goyalsau wrote:Thanks Fabio , I way i was doing was very time consuming & lengthy at the same time, According to my solution answer is 99 But the way Beat_gmat_09 solved answer is 81, There is a formula in permutation that i know but don't know exactly when to use it, or not to use it, Please explain in detail...
Sorry for the delay, goyalsau. (I was having lunch.)
Well...
> ... very time consuming & lengthy at the same time...
Yep, I agree, but this problem´s difficulty is really above-average, therefore separating in cases may get things a bit slower, but also a bit safer. I´ll show you the way I would solve it (below) and, to be frank, it´s not "much better" in these same matters but... it´s the way I would do it and, therefore, the way I would suggest my students to solve it.
(My solution will APPEAR to be HUGE, but I am EXPLAINING everything here, so it could be "compacted" without loss of any care, for sure!!)
> the way Beat_gmat_09 solved answer is 81...
Simply put, I did not check all your calculations, now I did it throughly and I must say his number is correct (81), although he used a "plug-in-formula" that I didn´t know beforehand (and I do not intend to memorize, by the way), because I believe that the reasoning itself is at least as important as being able to find the answer, otherwise a small modification in the question stem makes us "unable" to plug-in and we are left without any hope...
(I edited this last paragraph because now I understood that his "plug-in formula" seems really suitable. Before, I simply said that I could not see where "repetitions" are involved, but what he says is that you may repeat say 2 balls in one urn and 2 balls in another one. The number "2" is allowed to be repeated, that is what he means!)
> There is a formula in permutation that i know but don't know exactly when to use it, or not to use it...
Well, I cannot be sure I will quote THE one you have in mind (yep, my human limitation here, LoL), but I guess the most common is n! over A! B! ... where we have n objects, A of them of "type A", B of them of "type B", etc. This formula is really useful, but it can be "avoided" using [combinations]+ [arrangements (in particular permutations)]. Anyway, I guess this is NOT the case, because we want to consider all balls as different ones (say different colors).
> Please explain in detail...
My pleasure! The problem IS really beautiful and extremely instructive, by the way. Let´s have a look at it closely!
(You will see where you got wrong simply comparing your solution to mine... it will be easy, because I made exactly what you did, but with more "control/clearness" over the process, I believe.)
Phase 1: Let us solve the problem considering all balls identical.
In this scenario, there is a "classical argument" that I believe all GMAT takers should know BY HEART, at it is the following:
Consider each b one of the (identical) balls and each x as a separator. Please look at the following "coding" that it will be evident how I created it:
Example 01: b b b b x (empty) x (empty) -- means: 4 balls in urn A, 0 balls in urn B and 0 balls in urn C
Example 02: b b b x b x (empty) -- means: 3 balls in urn A, 1 ball in urn B, 0 balls in urn C
Example 03: b b x (empty) x b b -- means: 2 balls in urn A, 0 balls in urn B, 2 balls in urn C
Example 04: (empty) x b x b b b --- means: 0 balls in urn A, 1 ball in urn B, 3 balls in urn C
Well, I guess you have enough examples to understand that the number of ways of distributing 4 identical balls between 3 urns (leaving urns empty allowed) is equal to the number of ways to choose (with combination) two positions between 6 possible ones (4+2) to allocate the separators!
This number is C(6,2) = 15 of course.
Phase 2: it is naive to think that the answer to our REAL question would be 15*P4 = 15*4! but if you thought so, your brain is already thinking right, but careless... why? Because it is NOT true that all 15 "identical" cases "hide" the same number of desired situations, therefore the last part of my solution is to separate 15 in some groups and decide, in each group, how many hidden desired scenarios are there!!
Explicitly: from the 15 cases we obtained in Phase 1:
(I) there are 3 of the 15 cases in which all four balls are situated in exactly one of the urns.
Sure: (first) all balls in urn A, (second) all balls in urn B AND (third) all balls in urn C.
Important: in case I we do not distinguish identical balls of different balls, for sure, because all of them are in the same urn, therefore all 3 cases you found are exactly all 3 cases that I found.
As Rahul likes to type (it is really nice), we have ..................................................................................3 turning out to be REALLY 3 (till now)
(II) there are 6 of the 15 cases in which 3 balls are in one of the urns and the last ball is in another of the urns.
Sure: You have 3 possibilities to choose which urn will receive 3 of the (identical) balls and then 2 possibilities to choose which of the remaining (2) urns will receive the 4th last ball.
Important: in case II we MUST distinguish identical balls of different balls, for sure. To do that, please imagine that urn A receives 3 balls and urn B receives the last one, look:
A_ b, b, b
B_ b
C_ empty.
That put, how many choices are REALLY there to choose
> the balls to put at urn A? Answer: C(4,3) = 4
> the single ball to put at urn B? Answer: C(1,1) = 1 LoL... (there is really no choice, it is the one that remained...)
Therefore the 6 cases of "identical" balls turns out to be REAL 6 *(4*1) = 24 cases .........................................6 turning out to be REALLY 24 (tiil now)
(III) there are 3 of the 15 cases in which 2 balls are in one of the urns and the last 2 balls are in another of the urns.
Sure: You have 3 possibilities to choose which urn will NOT receive ANY of the (identical) balls.
Watch out: the balls are identical, therefore if you say 3 options for the first urn to receive 2 balls and 2 options for another urn to receive the other 2 balls you got it wrong, because choosing in this order A and B would be the same as choosing in this order B and A... so if you approach like this, you should say 3*2 but get only half of it, therefore 3 as I said!
Important: in case III we MUST distinguish identical balls of different balls, for sure. To do that, please imagine that urn A receives 2 balls and urn B receives the last 2 ones, look:
A_ b, b
B_ b, b
C_ empty.
That put, how many choices are REALLY there to choose
> the balls to put at urn A? Answer: C(4,2) = 6
> the balls to put at urn B? single ball to put at urn B? Answer: C(2,2) = 1 LoL... (the same reason to laugh)
Therefore the 3 cases of "identical" balls turns out to be REAL 3 *(6*1) = 18 cases .........................................3 turning out to be REALLY 18 (tiil now)
(IV) there are 3 of the 15 cases in which 2 balls are in one of the urns and the last 2 balls are separated in the other 2 urns.
Sure: You have 3 possibilities to choose which urn will receive TWO of the (identical) balls (the other urns will receive one ball each urn, for sure).
Important: in case IV we MUST distinguish identical balls of different balls, for sure. To do that, please imagine that urn A receives 2 balls and urn B and C receives the last 2 ones, look:
A_ b, b
B_ b
C_b
That put, how many choices are REALLY there to choose
> the balls to put at urn A? Answer: C(4,2) = 6
> the single ball to put at urn B? Answer: C(2,1) = 2 (sure)
> the single ball to put at urn C? Answer: 1 (the one that remained, sure)
Therefore the 3 cases of "identical" balls turns out to be REAL 3 *(6*2) = 36 cases .........................................3 turning out to be REALLY 36
Now let us "sum up" to see what happened:
(I) 3 turning out to be REALLY 3 (till now)
(II) 6 turning out to be REALLY 24 (tiil now)
(III) 3 turning out to be REALLY 18 (tiil now)
(IV) 3 turning out to be REALLY 36
Summing up: 3+6+3+3 = 15 turning out to be REALLY 3+24+18+36 = 81 (the answer)
I really hope you understand and, more than that, that you LIKE it!
Best Regards,
Fabio.
