Seven boxes numbered 1 to 7 are arranged in a row.Each is to be filled

Tricky way

suppose we have to fill first n-2 boxes and number of ways [f(n-2)]to fill them is x+y, where x ways ended up with Black and y ways ended up with Blue.

Then number of ways[f(n-1)] to fill n-1 boxes = 2x+y, where x+y ways ended up with Black and x ways ended up with Blue. (Why?)
As number of ways to fill subsequent box of black is 2(blue or black), and number of ways to fill subsequent box of blue is 1(black).

Then number of ways[f(n)] to fill n boxes = 2(x+y)+x= 3x+2y, where (2x+y) ways ended up with Black and (x+y) ways ended up with Blue.

We can see that f(n) = f(n-1)+f(n-2)

In our question. number of ways to fill first box = 1+1=2 (x=1, y=1) Either blue or black

number of ways to fill first 2 boxes = 2*1+1 = 3

Now it follows the above pattern - f(n) = f(n-1)+f(n-2)

f(3) = 2+3=5;
f(4)= 3+5=8
f(5)= 5+8=13
f(6)= 8+13=21
f(7)= 13+21=34 (Answer)

Could you elaborate case 2 and 3 a little further. Whats the idea behind?

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robinbjoern

Case 2 is when we need to arrange 2 Blue and 5 Black boxes

Let's place 5 black boxes with some gaps between adjacent black boxes

_ _ _ _ _

Now we have 6 spaces left for 2 blue boxes as mentioned below

B _ B. _ B _ B _ B _ B

We need to fill 2 of these 6 spaces available for Blue so we choose 2 places out of 6 in 6C2 ways =n 15


We repeat the same exercise in case 3 as well

Great way to look at it! Thanks for your time

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