My Math Forum place balls in boxes

 April 26th, 2014, 03:34 AM #1 Member   Joined: Apr 2013 Posts: 52 Thanks: 0 place balls in boxes Hello!!!! With how many ways can we place n similar balls in k boxes,which are numbered with the numbers 1,...,k? The same question,when the balls are also numbered.
 April 26th, 2014, 08:15 AM #2 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 Thanks: 60 Math Focus: सामान्य गणित 1 way n! / (n - k)! ways Thanks from evinda
April 26th, 2014, 08:19 AM   #3
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Quote:
 Originally Posted by MATHEMATICIAN 1 way n! / (n - k)! ways
How did you find it?

 April 26th, 2014, 08:50 AM #4 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 Thanks: 60 Math Focus: सामान्य गणित sorry i misunderstood your question. first case : ways : k! / {(k-n)!n!} second case : ways : k! / (k - n)!
April 26th, 2014, 09:03 AM   #5
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Quote:
 Originally Posted by MATHEMATICIAN sorry i misunderstood your question. first case : ways : k! / {(k-n)!n!} second case : ways : k! / (k - n)!
But how did you find it? I still haven't understood it.

 April 26th, 2014, 09:23 AM #6 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 Thanks: 60 Math Focus: सामान्य गणित you have k empty boxex and n balls ( n < k ) for the first ball : you can place it in any one of k boxes for the second ball : you can place it in any one of k-1 boxes (as one box is filled with first ball) for the third ball : you can place it in any one of k-2 boxes (as two boxes is filled with first and second balls) . . . . finally, for the nth ball : you can place it in any one of k-(n-1) boxes (as (n-1) boxes is filled with (n-1) balls) then total number of ways to fill up the boxes = k(k-1)(k-2) . . . . . (k-(n-1)) multiplying and dividing by (k-n)! = k(k-1)(k-2) . . . . . (k-(n-1))(k-n)! / (k-n)! = k! / (k-n) ! this gives number of arrangement for numbered balls if the balls are similar, the position of balls with respect to each other would not matter the balls can be arranged among themselves in n! ways so for similar balls : number of arrangement = k! / {(k-n)!n!} Last edited by MATHEMATICIAN; April 26th, 2014 at 09:30 AM.
April 26th, 2014, 02:58 PM   #7
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Hello, evinda!

Quote:
 (a) In how many ways can we place $\displaystyle n$ similar balls in $\displaystyle k$ boxes, which are numbered with the numbers $\displaystyle 1,2,3,\cdots,k$? (b) The same question, when the balls are also numbered.

I will assume that some boxes may be empty.

(a) $\displaystyle n$ identical balls and $\displaystyle k$ numbered boxes.

I'll illustrate with a specific example.
We have: $\displaystyle \,n = 7$ balls and $\displaystyle k = 3$ boxes.

Place the 7 balls in a row, inserting a space before, after and between them.
$\displaystyle \qquad \_\,\circ\,\_\,\circ\,\_\,\circ\,\_\,\circ\,\_ \, \circ\,\_\,\circ\,\_\,\circ\,\_$

Distribute 2 "dividers" among the 8 spaces.

$\displaystyle \quad \circ\,\circ\,|\,\circ\,|\,\circ\,\circ\,\circ\, \circ$ represents $\displaystyle (2,1,4).$

$\displaystyle \quad \circ\,\circ\,\circ\,\circ\,\circ\,|\,|\,\circ\, \circ$ represents $\displaystyle (5,0,2).$

$\displaystyle \quad |\,|\,\circ\,\circ\,\circ\,\circ\,\circ\,\circ\, \circ$ represents $\displaystyle (0,0,7).$

There are: $\displaystyle \,8^2 \:=\:64$ ways.

In general, the number is: $\displaystyle \,(n+1)^{k-1}$

 April 27th, 2014, 12:43 AM #8 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 879 Thanks: 60 Math Focus: सामान्य गणित i misunderstood second time also

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