Word Problems regarding Probability This post is going to be lengthy. I am studying Probability to recall my knowledge about it before I take a class in Stats this college. The thing is, the textbook I have didn't provide any solution which could help me determine if my answers were correct or not. Anyway, here are the problems with their respective solutions I have made:
$1.$ In how many ways can a librarian arrange $2$ Biology and $5$ Math books in a shelf?

My attempt: $2$ Bio books $\times$ $5$ Math books = $10$ ways

$2.$ How many $2$-letter words can you form using letters $w,x,y,z$ without repeating letters?

My attempt: 4!/2! = 12

$3.$ How many ways can $5$ questions be answered if for every question there are $3$ possible answers?

My attempt: 5 x 3 = 15
15! is the answer, I guess.

$4.$ There are $3$ math books and $3$ history books that are to be arranged in a shelf. How many different ways can the books be arranged on the shelf if $2$ history books are also to be kept together and $2$ mathematics books are also to be kept together? The $2$ math books should be immediately followed by the $2$ history books, and vice versa.

I have no idea how to tackle this one. The load of words confuse me.
I'm guessing it's $5 \times 5$? Since both $2$ books for history and math are
to be kept together.

$5.$ Cinderella and her $7$ dwarves will eat in a round table. Happy wishes not to be seated opposite Grumpy. What's the probability that things will not work out for Happy?

My attempt: (7-1)! = 6!

Thank you in advance. Any help will mean a lot.

Ok, here we go!
I'll give you some answers and working, and leave some for you:


*

*This depends on the wording. If the books are all distinct, then there are $7! = 7*6*5*4*3*2*1 = 5040$ arrangements. But, if bio books are identical and math books are identical, there are$ \frac{7!}{5!*2!} = \frac{5040}{240} =$ 21.




*There are 4 options for the first letter, 3 for the second so since $4\times3$ = 12, you are correct.




*For the first question there are 3 options, second 3 options, 3rd 3 options... so the total will be $3 \times 3 \times 3 \times 3 \times 3$ = $3^5$ = 243 possibilities.



*Assuming you mean we have two math books followed by two history books or vice versa, we can place this block of 4 books amongst the 6 spaces we can order them in. Assuming Math books are identical and history books are identical, we have the following possibilities (blank spaces represent where we can place the other books):

(4-block)-- = 2 possibilities to place the 2 remaining books in the remaining spaces
-(4-block)- = 2 possibilities to place the 2 remaining books in the remaining spaces
--(4-block) = 2 possibilities to place the 2 remaining books in the remaining spaces
So total 6, but we can arrange it within the 4-block as history first then math or math first then history so multiply by 2: 12 is the answer.


*

*Firstly, this asks for the probability, not the possibility. I've given you some tips on the other one so I'll leave this for you to try and figure out, here's a hint:


First seat Happy and look then what are the possibilities left for Grumpy to seat.


N.B:
In case you want to learn, look up combinatorics - covering combinations, arrangements and permutations. Its a fascinating field.

Good luck!

