All of you might have encountered various scenarios wherein, you would be handling multiple cases be it in QA or DI-LR. It would be again, very common that you tend to forget the ‘counted’ cases or miss a few cases leading to the wrong answer and/or waste a significant amount of time re-doing the problem. So, it might sound extremely superficial or even irrelevant but, ‘how to count’ is one of the questions that goes unanswered in almost every aspirant’s preparation. With CAT 2015 giving you slightly less than 2 minutes on an average to solve a question plus some questions without answer options, you have to be extremely alert while dealing with these question types so that you crack them perfectly in the first go itself.

While counting is indeed a matter of finding the various permutations and combinations, it can be simplified if you understand the logic behind the same. In this post, I will touch a bit of theory and solve a few examples and then show you how to actually count cases in tests more efficiently. If you are already organized when it comes to solving questions, this would reinforce the belief. If you are struggling while dealing with simple one line questions which involve a countable number of cases, this would be of help to you.

A good way to deal with counting is to move in a predefined manner: either from the lowest to the highest number or from the lowest alphabet to the highest alphabet and so on. For example, if you are dealing with scenarios involving tossing of 3 coins simultaneously and coming up with at most 2 heads, instead of going all out with the permutations and combinations formula (just because it was taught in your class/advised by a friend/was a ‘shortcut that would save time’/just like that) and then fumbling to get to the answer, you can simply write down the 8 possibilities and then figure out which ones you want to take. This is an extremely simple question and would probably not appear in CAT 2015 but you can use the method to solve complex questions as well.

To start with, we would represent the head by an H and the tail by a T. So, we will start with 3 Hs and go about replacing all the Hs one by one till we get all Ts. So, first case will always be HHH and then we will proceed as follows:

THH | Replace the first H by T |

HTH | Replace the middle H by T |

HHT | Replace the last H by T |

TTH | Replace the first two Hs by Ts (keep the sequence in mind, 1 takes precedence over 2, 2 takes precedence over 3, 1 and 2 take precedence over 1 and 3, 1 and 3 take precedence over 2 and 3 |

THT | Replace the first and the last Hs by Ts |

HTT | Replace the second and the last Hs by Ts |

TTT | Replace all Hs by Ts |

What I am doing here is that, increasing the number of tails one at a time by following the rule of precedence and then looking at individual cases. This helps me get more organized while writing down the cases.

Similarly if A, B, C, D and E are playing a competition such that each game involves 3 teams and each team plays the maximum possible number of games without repeating any of the previously formed groups, is better to do it in an organized manner. Basically, you know that there should be distinct groups of 3 teams such that all the five teams participate equally which would come to ^{5}C_{3} = 10 matches. Now, instead of randomly making groups of 3, it would be better if we fixed two teams and then went about adding the third. Also, as the order is not important, once we use a higher position team, we will not use it again when we move to the next pair (basically, if we have selected ABC, we will not consider BCA as a different team).

This means that we have to pick three teams going from left to right. So, A cannot be the second or third team that can be chosen and E cannot be the first or the second team that can be chosen. Once you understand this line, you can form the groups of 3 in an extremely technical manner without getting lost in the combinations.

The groups of the first two teams could be:

AB, AC, AD, BC, BD, CD ie. 6 cases. Remember that I have not considered E as it could only be the third team. Now, we will complete these groups of two by adding all the possible third elements (again in a sequential manner) and get the following: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE, a total of 10 cases. Once we jot these down, we have a skeleton ready for the entire set. With practice, it should not take you more than 30 seconds to finish this step.

Let’s take an example that appeared in CAT 2007:

Suppose you have a currency named Miso in three denominations, 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

(a) 17 (b) 16 (c) 18 (d) 15 (e) 19

Answer: If you solve this using equations and random substitution, you can go wrong. In a section where solving 10 questions was the key to getting an awesome score, accuracy would have been crucial. To begin with, we will start with the largest value in this case which is 50 Misos. You can have either 0, 1 or 2 notes of value Misos 50. So, we can fix it at 0, 1 and 2 and then fix the second largest denomination of 10 (ranging from 0 to 10, depending on the predecessor) and we should be done post that as the number of 1s would be dependent on the number of 50s and 10s and so, would automatically get fixed.

Case I: 0 notes of 50 Misos:

We will have to fix the number of 10 Misos notes here. As it can range from 0-10, we have 11 cases.

Case II: 1 note of 50 Misos:

Again, we know that the number of 10 Misos notes would range from 0-5 and so, we have 6 cases.

Case III: 2 notes of 50 Misos:

As there cannot be a 10 Misos note here, we have only one possibility of zero 10 Misos notes and so, only 1 case.

So, total of 11+6+1=18 cases. So, **option (C)**

Let’s take a bit more complex example: What is the number of 3s that appear in the first 1234 whole numbers?

This involves writing down the numbers in a simpler manner and then use our knowledge of permutations and combinations. The trap here is in the form of a minor detail mentioned as the whole number which implies that we start with 0 and end at 1233 (as 1233 will be the 1234^{th} whole number).

The common mistake made by candidates is that, they solve it using single digit, two digit, and three digit numbers individually. We can club the first 999 numbers and then deal with the rest individually. This is because, if we write the numbers as: 001, 002, 003, 004 till 999, we are essentially exhausting all the possible scenarios (as was the case with the heads and tails example). If you see the heads and tails example carefully, you would realize that in the 8 cases, heads and tails **appear equally frequently at all the three positions** (4 times each). This is a very important observation. So, in this case, all the digits occur equally in terms of frequency across all the 1000 cases for each position. As we have 10 digits, we will have 1/10^{th} probability that a digit takes a particular position in a particular case and so, 100 possible cases for each position and so, 100+100+100=300 incidences in total.

For the remaining cases from 1000 to 1233, you have to essentially find the number of 3s in the first 233 natural numbers. Again, you can split the same from 1 to 200 and then 1 to 33. 1 to 100 will have (1/10)*100*2=20 incidences which will be repeated from 101 to 200 (as the first digit does not involve a 3) and so, will be equal to 20 again. For the last 33 incidences, you can count them directly: 3, 13, 23, 30, 31, 32, 33 = 8 incidences.

So, total of 300+40+8=348 times.

If you know the trick to do this, you can simply write 340 first and then go about counting the last 8. It should not take more than a minute to solve this one.

The applications of the ‘how-to-count’ concept are immense and it is useful across almost all the topics as you can see in the problems for practice part.

**Problems for practice:**

1) Find the number of scalene triangles having integral sides that could be formed of perimeter less than or equal to 18 units.

2) What will be the sum of all the odd numbers that could be formed by using the digits 1, 2, 3 and 4 without repetition?

**Additional tip**: To get your brain trained to deal with arrangements and counting, I would highly recommend solving Su-Do-Ku and Kakuro puzzles (the latter has some brilliant applications in arrangement based LR sets as well so you might want to give it a shot).