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1 solution
I need help. Can you tell me why zone D in both the cases has 2 and 3 options respectively, instead of 1 and 2 ?
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Because, i. After C and E have been filled with distinct colours, D can be filled with the same colour as B ... or with the colour that's left out so 2 ways. ii. After C and E have been filled with the same colour, D can be filled with one of the 2 remaining colours, or with the same colour as B ... so 3 ways.
Another way of looking at it is in terms of colours used: i. 5 colours: 5 colours for 5 parts = 5P5 = 5! = 120 ii. 4 colours: Pick colour to be repeated in 5 ways * pick pair to have the same colour (C-E or B-D) in 2 ways * 4 Colours for 3 parts in 4P3 ways = 5 2 24 = 240 iii. 3 colours: Colour for C-E in 5 ways * colour for B-D in 4 ways * 3 colours for 1 part in 3 ways = 5 4 3 = 60
So answer is 120 + 240 + 60 = 420
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went the same way !
The first 2 cases I got the same.. for the last case i did this: out of 5 coolours i have to choose 3 AABBC type id get 5 4 3. now, i have to arrange these colours into the figure, so colour B can go in the vertical slots and A in the horizontal, or vice versa. this gives us 2 possibilities.. so the last case I got 120, and the answer 480. where am i going wrong?
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The mistake you're making is in the 5 4 3. Since AABBC and BBAAC is the same, because u're further assigning them to the 2 pairs, u should calculate the no. of ways of picking the 3 colours as 5 4 3/2 = 30.
Another way to took at this is, to pick 3 colours from 5 in 5C3 ways i.e. 10 ways. Then u pick 1 out of these 3 colours to NOT be repeated, in 3 ways. So effectively, picking 3 colours from 5, with 2 of them repeated can be done in 3 * 5C3 i.e. 3*10 = 30 ways.
And then, the assignment brings a factor of 2, to give u the total no. of ways as 2*30 = 60.
Is it possible to build a recursion if there were more zones?
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yes it is
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Can you please explain how?
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@Kenneth Wang – for each recursive level,
you must mark down the color you just used and disable its availability for the adjacent zones before you can call the recursion,
and you have to enable it back after you done with the recursive call before you move on to the next color
apply for any number of zones
I am so so so so stupid. I had the exact same solution as you, but I thought D could have only 1 color instead of 2 colors after E and C were distinct colors.
Since you can use multiple times same color. A-options=5, B,D-options=4, C,E-options=3 so 5.4.4.3.3=720 and B,D and C,E can be exchanged. Hence, 720x2-1440 different ways
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The mistake you are doing here is this : you have included the cases where B (or D) will have same colour as C (or E)
Let's start coloring from zone A - there are 5 possible colors. Now color zone B - 4 options. Now we have two options:
Zones C , E have distinct colors: then 3 ⋅ 2 options for their coloring and 2 for the coloring of zone D .
Both C , E have the same color: then 3 options to color them and 3 to color zone D
So, in total, we have: 5 ⋅ 4 ⋅ ( 3 ⋅ 2 ⋅ 2 + 3 ⋅ 3 ) = 2 0 ⋅ 2 1 = 4 2 0