Checking Basic Combinatorics Homework - 8 Cooks & 4 Restaurants

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Homework Statement



Here is some basic combinatorics, I need someone to check it for me, please before the lecturer :)
Sorry for the stupid questions, hope I've made myself clear with the explanations of the answers given.


(1)(a) If 8 cooks are to be divided among 4 restaurants, how many divisions are possible?
(b) What if each restaurant must receive precisely 2 cooks?
(c) How many possible ways we can pair these 8 cooks up among themselves?
(d) How many possible ways we can distribute 8 identical bowls into the 4 restaurants?


Homework Equations





The Attempt at a Solution


a) is it 4^8?

b) for the first restaurant we have (8 2) and for the next 3 we have to decrease 2 cooks so we have (8 2) * (6 2) * (4 2) * (2 1)

c) 8! ?

d) V 4 8 = 1680?

2
a) S = {E,F}
b)E={head}
c)F={heads>tails}
d)E U F ={head,tail}
e)E n F
f)(E U F)c = {tails>heads} ? isn't it the De Morgan's where EcF = (E U F)c?
g)= {tails>heads}
h) impossible? like logic's (0 and 1) and 1 = 0?

Thank you !
 
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Your answers to 1a and 1b are correct - assuming that it is okay for a restaurant to have no cooks.
1c is the same as 1b - because I doubt that they mean to make copies of the cooks.
I'm getting 165 for 1d.
 
1c is not the same as 1b, because once you've paired up the cooks there are a couple different ways you can put each pair in a restaurant (and that should be enough to figure out what the correct solution is).
 
.Scott said:
Your answers to 1a and 1b are correct - assuming that it is okay for a restaurant to have no cooks.
i would have guessed it is not ok for a restaurant to have no cooks, but I agree it is unclear.
The answer to 1b in the OP is incorrect. Why is the final factor (2 1) when all the others are (n 2)?

1d is an interesting question. The easiest is to imagine the 8 bowls set out in a line and you have to place three partitions in the line. Those to the left of the first partition go to the first retaurant, etc. The next step is to realize that 8 bowls and 3 partitions make 11 things, of which any 3 can be partitions. can you get it from there?

There seems to be a second question missing in the OP. Anyway, it's better to put it in a separate thread.