Difference between revisions of "Ball-and-urn"
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It is used to solve problems of the form: how many ways can one distribute <math>k</math> indistinguishable objects into <math>n</math> bins? We can imagine this as finding the number of ways to drop <math>k</math> balls into <math>n</math> urns, or equivalently to drop <math>k</math> balls amongst <math>n-1</math> dividers. The number of ways to do such is <math>{n+k-1 \choose k}</math>. | It is used to solve problems of the form: how many ways can one distribute <math>k</math> indistinguishable objects into <math>n</math> bins? We can imagine this as finding the number of ways to drop <math>k</math> balls into <math>n</math> urns, or equivalently to drop <math>k</math> balls amongst <math>n-1</math> dividers. The number of ways to do such is <math>{n+k-1 \choose k}</math>. | ||
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| + | == Reasoning (One of Several) == | ||
| + | If you could only put one ball in each urn, then there would be <math>{n \choose k}</math> possibilities; the problem is that you can repeat urns, so this does not work. You can, however, reframe the problem as so: imagine that you have the <math>n</math> urns (numbered 1 through <math>n</math>) and then you also have <math>k-1</math> urns labeled "repeat 1st", "repeat 2nd", ..., "repeat <math>k-1</math>st". After the balls are in urns you can imagine that any balls in the "repeat" urns are moved on top of the correct balls in the first <math>n</math> urns. There is a one-to-one correspondence between the non-repeating arrangements in these new urns and the repeats-allowed arrangements in the original urns. | ||
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| + | For a simple example, consider 4 balls and 5 urns. The one to one correspondence between several of the possibilities and the "repeated urns" version is shown. Since there are 4 balls, these examples will have three possible "repeat" urns. For simplicity, I am listing the numbers of the urns with balls in them, so "1,1,2,4" means 2 balls in urn 1, 1 in urn 2, and 1 in urn 4. The same is true for the "repeat" urns options but I use the notation <math>r_1</math> etc. | ||
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| + | *1,2,3,4 <-> 1,2,3,4 (no repeats) | ||
| + | *1,1,2,4 <-> 1,2,4, <math>r_1</math> | ||
| + | *1,2,2,2 <-> 1,2, <math>r_2</math>, <math>r_3</math> | ||
| + | *4,4,5,5 <-> 4,5, <math>r_1</math>, <math>r_3</math> | ||
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| + | Since the reframed version of the problem has n+k-1 urns, and k balls that can each only go in one urn, the number of possible scenarios is simply <math>{n+k-1 | ||
| + | \choose k}</math> | ||
== Problems == | == Problems == | ||
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*[[2007 AIME I Problems/Problem 10]] | *[[2007 AIME I Problems/Problem 10]] | ||
*[[1986 AIME Problems/Problem 13]] | *[[1986 AIME Problems/Problem 13]] | ||
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[[Category:Combinatorics]] | [[Category:Combinatorics]] | ||
Revision as of 21:24, 15 February 2014
The ball-and-urn technique, also known as stars-and-bars, is a commonly used technique in combinatorics.
It is used to solve problems of the form: how many ways can one distribute 
indistinguishable objects into 
bins? We can imagine this as finding the number of ways to drop balls into urns, or equivalently to drop balls amongst 
dividers. The number of ways to do such is 
.
Reasoning (One of Several)
If you could only put one ball in each urn, then there would be 
possibilities; the problem is that you can repeat urns, so this does not work. You can, however, reframe the problem as so: imagine that you have the urns (numbered 1 through ) and then you also have 
urns labeled "repeat 1st", "repeat 2nd", ..., "repeat st". After the balls are in urns you can imagine that any balls in the "repeat" urns are moved on top of the correct balls in the first urns. There is a one-to-one correspondence between the non-repeating arrangements in these new urns and the repeats-allowed arrangements in the original urns.
For a simple example, consider 4 balls and 5 urns. The one to one correspondence between several of the possibilities and the "repeated urns" version is shown. Since there are 4 balls, these examples will have three possible "repeat" urns. For simplicity, I am listing the numbers of the urns with balls in them, so "1,1,2,4" means 2 balls in urn 1, 1 in urn 2, and 1 in urn 4. The same is true for the "repeat" urns options but I use the notation 
etc.
- 1,2,3,4 <-> 1,2,3,4 (no repeats)
- 1,1,2,4 <-> 1,2,4,
- 1,2,2,2 <-> 1,2,
, 
- 4,4,5,5 <-> 4,5, ,
, 
Since the reframed version of the problem has n+k-1 urns, and k balls that can each only go in one urn, the number of possible scenarios is simply 
