A very interesting problem (1 Viewer)

no_arg

Member
Joined
Mar 25, 2004
Messages
67
Gender
Undisclosed
HSC
N/A
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
 

no_arg

Member
Joined
Mar 25, 2004
Messages
67
Gender
Undisclosed
HSC
N/A
You would imagine that this (ie 77) is the correct answer but it is not!!
Think about it!!
 

Sober

Member
Joined
Dec 6, 2005
Messages
215
Gender
Male
HSC
2003
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.
 
Last edited:
I

icycloud

Guest
Sober said:
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 :p Those interested might want to look up the "Polya Burnside Lemma" (university mathematics).
 
Last edited by a moderator:

no_arg

Member
Joined
Mar 25, 2004
Messages
67
Gender
Undisclosed
HSC
N/A
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
 
Last edited:

ice ken

Member
Joined
Jul 4, 2006
Messages
337
Location
adel
Gender
Male
HSC
2006
:rofl: it sounds more like a umat question than a maths 2 question
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
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.
 

jemsta

I sit here alone
Joined
Apr 6, 2005
Messages
5,711
Location
O.P
Gender
Male
HSC
2005
random guessing are we
im not gonna even attempt this question coz the probability of me gettin it right is close to 0
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
6x9=54

Is 2 and a half hours quick enough for you? :p
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top