CAT Coaching Onlne - Links across topics in Math

Some simple interesting patterns emerge when we look across topics. Very often, thinking about these patterns helps us wind our heads around one or the other topic. In this post, we discuss a few of those links. Nothing profound, but just an outline to help remember simple ideas

Simple Interest Compound Interest and Progressions.

The amounts at the end of each year form an AP if interest is calculated on a simple interest basis, and form a GP if interest is calculated on a compound interest basis.

For instance if amount invested is Rs. 1000 and interest rate were 10%, then amount at the end of yr 1, yr 2, yr 3,.. would be Rs. 1100, 1200, 1300,...Each year's amount is previous year's amount + Rs. 100

The same amount of Rs. 1000 invested at same 10% but on a compound interest basis would give us amounts at the end of yr 1, yr 2 of Rs. 1100, 1210, 1331...Each year's amount is previous year's amount * 1.1. This is the basis for the formula for amount on principal invested in Compound Interest basis

Combinatorics - This can be linked with almost any topic
See if you can spot some idea in some other topic that is 'sitting inside' the following combinatorics questions

1. Consider a shelf that has 4 copies of a book and 3 copies of a painting. In how many ways can we select at least one article from this shelf?

2. Consider a shelf that has 4 different books and three different paintings. In how many ways can we select at least one article from this shelf?

3. In how many ways can be place 5 different toys into 3 different boxes such that all 5 toys are allotted and no box is empty?

How else these can be interpreted is given below.

1. Consider a shelf that has 4 copies of a book and 3 copies of a painting. In how many ways can we select at least one article from this shelf?
How many factors greater than 1 does the number 2^4 & 3^3 have?

2. Consider a shelf that has 4 different books and three different paintings. In how many ways can we select at least one article from this shelf?
How many non-empty subsets does the set {1, 2, 3, 4, 5, 6 7} have?

3. In how many ways can be place 5 different toys into 3 different boxes such that all 5 toys are allotted and no box is empty?
How many onto fucntions can be defined from the set {1, 2, 3, 4, 5} to the set {a, b, c}?

Weighted averages and Mixtures.
Class A has 40 students whose average mark is 40, class B has 60 students whose average mark is 50. What is the overall average? Class A has an average mark of 40 and class B has a an average mark of 50. If the overall average is 46, what is the ratio of students in class A and class B? These two questions are two sides of the same coin. Sometimes we teach one in weighted averages and the other in mixtures.

Mathematicians are supposed to love connections. There are far more powerful connections across topics than the ones I have mentioned above. It is instructive to observe even these simple connections.

CAT Online Coaching - Permutation and Combination, Fixing the Errors II

This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to questions 4, 5 and 6. The questions and the incorrect solutions here

4. In how many ways can we select 2 cards from a card pack such that both belong to the same suit?

Given solution

The first card can be selected in 52 ways, as it can be any card in the pack. Now, there are 12 different ways of selecting the second card. Depending on the suit of the first card, the choice set we have for the second card gets limited. If the first card were a club, then the second card must be from among the 12 remaining clubs and so on.

So, total number of outcomes = 52 * 12.

Bug in the solution:

We are double-counting when we do this. If we select 2H as the first card and 5D as the second card; this is same as selecting 5D as the first card and 2H as the second card. However, in our current approach, we will end up counting both of these.

Correct solution:

The correct approach would be to select one of the four suits first and then select 2 cards from 13 in that suit. Or, the number of ways of doing this = 4 * ¹³C₂ = 52 * 6.

5. When a coin is tossed 10 times, what is the probability of getting more heads than tails?

Given solution

When a coin is tossed 'n' times, the probability of getting more heads than tails = Probability of getting more tails than heads. So, we do not need to really compute this. Probability = 1/2.

Bug in the solution:

We are not counting the scenarios where there are an equal number of heads and tails. If we threw the coin an odd number of times, the probability of getting more heads than tails will be ½.

Correct solution:

Number of heads could be 6, 7, 8, 9, or 10. Probability = ( ¹⁰C₆ + ¹⁰C₇ + ¹⁰C₈ + ¹⁰C₉ + ¹⁰C₁₀)/ 2¹⁰

The other way of thinking about this would be ½ (1 – ¹⁰C₅)/ 2¹⁰. If we removed the scenario where the number of heads and tails are equal, the probability of more heads than tails should be equal to probability of more tails than heads.

6. In how many ways can we arrange 3 boys and 3 girls on a straight line such that no two boys stand next to each other

Given solution

We can arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total number of possibilities = 2 * 3! * 3! = 72 ways.

Bug in the solution:

We are not accounting for arrangement such as BGBGGB and BGGBGB. Note that the question says that no two boys should stand next to each other. Two girls can stand next to each other.

Correct solution:

There are 4 possible outlines – BGBGBG, GBGBGB, BGBGGB or BGGBGB. So, total number of possibilities = 3! * 3! * 4 = 144. 

CAT Online Coaching - Permutation and Combination, Fixing the Errors

This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to the first three questions.

1. Five boys need to be allotted to 4 different rooms such that each boy is allotted a room and no room is empty. In how many ways can this be done?

Given solution

Let the boys be A, B, C, D and E. Let the rooms be 101, 102, 103 and 104. Now, we know that exactly one room will have two occupants. First, let us try to send 4 boys to 4 rooms, we can worry about the fifth occupant later on. Let us select 4 out of the 5 boys first. This can be done in 5C4 ways. Now, these 4 can be allotted to 4 different rooms in 4! ways. So, 4 boys in 4 rooms can be done in 5C4 * 4! = 5 * 24 = 120 ways.

Now, the fifth boy has to go into one of the rooms. He can do this in 4 ways as there are 4 different rooms available. So, total number of outcomes = 120 * 4 = 480.

Bug in the solution:

We end up double-counting here.

A, B, C and D could be allotted rooms 101, 102, 103 and 104 in that order. Post this, E could “double-up” with A. So, we would have A and E in 101, B in 102, C in 103 and D in 104.

In another scenario, E, B, C, D could be allotted 101, 102, 103 and 104 in that order. Post this, A could “double-up” with E. So, we would have A and E in 101, B in 102, C in 103 and D in 104.

The two scenarios mentioned above are identical. However, we end up counting both. This is the reason for the double count.

Note that in our method, we end up having a ‘first’ occupant for a room and a ‘second’ occupant. We say, A goes into room number 101 and then E ‘joins’ him. The moment you do that, ‘order’ creeps in. We end up factoring in order when we shouldn’t.

Awesome, isn’t it.

Correct solution:

The boys should be allotted into rooms as 2 + 1 + 1 + 1. As in, exactly one room has 2 occupants. Some two guys have to be room-mates. Let us first select these two room-mates. This can be done in ⁵C₂ ways. After we have selected these two room-mates, we practically need to just allot these 4 ‘groups’ into 4 rooms. This can be done in 4! Ways.

Total number of options = ⁵C₂ * 4! = 10 * 24 = 240 ways.

2. How many 4-digit numbers with 4 distinct digits are possible?

Given solution

Let the 4-digit number be 'abcd'. Now,

'd' can take 10 possible values - 0 to 9. 

'c' can take 9 possible values - 0 to 9 except d

'b' can take 8 possible values - 0 to 9 except d and c

'a' can take 6 possible values - 1 to 9 except d, b and c

So, total number of possibilities = 6 * 8 * 9 * 10 = 4320.

Bug in the solution:

The simple way of stating the bug is as follows “But one of b, c, d might have been zero”.

When we say a can take 6 possible values. We eliminate 0, b, c, and d from the list 0 to 9. If b, c or d were zero, then that in that case, a will still have 7 possible options.

In this question, the leading digit has a constraint. So, start with that. No point factoring in the constraint in the last step. Get that out of the way first.

Correct solution:

Let the 4-digit number be 'abcd'. Now,

'a' can take 9 possible values - 1 to 9. 

'b' can take 9 possible values - 0 to 9 except a

'c' can take 8 possible values - 0 to 9 except a and b

'd' can take 7 possible values - 1 to 9 except a, b and c

Total number of options = 9 * 9 * 8 * 7.

3. In how many ways can we rearrange the letters of the word TWOIIM such that the vowels appear together?

Given solution

Let us take the three vowels together and call it as X. Now, we need to rearrange the letters of the word TWMX. This can be done in 4! ways. Or, there are 24 ways of rearranging the letters of the word TWOIIM with vowels appearing together.

Bug in the solution:

We do not account for the different variations that ‘X’ can come in. Now, the three vowels have been replaced with X. Or, X is nothing but OII in some order. The key point here is that X could be OIO, IOO or OOI. X can take three different forms. The solution has overlooked this.

Correct solution:

Let us take the three vowels together and call it as X. Now, we need to rearrange the letters of the word TWMX. This can be done in 4! ways.

Now, the three vowels have been replaced with X. Or, X is nothing but OII in some order. The key point here is that X could be OIO, IOO or OOI. X can take three different forms.

So total number of possibilities = 4! * 3 = 72 ways.

Permutation and Combination: View the entire Chapter for free

Permutations and Combinations is a wonderful topic to prepare for and is a fun topic if done the right way. The entire chapter on Combinatorics from the book on Quantitative Aptitude for CAT is available for free here .

As an author, I quite enjoyed the challenge of creating questions for this topic and was pleased with the way the chapter turned out as well. This is considered a tricky topic by many CAT aspirants and so I was keen to have this as the free sample topic.

Please send any feedback you have on this topic (and/or the book) to prep@2iim.com.

The book can be bought at Flipkart through this link .