Can someone describe why this answer is incorrect?

[cdux]

"90 students are split randomly into 3 groups of 30. What is the chance for 3 particular students A, B, C to end up each in a different group?"

One of the right answers is (90*60*30/(90*89*88) but I wonder if my type of solution can be corrected for the same result.

 

[cdux]

PS. I hadn't initially mentioned the part about "particular students A, B, C". I wonder if it's part of the mistake..

probably not though.

 

[cdux]

OK let me rephrase / alter the question since it may be too confusing:

Is there a way to solve that question by chiefly using Combinatorics? (and not just step-by-step probability logic)

I can see that the denominator of the '(90*60*30/(90*89*88)' solution can maybe be translated to k-permutations of n, but what about the numerator? Is it forced to only be a case of the basic principle of enumeration or is there a more concise formula to express it?

 

[bobby2k]

Is there a way to solve that question by chiefly using Combinatorics? (and not just step-by-step probability logic)

Hi, I used combinatorics like this:

$\Large \frac{{87 \choose 29} * {3 \choose 1}}{{90 \choose 30}}*\frac{{58 \choose 29} * {2 \choose 1}}{{60 \choose 30}}*\frac{{29 \choose 29} * {1 \choose 1}}{{30 \choose 30}}$

(In the first group we choose 29 out of the 87 other, 1 one out of our 3, in the second we choose 29 out of the 58 remaining of the others, and one of our 2 that is left, and the third fraction is actually just one.)

If you use the definition of binomal coefficient, factorial, and you use algebra to short numerators and denominators, you will get the desired answer.

I tried to put your first answer in the calculator, and it does not seem to give the same answer as the other answer. I think that in order to answer why it is a wrong answer, you need to explain how you thought when you got it.

 

[cdux]

Hi, I used combinatorics like this:

$\Large \frac{{87 \choose 29} * {3 \choose 1}}{{90 \choose 30}}*\frac{{58 \choose 29} * {2 \choose 1}}{{60 \choose 30}}*\frac{{29 \choose 29} * {1 \choose 1}}{{30 \choose 30}}$

(In the first group we choose 29 out of the 87 other, 1 one out of our 3, in the second we choose 29 out of the 58 remaining of the others, and one of our 2 that is left, and the third fraction is actually just one.)

If you use the definition of binomal coefficient, factorial, and you use algebra to short numerators and denominators, you will get the desired answer.

I tried to put your first answer in the calculator, and it does not seem to give the same answer as the other answer. I think that in order to answer why it is a wrong answer, you need to explain how you thought when you got it.

Thank you. Excellent answer.

And so involved. I don't know why they include them as 1/5th (or less) of a curriculum. This is so interesting (and complex) as a subject that it deserves 6 months on its own.

Though I wish it was possible to eliminate that "90->60->30" succession from the arithmitic. For some reason it sounds unintuitive (though it may be just me).

I had also heard of another impressive (and correct) answer that starts with the denominator being all possible arrangements of the students: 90!/(30!*30!*30!).

My attempt started with "all possible arrangements of the 90 students in 30-student groups" as denominator and then I was trying to eliminate having "either 2 or 3 of the same 3 in the same class" on the enumerator and then reversing it to get "neither 2 or 3 in the same class" (it was also confusing that 0 of them would be impossible because that would mean at least 2 would be in another group so I didn't know if it should be included).

I suspect that I did a mistake at least with the denominator because it might produce more combinations than possible (judging from correct answers) but I'm not sure..

PS. By the way, CASIOs (regular ones) do that calc easily with nCr.

 

[bobby2k]

Though I wish it was possible to eliminate that "90->60->30" succession from the arithmitic. For some reason it sounds unintuitive (though it may be just me).

Hey again! :)

Why is it not logical? The 90-60-30 comes from the fact that the first you have 90 people, after you have created the first group you have 60 people, then 30.(in the binomal coefficient in the denominators)

I had also heard of another impressive (and correct) answer that starts with the denominator being all possible arrangements of the students: 90!/(30!*30!*30!).

Really?, I tried evaluation this expression, but I did not get the same answer.

My attempt started with "all possible arrangements of the 90 students in 30-student groups" as denominator and then I was trying to eliminate having "either 2 or 3 of the same 3 in the same class" on the enumerator and then reversing it to get "neither 2 or 3 in the same class" (it was also confusing that 0 of them would be impossible because that would mean at least 2 would be in another group so I didn't know if it should be included).

I suspect that I did a mistake at least with the denominator because it might produce more combinations than possible (judging from correct answers) but I'm not sure..

I am not sure about everything you do, but a lot in the expression in your first post seems unnatural.
If you want to calculate the probability by excluding 3 in one group, and 2 in one group, you can calculate it like this:

$\Large 1-\frac{{3 \choose 1}*{87 \choose 27}}{{90 \choose 30}}-\frac{{3 \choose 1}*{3 \choose 2}*{2 \choose 1}*{87 \choose 28}*{59 \choose 29}*{30 \choose 30}}{{90 \choose 30}*{60 \choose 30}*{30 \choose 30}}$

This expression gives the same numerical answer as(90*60*30/(90*89*88))
The first fraction excludes 3 beeing in the same group, and the second excludes 2 beeing in the same group. You have to account for the fact that you can choose any of the 3 persons(when 2 are in the same group), and any of the 3 groups. I think this is more complicating than calculating it directly.

However, even though you asked for a combinatorical approach, I am curious if there is some cool shortcut we can use to get (90*60*30/(90*89*88)), but I do not see it right now.

 

[D H]

However, even though you asked for a combinatorical approach, I am curious if there is some cool shortcut we can use to get (90*60*30/(90*89*88)), but I do not see it right now.

There's no reason to resort to combinatorics to solve this problem. Denote g(x) as the group to which person x is assigned. The problem is asking P(g(C)≠g(B) ∧ g(C)≠g(A) ∧ g(B)≠g(A)). This can be rewritten as P(g(C)≠g(B) ∧ g(C)≠g(A) | g(B)≠g(A)) * P(g(B)≠g(A)), the product of the probability that C is not in the same group as either A or B given that A and B are in different groups and the probability that A and B are in different groups. These two probabilities are easily computed.

1. P(g(C)≠g(B) ∧ g(C)≠g(A) | g(B)≠g(A))

Let's start by assigning A and B to two different groups. (That's the given conditional.) This probability asks what the probability that C is in the third group is. There are 88 people to assign to groups after having put A and B in two different groups. Their are 30 people who can be assigned to that third group. Assuming the probability that the probability that C will be assigned to this group is the same as any other person, we have P(g(C)≠g(B) ∧ g(C)≠g(A) | g(B)≠g(A)) = 30/88.

2. P(g(B)≠g(A))

A similar line of reasoning yields that P(g(B)≠g(A)) = 60/89.Finally, P(g(C)≠g(B) ∧ g(C)≠g(A) ∧ g(B)≠g(A)) =(30/88)*(60/89).