A probability problem by Krishneel Donz

in the simplest possible format, probability is done by multiplication - the middle one has 3 options - the one attached to it has 2 options - (that's 6 so far) - the next two have 4 options, assuming the middle is A and the next is B, then they can be AA, CC, AC, CA - the last one's probability depends on the preceeding, it has 2 options if AA or CC, and 1 option if AC or CA - the propability can be then calculated by splitting those 4 options in the folowing form:

probability=(3x2x2x1)+(3x2x2x2)=36

Just note that any permutation of the valid coloring is also a valid coloring. Hence number of valid coloring should be divisible by $3!=6$ . This leaves only one choice.

Circle $E$ can be paint of $3$ colors, $C, B$ and $D$ of $2$ colors, in total $24$ ways then for select which will be the color of $A$ are to cases:

$1)$ $B$ and $D$ are the same color $\Rightarrow$ there are $2$ ways.

$2)$ $B$ and $D$ are different colors $\Rightarrow$ there is $1$ way.

For first case are $12 \times 2=24$ ways and for the second are $12\times 1=12$

$\therefore$ are $\boxed{36}$ different ways

3*2=6........what` divisible by 6 and just less than 40-->36

It all starts with the middle disc. For the middle disc there are 3 ways of coloring it. And corresponding to the middle disc only one out of the 3 discs that are not joined with the middle one should have the same color as that of the middle one. hence it follows like this ; Middle disc; 3 Disc connected to middle one; 2 Disc connected to above one in either direction {up or left}; 3 The last disc remaining ; 2 Hence the answer is 36