[Guest]

Susan has four red marbles and eight blue marbles. She arranges her twelve marbles randomly, in a ring.

Determine the probability that no two red marbles are adjacent.

[Guest]

If we consider the marbles in a line there are 12C4 combinations = 495.

How many of these are valid? Begin by placing the 4 red marbles spaced out in a line. It is necessary that at least 1 blue marble separates them all. So take 4 blue marbles and put 1 in between each red one (and one at the end which would be in between the first and last). You now have 4 blue marbles left which you can place randomly somewhere between the red ones. The number of combinations with repetition (according to wiki) is in this case: (4+4-1)!/4!(4-1)! = 35. The probability that of no to adjacent red marbles is 35/495 = 7/99.

[Guest]

Consider the following three cases

Case 1 - BRRB:

The number of such combinations is

12 * C(8,2) - Choose the slot for the RR pair (12)

Two Bs on either side

Choose two slots from the remaining 8 for the Rs - C(8,2)

But this includes combinations where the other two Rs combine to form a

RR pair. Remove the duplicates

Choose a slot from the remaining 8 for the other RR pair - C(8,1)

So the total = 12 * C(8,2) - 12 * C(8,1)

Case 2 - BRRRB:

12 * C(7,1) - Choose the slot for the RRR pair (12)

Two Bs on either side

Choose one slot from the remaining 7 for the R - C(7,1)

Case 3 - BRRRRB:

12 - Choose the slot for the RRRR pair (12)

So the total number of combinations where R&R are adjacent =

X = 12*C(8,2) - 12*C(8,1) + 12*C(7,1) + 12

= 336 - 96 + 84 + 12

= 336

The total combinations =

Y = C(12,4)

= 495

So the probability that no two red marbles are adjacent is

= 1 - (X/Y)

= 1 - (336/495)

= 1 - (112/165)

= 53/165

[superprismatic]

If we consider the marbles in a line there are 12C4 combinations = 495.

How many of these are valid? Begin by placing the 4 red marbles spaced out in a line. It is necessary that at least 1 blue marble separates them all. So take 4 blue marbles and put 1 in between each red one (and one at the end which would be in between the first and last). You now have 4 blue marbles left which you can place randomly somewhere between the red ones. The number of combinations with repetition (according to wiki) is in this case: (4+4-1)!/4!(4-1)! = 35. The probability that of no to adjacent red marbles is 35/495 = 7/99.

You get 35 when the chain ends with a red, another 35 if it begins with one, and another 35 if neither the beginning or the end has a red. So, the answer is 105/495=7/33.

[Guest]

Consider the following three cases

Case 1 - BRRB:

The number of such combinations is

12 * C(8,2) - Choose the slot for the RR pair (12)

Two Bs on either side

Choose two slots from the remaining 8 for the Rs - C(8,2)

But this includes combinations where the other two Rs combine to form a

RR pair. Remove the duplicates

Choose a slot from the remaining 8 for the other RR pair - C(8,1)

So the total = 12 * C(8,2) - 12 * C(8,1)

Case 2 - BRRRB:

12 * C(7,1) - Choose the slot for the RRR pair (12)

Two Bs on either side

Choose one slot from the remaining 7 for the R - C(7,1)

Case 3 - BRRRRB:

12 - Choose the slot for the RRRR pair (12)

So the total number of combinations where R&R are adjacent =

X = 12*C(8,2) - 12*C(8,1) + 12*C(7,1) + 12

= 336 - 96 + 84 + 12

= 336

The total combinations =

Y = C(12,4)

= 495

So the probability that no two red marbles are adjacent is

= 1 - (X/Y)

= 1 - (336/495)

= 1 - (112/165)

= 53/165

Correction:

Case 1 - BRRB

= 12*C(8,2) - (12*C(7,1))/2

Choose two slots from the remaining 7 slots (and not 8)

And to remove the duplicates (which is present in both the combos) remove "(12*C(7,1))/2"

X = 12*C(8,2) - (12*C(7,1))/2 + 12*C(7,1) + 12

= 336 - 42 + 84 + 12

= 390

Y = 495

P = 1 - X/Y

= 105/495 = 7/33