There are 12 chameleons: 6 are blue, 6 are green.

They randomly form four groups of three. In any group, if there is only one chameleon of a particular color, it changes to the color of the other two. Otherwise, chameleons don't change color.

After this random grouping, if the probability that there will be nine blue chameleons and three green ones is $\dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers, what is $a+b$ ?

The answer is 86.

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I was wondering that very thing when I put this problem together. I finally did a Monte Carlo simulation before posting the problem, randomly picking chameleons and checking how many were "9|3 combos". That ratio was 9/77.

I also decided that the solution was likely correct once @Ivan Koswara solved it, as he has a good reputation for getting problems like this right! :-)

I convinced myself that you didn't need the $3!$ because its already accounted for in the way we counted. We already presorted the groups, assuming that the first had a "certain" green one, (of the three left), the second had another given green one (of the two left). and the third had the last one.

If we hadn't done this the above equation would have been:

$n = \frac{3\binom{6}{3}\times2\binom{6}{2}\times\binom{4}{2}}{3!}= 1800$

which would have yielded the same result.

Does that make sense, @Calvin Lin ?

Geoff Pilling
- 4 years, 7 months ago

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Ah, yes that's why.

In my interpretation of your counting of $n$ , when you pick the 2nd and 3rd groups, you only determined the two blue, but didn't account for which green is added. IE there should be ${3 \choose 1 } \times { 6 \choose 2 }$ ways, which gives the equation in your comment.

Can you clarify that in your solution?

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@Calvin Lin – Done! ............

Geoff Pilling
- 4 years, 7 months ago

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@Geoff Pilling – can you clarify on why is there a constant 3 before 6 choose 3 and 2 before 6 choose 2?

Jeffrey Zhang
- 2 years, 8 months ago

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@Jeffrey Zhang – Well it all comes down to how you count. You can either have the 3 and 2 on the top, and include the 3! on the bottom or you can take away all three. It comes out to the same solution in the end. Calvin and I were discussing this very same thing about 2 years back. It's a little confusing... :-/

Geoff Pilling
- 2 years, 8 months ago

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@Geoff Pilling – Thank you for replying even when the problem is a year old haha.

I was mainly just straight up confused about the equation. But now my understanding is: you need 3* (6 choose 3) because there are 3 ways we can put this group among the 3 groups, similar f or 2 * (6 choose 2) and for (4 choose 2). And then you need to divide the number of ways to arrange the 3 groups which is 3!. And we don't need to care about the last group as they will already be determined. Is this the reasoning?

Jeffrey Zhang
- 2 years, 8 months ago

Uhm... doesn't "after any color changes have taken place" imply that if none of the chameleons changed color, then the grouping will not be counted?

Manuel Kahayon
- 4 years, 7 months ago

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Good point, Manuel... Lemme see if I can think of a better way to word it.

Geoff Pilling
- 4 years, 7 months ago

I don't understand the n equation. Especially the 3!. I understand that there are (6c3)×(3c1)(6c2)×(2c1)(4c2) ways of getting (3,1,1,1) green chameleons, but why divide by 3! ? Is it because of the three groups of 1 green/ 2 blue which are undistinguished?

Maxence Seymat
- 2 years, 4 months ago

Name the blue chameleons $B_1, B_2, B_3, \ldots, B_6$ , and the green chameleons $G_1, G_2, G_3, \ldots, G_6$ .

First, to have 9 blue chameleons, it means we need to convert three green ones to blue. However, to convert a chameleon to a color, we need two of that color and one of another color in a group. Thus, to convert a green chameleon to blue, we need two blue chameleons to convert it. Since we need to convert three green chameleons to blue, we need six blue chameleons to convert them all, exactly enough. So there are three groups where green chameleons get converted to blue ones, and the remaining one is all green.

Consider the group containing $B_1$ . It must contain one other blue chameleon and one green chameleon. Choosing the blue chameleon has $\frac{5}{11}$ probability, then choosing the green chameleon has $\frac{6}{10}$ probability (after choosing that blue one, we have 6 out of 10 that are green); we multiply by 2 because we could have picked them in either order (blue first then green, or the other way around).

WLOG it's not $B_2$ that's chosen (otherwise rename chameleons). Consider the group containing $B_2$ . Again, we need one blue chameleon, with probability $\frac{3}{8}$ (3 out of 8 chameleons remaining are blue), and one green with probability $\frac{5}{7}$ , and again we multiply by 2.

Likewise with the third group. We fix one blue chameleon $B_3$ ; there's $\frac{1}{5}$ probability to get another blue, followed by $\frac{4}{4}$ probability to get a green; we multiply by 2 again. The remaining three green chameleons automatically form the last group.

In total, that's $\frac{5}{11} \cdot \frac{6}{10} \cdot 2 \cdot \frac{3}{8} \cdot \frac{5}{7} \cdot 2 \cdot \frac{1}{5} \cdot \frac{4}{4} \cdot 2 = \frac{9}{77}$ chance of getting the picks right.

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Ah.... Nice solution, @Ivan Koswara !

Geoff Pilling
- 4 years, 7 months ago

@Ivan Koswara i'm confused as to why we aren't starting with [6/12 * 5/11 * 6/10] *3 , why are we skipping B1, B2, B3, and jumping to the second blue and green for the first three groups?

Hana Elhattab
- 4 years, 4 months ago

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We look for the group containing B1, so the probability that B1 is in the group is not 6/12; it's 1.

Ivan Koswara
- 4 years, 4 months ago

Struggled with this one.. used a longer approach... the solutions by Geoff and Ivan are unquestionably the way to go... still...for what its worth... posting the way I solved...

There are 5 basic sets that can be formed

1) 3B, 3B, 3G, 3G

3B, 3G each can be arranged in 1 way and the 4 groups can be arranged in 4!/2!2! = 6 ways so total of 6 ways

2) 2B1G, 2B1G, 2B1G, 3G

2B1G each can be arranged in 3!/2! = 3 ways so 27 ways and the 4 groups can be arranged in 4!/3!= 4 ways so total of 108 ways

3) 2G1B, 2G1B, 2G1B, 3B

2G1B each can be arranged in 3!/2! = 3 ways so 27 ways and the 4 groups can be arranged in 4!/3!= 4 ways so total of 108 ways

4) 2B1G, 2B1G, 2G1B, 2G1B

2B1G AND 2G1B each arranged in 3!/2! = 3 ways so 81 ways and the 4 groups in 4!/2!2! = 6ways so total of 486 ways

5) 3B, 3G, 2B1G, 2G1B

2G1B, 2B1G each arranged in 3 ways and 3B, 3G each in 1 way so 9 ways and the 4 groups arranged in 4! = 24 ways so total of 216 ways

Total number of ways = 6 + 108 + 108 + 486 + 216 = 924

The only way to form 9 Blue and 3 Green is set number 2 to wit 2B1G, 2B1G, 2B1G, 3G in 108 ways

Probability = 108/924 = 9/77

Great problem Geoff!

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The favourable cases are not hard to find. The total number of cases can be found using 12C3×9C3×6C3×3C3.

Kushagra Sahni
- 4 years, 7 months ago

Thanks, @Satyen Nabar ! Its always interesting to see different approaches to the problem... :0)

Geoff Pilling
- 4 years, 7 months ago

Satyen Babar most understandable answer.... i had a doubt, they haven't made the distinction between any two blue or any two green chameleons so there are in total only 5 ways to form the sets because the chameleons aren't distinct....and since one of the given arrangement favours the condition shouldn't the probability be 1/5?

Shamant Basidoni
- 4 years, 2 months ago

A slight variation on Geoff’s solution: there are ${6}\choose{2}$ ways to choose the first two chameleons and ${4}\choose{2}$ ways to choose the second two. There is only one way to choose the remaining two. Multiplying this we get the number of ways to split the blue chameleons into 3 groups consisting of two fellows, but the order is preserved so we divide by 3!. Then, there are ${6}\choose{3}$ ways to choose the companions for our pairs of blue chameleons, and there are 3! ways to combine green singles and blue pairs. Finally, we should divide this by the total number of ways to split chameleons into 4 groups by 3, which can be written as $\frac{\binom{12}{3} \times \binom{9}{3} \times \binom{6}{3}}{4!}$ . The resulting formula is

$\displaystyle \frac{\frac{\binom{6}{2}\times\binom{4}{2}}{3!}\times\binom{6}{3}\times3!}{\frac{\binom{12}{3} \times \binom{9}{3} \times \binom{6}{3}}{4!}} = \frac{9}{77}$

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Let each blue and each green chameleon be represented by $B,\:G$ , respectively and name the four chameleon groups $G_1,\:\ldots,\:G_4$ . Each group has three distinct positions to a total of 12 positions $p_i$ :

$\underset{G_1}{\underbrace{p_1,\:p_2,\:p_3}}, \:\ldots,\: \underset{G_4}{\underbrace{p_{10},\:p_{11},\:p_{12}}}$

When the chameleons form groups randomly, it is the same as if they each choose one of the 12 positions randomly - if we align these positions as above, that is equivalent to forming a 12-letter word consisting of six letters $B,\:G$ each. The total number of different groups the chameleons can form is equal to the number of possibilities to choose 6 of 12 positions for $B$ , that is $\binom{12}{6}=914$

To end up with nine blue chameleons, we need to convert three chameleons from green to blue. That is only possible with three groups $BBG$ (in any other order) and one group $GGG$ . Now let's count the number of words that satisfy these conditions:

- We have four choices in which $G_i$ to place $GGG$
- Each of the other three groups consists of either $BBG,\:BGB$ or $GBB$ - remember we consider the positions to be distinct!

The probability to end up with nine blue chameleons and three green ones is $P=\frac{4\cdot 3^3}{914}=\frac{9}{77}\quad\Rightarrow\quad a+b=9+77=\boxed{86}$

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So we want to pick a $B,B,R$ (This can be done in 3 distinct ways).

Then, We again want to pick a $B,B,R$ ( 3 distinct ways to pick)

Then, We again want to pick a $B,B,R$ ( 3 distinct ways to pick)

After this process the last pair would automatically be $R,R,R$ (1 way to pick)

This whole picking process (of groups) can be done in 4 different ways.

Note: Here we are picking groups so ${BRB}$ and ${BBR}$ and ${RBB}$ are same.

For example: ${RRR},{BRB},{BBR},{RBB}$

So our probability is :

$4 * 3 * \frac{6}{12} * \frac{5}{11} * \frac{6}{10} * 3 * \frac{4}{9} * \frac{3}{8} * \frac{5}{7} * 3 * \frac{2}{6} * \frac{1}{5} * \frac{4}{4} * 1 * 1$

= $\boxed{ \frac{9}{77} }$

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The only way you can end up with nine blue chameleons are if there are 3 groups with one green and two blues each, and one group with three green ones in it.

The total number of ways to split the twelve chameleons into four groups of three is:

$N = \frac{12!}{(3!)^4\times4!} = 15400$

The 12! on the top is the total number of ways to pick the 12 chameleons. The 4! accounts for the number of ways you can rearrange the 4 groups. And the four 3!'s account for the different ways each of the four groups could have been arranged.

And the number that have the combinations described above is given by:

$n = \frac{3\binom{6}{3}\times2\binom{6}{2}\times\binom{4}{2}}{3!} = 1800$

This is because there are $\binom{6}{3}$ ways to pick 3 green ones in one group, then $\binom{6}{2}$ ways to choose the one green and two blues in the next group, and finally $\binom{4}{2}$ ways of picking one green and two blues in the third group. The fourth group is just the left overs so only one choice there once the other three groups are chosen.

So the probability we will end up with a combination that will give us nine blue chameleons is given by:

$P = n/N = 1800/15400 = 9/77$

$9+77 = \boxed{86}$