Struggling with Permutations and Combinations?

[F.B]

I just started permutations and combinations and I am already stuck so can you please help me understand this.

This first question i keep doing something so can you check my work.

Solve the equation for n

P(n,5)=14 x P(n,4)

Heres what i did.

n!/(n-5)!=14 x n!/(n-4)!
1/(n-5)=14
14n-70=1
n=71/14
This is the answer i get but the book has a different answer. They have 18 as an answer.

My second question is:

A standard deck of cards has had all the face cards (jack,queen and king) removed so that only the ace through 10 of each suit remain. A game is played in which two cards are drawn (without replacement) from this decj and a six sided die is rolled. For the purpose of this game the ace is considered 1.

a)Determine the total possible number of outcomes for this game.
b)Find the probability of drawn one even card and rolling an even number
c)two even cards are drawn and even number is rolled.

I can't do b) and c) because i can't do a) so i need help with this one.

My third question is

A combination lock opens the right combination of 3 numbers from 00 to 99 is entered. The same number may be used more than once.

Heres how i think its done.

100!/(100-3)! x 3!
but somehow i think this is wrong.

My last question is this.

Twelve students have signed up to server on the yearbook committee.
a) How many ways can the staff adviser choose three people to act publisher, lead photographer, and editor, respectively.
b) In filling these three positions, what is the probability that Francesca is selected as publisher and Marc is chosen as editor?

I tried doing this:
a)12!/(12-3)!
Is this right.
b) i have no idea how to do it so please help me.

I really need to learn this so can anyone please help me or teach me how to do these types of questions.

 

[Tide]

1) You didn't handle your factorials correctly. E.g. the left side should be n(n-1)(n-2)(n-3)(n-4) and similarly for the right side. You should end up with n - 4 = 14 after cancellations.

2) There are $6 \binom {40}{2}$ possibe outcomes since only 40 cards remain after removing the face cards.

3) HINT: There are 100 choices for EACH number in the combination.

4) HINT: You know from (a) how many possibilities there are. If the positions of Francesca and Mark are designated then they can paired with any of 10 other people to fill the remaining position.

 

[F.B]

For number 2 there is only suppose to be 9390. I don't think your answer works unless I am using my calculator wrong.

 

[HallsofIvy]

The answer for 2 is 9360, not 9390. There are 40 possiblities for the first card drawn, then 39 for the second card drawn and 6 possibilities for the roll of the die: (40)(39)(6)= 9360.