Mixing Colours

Probability Level 5

We begin with three colours (i.e. red, yellow, blue). Let this set be called Generation 1, or GEN-1. We mix each pair of colours to form GEN-2, which has six colours (i.e. red, yellow, blue, orange, purple, green). Suppose this process continues while obeying three rules:

1) If two colours are selected, there is no difference in the order of mixing. (Red + Blue = Blue + Red)

2) Each colour is its own identity. (Red + Red = Red)

3) For all generations after GEN-1, no colour created in their respective generation can be mixed until the next iteration.

(Orange, Purple, Green are created in GEN-2; thus, they cannot mix until forming GEN-3)

Determine the number of colours in the set GEN-6 (after mixing).


The answer is 73114281.

Steven Zheng by Steven Zheng , 26, China

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1 solution

Let the number of GEN- i i colours be n i n_i ; GEN- 1 1 , n i = 3 n_i=3 (R,Y,B) is given.

Now, let us consider GEN- 2 2 : The total number of mixes is N 2 = n 1 2 = 3 2 = 9 N_2=n_1^2=3^2=9 , but other than the three single-colour mixes that produce the n 1 n_1 GEN- 1 1 colours ( R , Y , B R,Y,B ), the remaining 6 6 produce only 3 3 new colours ( O , P , G O,P,G ). This is because two of these six mixes produce only one colour (eg. R + Y = Y + R = O R+Y=Y+R=O ). Therefore, the number of GEN- 2 2 colours n 2 = n 1 + 1 2 ( n 1 2 − n 1 ) n_2=n_1+\frac{1}{2}(n_1^2-n_1) .

Now considered GEN- 3 3 . N 3 = n 2 2 = 6 2 = 36 N_3=n_2^2=6^2=36 . There are n 2 = 6 n_2=6 single-colour mixes reproducing GEN- 2 2 colours. But the remaining 30 30 two-colour mixes do not produce 15 15 new colours. This is because there are six two-GEN- 1 1 -colour mixes that produce the three ( n 2 − n 1 ) (n_2-n_1) new GEN- 2 2 colours ( O , P , G ) (O,P,G) , while these three colours have been counted in the single-colour mixes. Therefore, the two--colour mixes only produce 30 2 − 3 = 12 \frac{30}{2}-3=12 new GEN- 3 3 colours. Therefore, n 3 = n 2 + n 2 2 − n 2 2 − n 2 + n 1 = n 1 + 1 2 ( n 2 2 − n 2 ) = 3 + 1 2 ( 36 − 6 ) = 18 n_3=n_2+\frac{n_2^2-n_2}{2}-n_2+n_1=n_1+\frac{1}{2}(n_2^2-n_2)=3+\frac{1}{2}(36-6)=18 .

By induction, we have: n i = n 1 + 1 2 ( n i − 1 2 − n i − 1 ) n_i=n_1+\frac{1}{2}(n_{i-1}^2-n_{i-1}) ⇒ n 3 = 156 , n 5 = 12093 , n 6 = 73114281 \Rightarrow n_3=156, \quad n_5=12093, \quad n_6=\boxed{73114281}

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i have a problem with this question... purple = red + blue; green = blue + yellow; so purple + green = red + blue + blue + yellow = red + blue + green....by 1) and 2)

I think the extra information given here is actually a distraction if all you wanted was to solve the recurrence relation

Leslie Lenth - 6 years, 3 months ago

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