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

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

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New post 04 May 2008, 08:26
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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.
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Re: PS groups [#permalink]

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New post 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
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Re: PS groups [#permalink]

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New post 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
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