A very interesting problem (1 Viewer)

[no_arg]

Here is a spooky little question for everyone preparing for the trials.

In how many different ways can six identical black marbles and six idententical white marbles be arranged in a circle?

Think carefully abvout your answer

 

[housemouse]

11 factorial?????

 

[j3123]

11!/6!6!

 

[no_arg]

You would imagine that this (ie 77) is the correct answer but it is not!!
Think about it!!

 

[SoulSearcher]

I'll take a wild stab at 11!2!/5!6!, but my permutations and combinations are screwed up at the moment.

 

[Sober]

Unless you consider the same pattern going clockwise and anticlockwise to be identical then I see no reason why 77 is not the correct answer, (12 C 6) / 12 = 77.

 

[icycloud]

Unless you consider the same pattern going clockwise and anticlockwise to be identical then I see no reason why 77 is not the correct answer, (12 C 6) / 12 = 77.

Think about the case with 2 white marbles and 2 black marbles, and try applying your logic

p.s. no_arg, I believe this type of question is beyond the 4U course Those interested might want to look up the "Polya Burnside Lemma" (university mathematics).

 

[no_arg]

Question is drawn from a 2004 selective school trial!
Though there is no guarantee that the solution was correct.
Either way it is a basic counting problem and hence in the syllabus

 

[hyparzero]

Obviously the answer is somewhere around 25

 

[Riviet]

Obviously the answer is somewhere around 25

How is it obvious?

 

[ice ken]

:rofl: it sounds more like a umat question than a maths 2 question

 

[echelon4]

11!/5!.6! ???

 

[wogboy]

Generally for n balls (n is even), half of which are white and half are black, arranged in a circle, each arrangement fits into one or more of the following categories:

x_1 periodic
x_2 periodic
...
x_n periodic

(periodic means every sequence repeats itself)

where x_1,..,x_n are the even factors of n (if odd factors of n were included, the no. of black balls wouldn't equal the no. of white balls)

Definitions:
arrangements = no. of ways the balls can be placed, forgetting that they're in a circle (i.e. if they were in a line).
placements = no. of ways the balls can be placed, taking into account they're in a circle.
strict periodicity = the lowest periodicity (every how often the sequence repeats) an arrangement of balls has.

let p(n) = no. of possible placements for n balls, half of them black & half white.

I think the fundamental error alot of people are making is that going from arrangements -> placements, they're simply dividing by n. What should really be done is that the each arrangement be divided by it's strict periodicity instead, which may be n or some even factor of n.

For the case n = 2, we have the following possibilities:
2 periodic -> 2C1 = 2 arrangements
2 strict periodic -> 2 arrangements
-> p(n) = 2/2 = 1 placement

For n = 4, we have the possibilities:
2 strict periodic -> 2 arrangements (as before)
4 periodic -> 4C2 = 6 arrangements
4 strict periodic -> 4 arrangements
-> p(n) = 2/2 + 4/4 = 2 placements

For n = 6, we have:
2 strict periodic -> 2 arrangements (as before)
6 periodic -> 6C3 = 20 arrangements
6 strict periodic -> 18 arrangements
-> p(n) = 2/2 + 18/6 = 4 placements

For n = 8,
2 strict periodic -> 2 arrangements (as before)
4 strict periodic -> 4 arrangements (as before)
8 periodic -> 8C4 = 70 arrangements
8 strict periodic -> 64 arrangements
-> p(n) = 2/2 + 4/4 + 64/8 = 10

For n = 10,
2 strict periodic -> 2 arrangements (as before)
10 periodic -> 10C5 = 252 arrangements
10 strict periodic -> 250 arrangements
-> p(n) = 2/2 + 250/10 = 26

For n = 12,
2 strict periodic -> 2 arrangements (as before)
4 strict periodic -> 4 arrangements (as before)
6 strict periodic -> 18 arrangements (as before)
12 periodic -> 12C6 = 924
12 strict periodic -> 900
-> p(n) = 2/2 + 4/4 + 18/6 + 900/12 = 80

So my answer is 80. I hope it's right.

 

[no_arg]

Close!

 

[robbo_145]

ill have a quick stab and say 60

 

[jemsta]

random guessing are we
im not gonna even attempt this question coz the probability of me gettin it right is close to 0

 

[Ennaybur]

it is 6!6!6!

 

[robbo_145]

not random guessing, just can't be bothered posting up my working as it has a diagram...and im lazy

 

[sigurd.serpent]

quick whats 6x9
of course the answer is 42
but seriously i think the answer is 72

 

[Riviet]

6x9=54

Is 2 and a half hours quick enough for you?