
#1
Sep1110, 03:03 AM

P: 369

Let's consider a simple case with 2 balls and 3 boxes. Assuming all balls are the same and empty box is allowed. In addition, each box can take any number of ball. How many ways are there to place the balls into the boxes?
Here is my way to solve the problem. For the first ball, there are 3 ways to do so. For the second ball, still 3 ways. So total 9 ways to place 2 balls into 3 boxes. For the general way, my conclusion is [tex]m^n[/tex] But the answer is 2 0 0 0 2 0 0 0 2 1 0 1 0 1 1 1 1 0 There are only six ways. So what's wrong with my analysis? And what's the correct expression for this case? 



#2
Sep1110, 12:42 PM

P: 403

You are overcounting. When you say:
1100, 0110, 0011, 1010, 1001, 0101 In the general case, this is counting the number of lenght n binary sequences with exactly k 0's and there is a wellknown expression for this. 



#3
Sep1410, 12:49 PM

P: 34

You're looking for the number of multinomial coefficients. It's given explicitly on wikipedia.
Anyways, the answer you're looking for is [tex]{m+n1} \choose {n}[/tex] where [tex]m[/tex] is the number of boxes and [tex]n[/tex] is the number of balls. 



#4
Sep1410, 02:04 PM

P: 21

number of ways to place n balls into m boxes
we can colocated the first ball at box 1, box 2 or box 3. Then you can put the second ball at box 1 box 2 or box 3, so, for one ball, you have 3 options, and are 2 ball. to generalize, if you have n balls, and m positions for each ball, the posibles combinations are equal to product m.n and you can verify than 2.3=6




#5
Sep1610, 08:40 AM

P: 369





#6
Sep1610, 09:01 PM

P: 21

answer is (m+n1)¡ just like somebody sad.
n¡*(m1)¡ 



#7
Sep1610, 09:02 PM

P: 21

answer is (m+n1)¡ just like somebody sad.
. n¡*(m1)¡ 



#8
Sep1610, 09:03 PM

P: 21

answer is (m+n1)¡/n¡*(m1)¡ just like somebody sad.
. 



#9
Sep1710, 05:42 AM

P: 369




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