Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 26 May 2017, 14:13

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

how many ways can you place 9 marbles in 3 hats so that a)

Author Message
Current Student
Joined: 31 Aug 2007
Posts: 369
Followers: 1

Kudos [?]: 143 [0], given: 1

how many ways can you place 9 marbles in 3 hats so that a) [#permalink]

Show Tags

04 May 2008, 08:26
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

how many ways can you place 9 marbles in 3 hats so that a) each hat has at least 1 marble b) a hat may not have any marbles.
VP
Joined: 10 Jun 2007
Posts: 1443
Followers: 7

Kudos [?]: 285 [0], given: 0

Show Tags

05 May 2008, 13:47
young_gun wrote:
how many ways can you place 9 marbles in 3 hats so that a) each hat has at least 1 marble b) a hat may not have any marbles.

I'll take a shot...

You have 9 marbles total, and all marbles are exactly the same.
You want at least one marble in each hat...so let's work it out.

Say you have Hat (A, B, C)
If hat A gets 1 marble, you have
(1, 1, 7)
(1, 2, 6)
(1, 3, 5)
(1, 4, 4)
(1, 5, 3)
(1, 6, 2)
(1, 7, 1)
Total of 7

If hat B gets 2, you have
(2, 1, 6)
(2, 2, 5)
(2, 3, 4)
(2, 4, 3)
(2, 5, 2)
(2, 6, 1)
Total of 6

If you see the pattern, you have...
If A=1, Total = 7 ways
If A=2, Total = 6 ways
If A=3, Total = 5 ways
...
If A=7, Total = 1 way
You know that A cannot be more than 7 because other hats will not get any marbles. So number of marbles in A stops at 7. Now add up the total to get the answer for (a)...
7+6+5+4+3+2+1 = 28 ways

For (b), you have hat A with 0 marbles, do the same thing, you get
(0, 1, 8)
...
(0, 8, 1)
Total of 8
Do the same if hat B has 0 and hat C has 0 marbles, you get a total of
8*3 = 24 ways
Current Student
Joined: 31 Aug 2007
Posts: 369
Followers: 1

Kudos [?]: 143 [0], given: 1

Show Tags

05 May 2008, 17:43
bkk145 wrote:
young_gun wrote:
how many ways can you place 9 marbles in 3 hats so that a) each hat has at least 1 marble b) a hat may not have any marbles.

I'll take a shot...

You have 9 marbles total, and all marbles are exactly the same.
You want at least one marble in each hat...so let's work it out.

Say you have Hat (A, B, C)
If hat A gets 1 marble, you have
(1, 1, 7)
(1, 2, 6)
(1, 3, 5)
(1, 4, 4)
(1, 5, 3)
(1, 6, 2)
(1, 7, 1)
Total of 7

If hat B gets 2, you have
(2, 1, 6)
(2, 2, 5)
(2, 3, 4)
(2, 4, 3)
(2, 5, 2)
(2, 6, 1)
Total of 6

If you see the pattern, you have...
If A=1, Total = 7 ways
If A=2, Total = 6 ways
If A=3, Total = 5 ways
...
If A=7, Total = 1 way
You know that A cannot be more than 7 because other hats will not get any marbles. So number of marbles in A stops at 7. Now add up the total to get the answer for (a)...
7+6+5+4+3+2+1 = 28 ways

For (b), you have hat A with 0 marbles, do the same thing, you get
(0, 1, 8)
...
(0, 8, 1)
Total of 8
Do the same if hat B has 0 and hat C has 0 marbles, you get a total of
8*3 = 24 ways

thanks for your response...in asking that question, i was hoping to find out whether there is a short-cut way to determine the number of ways you could split X objects amongst Y containers. I think I have seen a really elegant solution to such a problem on this forum however I cant seem to find it now...
Re: PS groups   [#permalink] 05 May 2008, 17:43
Display posts from previous: Sort by