student studying and eating Munchkins

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merlin2007

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An unfortunate student, who is allergic to chocolate, has 15 hours left to cram for exams in algebra, biology, chemistry, drama, and English. She puts 15 plain Munchkins and 4 chocolate Munchkins into a paper bag and starts studying algebra. An the end of every hour, she pulls a Munchkin at random from the bag. If it is plain, she eats it and continues to study the same subject. It it is chocolate, she throws it away and moves along to the next subject.

Before I put on the actual questions to this problem, I want to ask if anyone can verify that the total number of distinguishable ways in which the donuts can be consumed is 1471. My argument is as follows:

Because there are 14 times where she will pick a Munchkin (she only has 15 hours, so the last hour donut doesn’t matter), and there are more total donuts than there are positions to fill, we consider the various numbers of chocolate donuts that might be possible.

(14 C 0) + (14 C 1) + (14 C 2) + (14 C 3) + (14 C 4)

If you can verify the logic of this, it would increase my confidence for the rest of the problem, which I have solved, but I don't want to overcomplicate this post.

Thanks a lot.
 
Please put on the actual questions; no idea what you're doing...

> I want to ask if anyone can verify that the total number of distinguishable ways in which the donuts can be consumed is 1471.
Are these examples of "ways":
c c c c p p p p p p p p p p
c c c p c p p p p p p p p p
p p p p p p p p p p c c c c
p p p p p p p p p p p p p p

If so, then the problem can be simplified to:
in how many different ways can 14 balls be pulled from
a container with 15 white balls and 4 black balls?

Agree?
 
my apologies for not being clear

First, the ones you listed are ways ... ppppppp ccc is different from ccc pppppp.
Second, to clarify on the distinguishing - two plain donuts are not distinguishable from another, but a chocolate donut is distinguishable from a plain donut. The two questions associated with the problem are:

(a) What is the probability that she devotes precisely three hours to
each subject?
(b) What is the probability that she devotes three hours to one subject,
four hours to two subjects, and two hours to the remaining two
subjects?

The reason I asked for a check on the number of possible ways is that this is the denominator for these two questions. I computed a) to be 1/1471, since there is only one position for the chocolate donuts in which the subjects are clearly split into three hours each.

I am further making the assumption that the order amongst the subjects does not matter, although the problem is not entirely clear about this.

Denis - With regard to your simplification, I'm not entirely sure, because I'm not sure whether you are considering order, since you can have 13 whites and one black in 14 different orders in the original problem.
 
Re: my apologies for not being clear

merlin2007 said:
> (a) What is the probability that she devotes precisely 3 hours to each subject?
> (b) What is the probability that she devotes three hours to one subject,
> four hours to two subjects, and two hours to the remaining two subjects?

Ahhh...now it's quite clearer!

> I am further making the assumption that the order amongst the subjects
> does not matter, although the problem is not entirely clear about this.

Well, it sure does NOT matter: no subject is mentionned by name in a) or b).

> Denis - With regard to your simplification, I'm not entirely sure, because
> I'm not sure whether you are considering order, since you can have 13
> whites and one black in 14 different orders in the original problem.

Yes, I meant exactly that (didn't want to do lots of typing!):
1: 0w 1b 13w
2: 1w 1b 12w
3: 2w 1b 11w
....
14: 13w 1b 0w
Click to expand...

So, (with A,B,C,D,E being 5 subjects and p=plain, c=chocolate), she can
spend an hour on 1st 4 subjects, the rest on the 5th, this way:
Ac(1)Bc(2)Cc(3)Dc(4)Ep(5)Ep(6)...Ep(14)(15)
(the 15th plain being of no consequence, she can eat 2 after Dc(4),
since she'll sure be hungry after 5 hours!!)

Or spend full time on 1st:
Ap(1)Ap(2)....Ap(14)(15)

Are we on the same wave length???
 
We are united

I believe we have finally reached a mutual understanding of the problem. Now the question is whether my previous logic was correct in calculating the denominator; that is, the total number of distinguishable configurations. As I demonstrated earlier, I split the configurations into cases where there were 0,1,2,3,4 chocolates consumed at all.


Thank you for persisting with me.
 
My solutions for the rest of the problem.

a) This situation can happen in only one way; namely, when all four chocolate Munchkins are consumed, and at specific times (after every two plain ones). Therefore the probability is 1/1471 .

b) The key observation is that there are five “events” to be arranged, two pairs of which are interchangeable. Each distinguishable arrangement corresponds to a unique position for the chocolate Munchkins. The number of distinguishable ways (in terms of positions of Munchkins) in which the subjects can be arranged is then (5!)/(2!2!) = 30.
The probability is then 30/1471.

I'm particularly wary about the second one, because I'm not sure whether I'm counting the right thing, since I'm treating the subjects as distinct, in the sense that the ones that take 2 hours are different from those that take 4.
 
Re: My solutions for the rest of the problem.

merlin2007 said:
The number of distinguishable ways (in terms of positions of Munchkins) in which the subjects can be arranged is then (5!)/(2!2!) = 30.
The probability is then 30/1471.
...since I'm treating the subjects as distinct, in the sense that the ones that take 2 hours are different from those that take 4.
Click to expand...
I agree with your answer. The question does not say which subjects are to get how many hours, does it?
 
Thanks

The problem does not.

I want to thank you all for such effective and prompt communication. It's the sort of thing that builds faith in people's willingness to assist.
 
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