bag n balls
You have a bag with n balls. 50% are black, 50% white (n is even).
How many different ways can you pull out all the balls?
E.g.
n = 2 -> {BW, WB} = 2 ways.
n = 4 -> {BBWW, BWBW, BWWB, WBBW, WBWB, WWBB} = 6 ways
- steve brick
How many different ways can you pull out all the balls?
E.g.
n = 2 -> {BW, WB} = 2 ways.
n = 4 -> {BBWW, BWBW, BWWB, WBBW, WBWB, WWBB} = 6 ways
- steve brick
Labels: SharedPuzzle
posted by Rajesh Lal at
5:38 PM
8 Comments:
If every ball had a different color, it would be n factorial, but there are only two colors, so you divide that by the number of each color, which, in this case is half of n, or n/2, so...
n!/((n/2)(n/2))
n!/((n/2)^2)
The above solution is in error. Then divisor should be [(n/2)!]^2 not [(n/2)]^2
For finding # of permutations of the balls (order matters)it should be:
#P=n!/[(n/2)!*(n/2)!]
#P=n!/[(n/2)!]^2
e.g.
n=2
#P=2!/(1!)^2=2/1=2
n=4
#P=4!/(2!)^2=24/2^2=6
n=6
#P=6!/(3!)^2=720/6^2=20
Cam
I'll go for n!/(((n/2)!)^2)
n = 2 => 2!/((1!)^2) = 2
n = 4 => 4!/((2!)^2) = 6
n = 6 => 6!/((3!)^2) = 30
n = 8 => 8!/((4!)^2) = 70
That's quite a good one, thanks Steve.
Ooops! I should have refreshed before publishing. I hadn't seen Cam's result. At least we agree.
Infinite number of ways you can pull all the balls out of the bag. The number of balls is only specified as an even number (infinity is even as we all know...)
Karl, if there are only 6 balls in the bag, there isn't an infinite number of ways to pull htme all out.
Chris - true for 6 balls, but there are n balls - an even number amount but unknown quantity....?
That's why Cam and I used n (an even number) to stand in for whatever the quantity of balls actually is. Are you trying to wind me up ;)?
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