics on my math course and I'm having a hard time getting my head around these questions:
How many different permutations of the word CAROUSEL are there where there is a consonant at either end of the word
In a waiting room there are 14 seats and 8 people, one has a very bad cough and must sit at least one seat away from everyone else; how many ways can this happen?
Seven numbers are chosen from the integers 1 - 19 inclusive;
How many have a) at most 2 even numbers B) at least 2 even numbers.
I've tried to work through this but I can't seem to get the thought process going; the answers are at the back of the book but I'd like the steps explained please so I can reapply them later.
 A: For part a) you can choose any of four consonants to begin the word, and for each of these, there is a choice of three to end the word. The remaining letters can be jumbled up any way you like, and since there are six, this can be done in $6!$ ways. So the total in this case is $4\times 3\times 6!$.
Hint for part b)
Count separately the cases where (i) the ill person sits at either end of the line, in which case he "occupies" two spaces. The others can be jumbled up, with empty seats counted as identical elements...and (ii) where the ill person sits somewhere in the middle, so must "occupy" three spaces...
Can you continue?
A: For part C part a.) 
So for part C there are total of 19 numbers and 7 are chosen so therefore $${19 \choose7}$$ is the total number of possible selections made. At most means 2 or less even numbers right? 
So the ways of getting 3 or more even numbers is not required so the set we do not want is:
$${9 \choose7}{10 \choose0}+{9 \choose6}{10 \choose1}+{9 \choose5}{10 \choose2}+{9 \choose4}{10\choose3}+{9\choose3}{10\choose4}$$ 
which can be denoted as: $$\sum_{x=0}^{4}\binom{9}{7-x}\binom{10}{x}$$
Remember: There are 9 even numbers and 10 odd numbers in a set of integers from 1 to 19.
Therefore we minus the selections we do not need from the total number of possible selections by:
$${19 \choose7}-\sum_{x=0}^{4}\binom{9}{7-x}\binom{10}{x}=11082$$
So we can have 11082 possible selections of at most 2 even numbers from selection of 7 integers from 19 integers.
Hoped this helped you (sorry for long working) and please tell me if you have problems (I am also an IB HL math student as well).
Part B of the question should be a breeze for you if you know how to solve Part A.
I'll let you solve part B for now. :)
A: Seven numbers are chosen from the integers 1 - 19 inclusive; How many have a) at most 2 even numbers B) at least 2 even numbers.

There are $[\frac{19}{2}]=$9 even numbers between 1,..,19 and 19-9 = 10 odd.
a) If you have at most 2 even you have 0 even , or 1 even or 2 even. Those are exclusive so you can sum up the results . You can pick 0 even and 7 odd with $C(10,7)$ ways, 1 even and 6 odd with $C(9,1)*C(10,6)$ ways and 2 even and 5 odd with : $C(9,2)*C(7,5)$ ways . Sum those and you are done

