In how many ways can X distribute 20 chocolates among A,B,C : GMAT Quantitative Section
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 08:08

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 25 Oct 2004
Posts: 247
Followers: 1

Kudos [?]: 28 [0], given: 0

In how many ways can X distribute 20 chocolates among A,B,C [#permalink]

### Show Tags

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.
Intern
Joined: 12 Feb 2005
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

14 Mar 2005, 20:25
I would go with A. Because from II we cannot reach to an exact answer.
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 45 [0], given: 0

### Show Tags

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 ?
Director
Joined: 19 Nov 2004
Posts: 559
Location: SF Bay Area, USA
Followers: 4

Kudos [?]: 195 [0], given: 0

### Show Tags

14 Mar 2005, 22:25
This is a 2-part PS (without answer choices) not a DS
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 45 [1] , given: 0

### Show Tags

14 Mar 2005, 22:34
1
KUDOS
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
Manager
Joined: 25 Oct 2004
Posts: 247
Followers: 1

Kudos [?]: 28 [0], given: 0

### Show Tags

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.
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 325 [0], given: 0

### Show Tags

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

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.
Manager
Joined: 13 Oct 2004
Posts: 236
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

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
Manager
Joined: 11 Jan 2005
Posts: 57
Location: Mexico City
Followers: 1

Kudos [?]: 4 [0], given: 0

### Show Tags

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?
Manager
Joined: 13 Oct 2004
Posts: 236
Followers: 1

Kudos [?]: 14 [1] , given: 0

### Show Tags

15 Mar 2005, 13:08
1
KUDOS
caspace,
try the problem with smaller numbers and see if you can generalize a solution and apply it to the given problem.
Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

15 Mar 2005, 13:17
why would it be 19c3? when the it seems that ALL THE chocolates have to be distributed amongst them
VP
Joined: 13 Jun 2004
Posts: 1118
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 45 [0], given: 0

### Show Tags

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

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.

Ooooops, sorry Nocilis thanks
Senior Manager
Joined: 15 Mar 2005
Posts: 419
Location: Phoenix
Followers: 2

Kudos [?]: 26 [1] , given: 0

### Show Tags

16 Mar 2005, 11:32
1
KUDOS
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.
_________________

Who says elephants can't dance?

Manager
Joined: 25 Oct 2004
Posts: 247
Followers: 1

Kudos [?]: 28 [0], given: 0

### Show Tags

16 Mar 2005, 18:57
Prep gmat could you please explain how you arrived at the answer....
Manager
Joined: 15 Feb 2005
Posts: 246
Location: Rockville
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

17 Mar 2005, 06:27
why is this not a DS problem guys?
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 325 [0], given: 0

Re: PS: permutation - combinations [#permalink]

### Show Tags

17 Mar 2005, 08:31
Because it is marked as PS in the thread title.

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?
Manager
Joined: 13 Oct 2004
Posts: 236
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

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?.
Manager
Joined: 13 Oct 2004
Posts: 236
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

19 Mar 2005, 17:40
Swath,
http://www.gmatclub.com/phpbb/viewtopic ... highlight=
SVP
Joined: 03 Jan 2005
Posts: 2243
Followers: 16

Kudos [?]: 325 [0], given: 0

### Show Tags

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!
Manager
Joined: 25 Oct 2004
Posts: 247
Followers: 1

Kudos [?]: 28 [0], given: 0

### Show Tags

20 Mar 2005, 17:17
Thanks for the help ..............
20 Mar 2005, 17:17

Go to page    1   2    Next  [ 22 posts ]

Similar topics Replies Last post
Similar
Topics:
1 How many 1/2 inch cubes fit in a 12x12x12 cube? 5 28 Mar 2015, 15:28
2 How many integers x are there in the range 1<x<500 (inclusive) such th 3 10 Oct 2014, 00:16
how can i solve this problem without assuming x ? 4 07 Aug 2014, 02:56
how many ways can a committee of 3 be selected from 7 so tha 4 20 Feb 2014, 13:06
|x+1| + |x+2| + | x-1|= 20 3 30 Jul 2013, 20:26
Display posts from previous: Sort by