Checking Basic Combinatorics Homework - 8 Cooks & 4 Restaurants

In summary, the conversation is about basic combinatorics and involves questions on dividing 8 cooks among 4 restaurants, pairing up the cooks, and distributing 8 identical bowls among the restaurants. The conversation also touches on the concepts of sets and De Morgan's law. The solutions to the questions are 4^8 for (a), 8! for (c), and 8 bowls and 3 partitions for (d). The correctness of the answer for (b) is debatable due to the ambiguity of whether a restaurant can have no cooks. Another question is mentioned but it is recommended to start a separate thread for it.
  • #1
gl0ck
85
0

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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  • #2
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.
 
  • #3
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).
 
  • #4
.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.
 

Related to Checking Basic Combinatorics Homework - 8 Cooks & 4 Restaurants

1. What is the total number of ways to assign 8 cooks to 4 restaurants?

The total number of ways to assign 8 cooks to 4 restaurants is 70, since the formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of objects and r is the number of objects being chosen. In this case, n = 8 and r = 4, so the calculation is 8! / (4!(8-4)!) = 8!/ (4!4!) = (8*7*6*5*4*3*2*1) / (4*3*2*1)(4*3*2*1) = 70.

2. How many ways can 8 cooks be assigned to 4 restaurants if each restaurant must have at least 1 cook?

If each restaurant must have at least 1 cook, then the number of ways to assign 8 cooks to 4 restaurants is reduced to 56. This is because one cook will be assigned to each restaurant, leaving 4 cooks to be assigned to the remaining 4 restaurants. This can be calculated using the formula nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects being chosen. In this case, n = 4 and r = 4, so the calculation is 4! / (4-4)! = 4!/0! = (4*3*2*1) / 1 = 24. Therefore, the total number of ways is 4*24 = 96. However, this includes arrangements where the same cook is assigned to multiple restaurants, so we must divide by 4! to account for this overlap, resulting in 96/4! = 96/24 = 4. Therefore, the final answer is 96 - 4 = 56.

3. What is the probability that a random assignment of 8 cooks to 4 restaurants will have exactly 2 restaurants with 2 cooks each?

The probability that a random assignment of 8 cooks to 4 restaurants will have exactly 2 restaurants with 2 cooks each is 0.3125. This can be calculated using the formula nCr/p^n * q^(n-r), where n is the total number of objects, r is the number of objects being chosen, p is the probability of success, and q is the probability of failure. In this case, n = 8, r = 2, p = 2/8 = 0.25, q = 1 - p = 0.75. Plugging in these values, we get 8C2 * (0.25)^2 * (0.75)^6 = (8*7/2*1) * (0.25)^2 * (0.75)^6 = 28 * 0.0625 * 0.1172 = 0.3125.

4. How many ways can 8 cooks be assigned to 4 restaurants if each restaurant can only have a maximum of 3 cooks?

If each restaurant can only have a maximum of 3 cooks, then the number of ways to assign 8 cooks to 4 restaurants is 35. This can be calculated by listing out all the possible combinations of cooks in each restaurant and adding them up: (3,3,1,1), (3,2,2,1), (3,1,1,3), (2,2,2,2), (2,2,1,3), (1,1,1,5), (1,1,2,4), (1,1,3,3), (1,2,2,3), (1,1,4,2), (2,2,3,1), (1,3,3,1), (2,3,2,1), (2,1,3,2), (1,3,2,2). Therefore, the total number of ways is 15.

5. How many ways can 8 cooks be assigned to 4 restaurants if each restaurant can only have a maximum of 2 cooks and each cook can only work at 1 restaurant?

If each restaurant can only have a maximum of 2 cooks and each cook can only work at 1 restaurant, then the number of ways to assign 8 cooks to 4 restaurants is 0. This is because there are only 8 cooks available, and each restaurant can only have a maximum of 2 cooks, meaning that at least 2 restaurants

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