To be good with unusual or seemingly complex questions, you need to be good with basic rules of counting i.e. breaking the question, based on the conditions given, into easier manageable cases; sometimes work on piece-meal basis i.e. assign an object to a position and proceed like-wise, object by object i.e. dont expect some great unifying theory or approach that solves the entire question in one go.

Basically more complex a question seems, let go of standard monolithic approaches and fallback on rudimentary counting techniques. A example of this type ….

A circular pizza is cut into 5 sectors, each of a different size. The sectors are then distributed amongst 3 friends such that each friend receives atleast one sector and no friend receives any adjacent sectors. In how many ways can the pizza pieces be distributed amongst the 3 friends?

Task 1: How many ways can the 5 sectors can be distributed into 3 groups, with each group having atleast 1 sector?

Given the condition that no friend receives two adjacent sectors, now in how many ways can the above task be done?

Task 2: Are we just dividing the 5 sectors into 3 groups or are we even assigning the groups so formed to persons? Does this result in branching out of the tree? How many branches?

Task 3: What is it easier to begin with …. distributing which 2 pieces to a friend? Or distributing which 1 piece to a friend? And in how many ways can it be done?

Task 4: With pieces being given to one friend, we are left with smaller questions …. x pieces to be given to 2 friends. Since the scale is pretty small, can you just do it by enumeration, instead of any theory?

Learning #1: Working on the top-level broad contours of distributing the pieces … ignoring the specific details

Each of the three friends receives atleast 1 sector …. this takes care of 3 sectors …. and we are left with 2 sectors. Since the number is very small, we can reason it out by enumeration itself …. either both these sectors will go to one of the friends OR two of the friends will receive these sectors, each one getting one sector. If the former is done, then the ‘grouping’ of the 5 sectors will be as {3, 1, 1}. And in the latter case, it will be {2, 2, 1}

Note how we conveniently considered this as grouping i.e. we are currently ignoring all ‘order’ … which of the friend receives both the extra 2 sectors, OR which two friends receive the 2 sectors each, and which of the 5 sectors are the 2 sectors received by a friend. Right now, we are only working on the broader idea of ‘5 objects are grouped into 3 groups, each group having atleast one object’.

Learning #2: This question also has the condition that no friend can receive adjacent pieces. With this condition, is it possible that a friend can receive 3 out of the 5 pieces?

Now, this is a reasoning idea, it will not be covered in any course, because it is not some unknown theory, but something more plain, something that you need to ‘try yourself’ and not ‘learn’. Had the 5 pieces been in a row, selecting 3 pieces such that no two are adjacent, is possible. However, here the 5 pieces are in a circle … the first one is adjacent to the last one, unlike in a row … and hence try selecting 3 pieces such that no two are adjacent … see if you can make such a selection.

So, with the given condition of no friend getting adjacent sectors, the only way to distribute 5 sectors amongst 3 groups is {2, 2, 1}

Learning #3: Now, we will fill in the details.  Each of 2 friends receive 2 sectors and 1 friend receives 1 sector.

Which 2 friend are the ones who receive 2 sectors? And which two specific sectors does each of these two friends receive?

Which friend receives 1 sector of the pizza? Which sector does this friend receive?

A student good in problem solving will consider all the above questions and will ‘choose’ to go ahead with answering that question which is most easiest as of now. So, which of the above questions are the easiest to answer?

This will also not be taught by any course. Choosing 1 is very easy … which 1 friend out of 3? 3 ways …. which 1 sector out of 5? 5 ways.

Thus, we will fill in the details, starting with the friend who receives one sector …. start building the branching out of the tree.

The friend who receives 1 sector could be any one of the three …. so three branches. And the friend could receive any one of the five sectors. So each of the 3 branches will have further 5 branching-out i.e. a total of 3 × 5 = 15 branches as of now.

Learning #4: Now we are left with a smaller problem …. there are 4 sectors and each of 2 friends should receive 2 sectors. No friend receiving adjacent sectors.

Whichever 1 sector is given to the first friend, the remaining 4 sectors will be just one group i.e. it wont be the situation that 1 sector is there and there are gaps on either side of it and then 3 sectors together … all the 4 sectors will be together in one unit. Thus, our job is very easy now.

Choose any two out of p, q, r, s, such that the two chosen are not adjacent and the remaining two, the two not-chosen also are not adjacent.

Here again it is not about theory, it is just about trying to meet the conditions. By enumeration, but fast in your mind, lightening fast. If p is chosen, we cannot choose q … p & r will leave out q & s; p & s will leave q & r, not allowed; alternately, with q, one could only choose s. Thus, that there is just 1 way, p & r is one group of 2 sectors and q & s is the other group of 2 sectors.

And these two groups are to be assigned to two friends. This is basic, 2 objects 2 positions … 2! ways.

Each of these 2 ways, of distributing the remaining 4 sectors amongst 2 friends, is further branching out for every one of the 15 branches.

Thus, final answer = 15 * 2 = 30 ways.