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In how many ways can X distribute 20 chocolates among A,B,C

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In how many ways can X distribute 20 chocolates among A,B,C [#permalink] New post 14 Mar 2005, 15:30
In how many ways can X distribute 20 chocolates among A,B,C and D

(I) If each receives atleast one

(II) If X may chose to give no chocolates to whom he dislikes on that particular day.
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 [#permalink] New post 14 Mar 2005, 20:25
I would go with A. Because from II we cannot reach to an exact answer.
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 [#permalink] New post 14 Mar 2005, 20:30
I am really not sure about that one...on test day I would have chosen A

on B, the verb "may" : we don't know whether someone will receive chocolates or not...

Can someone tell me if the number of choices in statement A would be 20*19*18*17 ? :roll:
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 [#permalink] New post 14 Mar 2005, 22:25
This is a 2-part PS (without answer choices) not a DS
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 [#permalink] New post 14 Mar 2005, 22:34
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nocilis wrote:
This is a 2-part PS (without answer choices) not a DS


thanks nocilis , i know i don't need the answer but actually i am interested in the way to find the answer with statement 1 if it was required :wink:
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 [#permalink] New post 15 Mar 2005, 06:56
Yes, Nocilis is right. It is a PS with two sub parts and not a DS. I am sorry if the problem was not very clear. Could somebody help with the soln.
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 [#permalink] New post 15 Mar 2005, 07:08
Antmavel wrote:
nocilis wrote:
This is a 2-part PS (without answer choices) not a DS


thanks nocilis , i know i don't need the answer but actually i am interested in the way to find the answer with statement 1 if it was required :wink:


:lol: You do need the answer Antmavel. Nocilis was trying to tell you that this is NOT a DS question. This is TWO PS questions (without given choices). In other words there's no (D) choice, or (A) choice, for that matter. :P
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 [#permalink] New post 15 Mar 2005, 08:16
Both these problems are repetitive combination problem:
1) # of ways 20 chocolates can be distributed to 4 ppl so that each receives atleast 1 = 19C3
2) 0 can be included, then 23C3
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 [#permalink] New post 15 Mar 2005, 08:33
I thought that (I) was a permutation problem because order is not significant. i go 20P3.

What does everyone think?
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 [#permalink] New post 15 Mar 2005, 13:08
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caspace,
try the problem with smaller numbers and see if you can generalize a solution and apply it to the given problem.
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 [#permalink] New post 15 Mar 2005, 13:17
why would it be 19c3? when the it seems that ALL THE chocolates have to be distributed amongst them
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 [#permalink] New post 15 Mar 2005, 16:43
HongHu wrote:
Antmavel wrote:
nocilis wrote:
This is a 2-part PS (without answer choices) not a DS


thanks nocilis , i know i don't need the answer but actually i am interested in the way to find the answer with statement 1 if it was required :wink:


:lol: You do need the answer Antmavel. Nocilis was trying to tell you that this is NOT a DS question. This is TWO PS questions (without given choices). In other words there's no (D) choice, or (A) choice, for that matter. :P


Ooooops, sorry Nocilis :oops: thanks
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Altogether different answer [#permalink] New post 16 Mar 2005, 11:32
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Quote:
In how many ways can X distribute 20 chocolates among A,B,C and D

(I) If each receives atleast one

(II) If X may chose to give no chocolates to whom he dislikes on that particular day.


The first is a simple problem solving question, so SUFFICIENT.

In the second, X may choose, X may not choose. Even if this was not the dilemma, "to whom he dislikes" can be one or more people. Thus we do not know how many people would receive the chocolates, and how many don't. Thus, INSUFFICIENT.

I just tried "solving" the first question. My answer is different from yours.

For distributing 20 chocs amongst 4 people where each receives at least 1, distribute 1 choc each to everyone. The first 4 chocs can be distributed in 20*19*18*17 ways. Thus now everyone has 1 choc each, and 16 chocs remain. Now 16 chocs can be distributed amongst 4 people in 4^16 ways. This of course if the chocs are distinct. Total ways = 20*19*18*17*(4^16)

If not distinct, first dictribution of 4 similar chocs to 4 ppl can be done in one way only. After that 16 chocs can be distributed to 4 people in P(16, 4) ways. Thus total number of ways = P(16,4)

I tried reducing the scale and testing if it really works.

Assume 5 chocs and 3 people. My answer is P(3,2) = 6 for similar chocs

See the distribution for similar chocs (the number represents the total chocs - doesn't make any difference because all chocs are similar)

1 1 3
1 3 1
3 1 1
2 2 1
2 1 2
1 2 2

For dissimilar, assume 3 chocs (A, B, C) and 2 ppl. Answer should be 3*2*(2^1) = 12 ways.
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 [#permalink] New post 16 Mar 2005, 18:57
Prep gmat could you please explain how you arrived at the answer....
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 [#permalink] New post 17 Mar 2005, 06:27
why is this not a DS problem guys?
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Re: PS: permutation - combinations [#permalink] New post 17 Mar 2005, 08:31
Because it is marked as PS in the thread title. :-D

swath20 wrote:
In how many ways can X distribute 20 chocolates among A,B,C and D

(I) If each receives atleast one

(II) If X may chose to give no chocolates to whom he dislikes on that particular day.


Assuming all chocolates are different from each other:

(1) Each at least receive one means ABCD each gets one chocolate. The choices is P(20,4). Then we just need to see who gets the rest of 16 chocolates. Each chocolate faces a pool of four people to choose from, the choices is 16^4. Therefore total outcome is 16^4*P(20,4).

(2) 20^4

However it may be more reasonable to assume that each chocolate is NOT different from each other. Anybody know how to do this problem if this is the case?
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 [#permalink] New post 17 Mar 2005, 15:42
(1) Therefore total outcome is 16^4*P(20,4).

(2) 20^4

Are the above answers correct?. Does it matter if the chocolates are the same or different?. What if 4 different kinds of chocolates that add up to 20? What is the OA?.
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 [#permalink] New post 19 Mar 2005, 17:40
Swath,
Checkout the thread below:
http://www.gmatclub.com/phpbb/viewtopic ... highlight=
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 [#permalink] New post 19 Mar 2005, 20:40
Wow that was a great thread. So according to the formula: C(n+r-1,r)
For question 2, n=4, r=20 so the total outcome = C(23,20)=C(23,3) right?

Now how do we understand this formula?

We need to divide the 20 chocolates into four shares, so we need to insert three divisors. In other words, we would have 23 empty slots, and then we pick three slots for the divisors and all the other slots would be the chocolates. So the total number of ways would be C(23,3). Got you, thanks!
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 [#permalink] New post 20 Mar 2005, 17:17
Thanks for the help ..............
  [#permalink] 20 Mar 2005, 17:17
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