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Blog Series: Fundamental Counting Principle, Permutation and Combination.
Blog #1: Using Fundamental Counting Principle with examples.

Before I jump to what Combinatorics is and what kind of questions appear on the GRE, I will like to briefly discuss one obstacle I face every time I am teaching this to a class. During my short teaching career, I have seen multiple students fear this particular topic not because they cannot learn the right approach but because they cannot unlearn the wrong approach i.e. using formulas blindly. This is not entirely the student’s fault since most schools do not pay attention to building a student’s ability to reason and fail to inculcate in them the idea that Mathematics is not plugging in numbers into expressions to reach to the right solution. So if you are confused about this concept, I urge you to put past concepts aside and read the article as if you are completely unfamiliar with combinatorics.

Lets assume you are attempting a Multiple Choice Question (MCQ) with 3 choices. You have to choose one option.

____- Write a,b or c.

What is the total no. of ways in which you can answer the question? 3. How? You can either select option a or b or c. This means, you have 3 choices for one option.
Now lets take a look at a slightly more tough example. Lets say you have to mark two different choices, what is the total number of ways in which you can answer the question now?

____ and ___

To understand, first realise that for the first choice you can choose any of the three options. This means you have three choices to choose from for the first blank. Once you have chosen the first choice, you are left with 2 choices for the second since you need to opt for different choices.
Now, you can find out the number of total ways in many different ways but lets first take the simplest path i.e. by writing down all the possible arrangements.
a,b
a,c
b,a
b,c
c,a
c,b

Therefore, the total number of ways is six.

Now, we can solve this question by the Fundamental Counting Principle, the total no. of arrangements can be found out by multiplying the no. of choices available for each blank.
Number of total ways= (No. of choices for 1st blank) x (No. of choices for 2nd Blank)
= 3 x 2
= 6

In actual practise, most of the times it is not feasible to write down choices by hand because it is neither effective nor possible and therefore, we opt for the fundamental counting principle. To understand, consider there are 100 participants in an audition out of which 3 candidates will make it to the next round. What is the total no. of ways in which the participants can be selected?

1. What is the total no. of candidates for the first slot?
2. What is the total no. of candidates for the second slot?
3. What is the total no. of candidates for the third slot?

The number of candidates for the first slot is 100. Since the second slot is to be filled with a new candidate, there are 99 possible ways to fill the second slot and 98 for the last.

Thus, the total no. of possible arrangements = 100 x 99 x 98 = 970200
One cannot simply write so many arrangements by hand.

The formulae for calculating permutation and combination are neither required for GRE nor recommended. This is because often times the GRE will add some restrictions to each possibility and it will be harder to solve by using the formula. However, using the fundamental counting principle is almost always easy. Try using the principle on the following questions:

Questions

1. In an online game, Tim needs to select a team of one male and one female player. For the male player, Tim can choose from a pool of 5 players. For the female player, Tim can choose from another list of 9 players. How many different teams can he choose?
2. John is taking a standardised test that contains ten reading comprehension questions. Each question has 5 different choices. In how many ways can he answer the test?

1. Choices for male player= 5
Choices for first female player= 9

Total number of ways= 5 x 9
= 45

2. Choices for first question/ No. of ways he can answer the first question = 5
Choices for second question = 5
.
.
.
.
Choices for tenth question = 5

Total number of ways in which John can attempt the exam= 5 x 5 … x 5 = 5^10