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.
Best Answer
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?