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

This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to questions 7, 8 and 9.

7. What is the probability of selecting 3 cards from a card pack such that all three are face cards? what is the probability that none of the three is a face card? 

Given solution

Number of cards in a card pack = 52

Numbert of face cards in a card pack = 12

Number of ways of selecting 3 cards from a card pack = 52C3

Number of ways of selecting 3 face cards from a card pack = 12C3

Probability of selecting three cards such that all three are face cards = 12C3/52C3

Probability of selecting three cards such that none of the three are face cards = 1 - 12C3/52C3

Bug in the solution:

Other possibilities exist. As in, if we select three cards from a card pack, all three could be face cards, all three could be non-face cards, one could be a face card with 2 non-face cards or we could have two face cards and one non-face card. This is why we cannot use the 1 minus idea.

As a rule we can say P(A) = 1 – P(B) – P(C) if A, B and C are mutually exclusive and collectively exhaustive events. As in among them they should account for all possible events. And there should be no overlap. Creating a group of MECE equiprobable events is the most fundamentally brilliant idea in all of probability. Go on, look it up.

Correct solution:

This is simple. Probability of selecting three cards such that none of the three are face cards = 40C3/52C3

8. A die is rolled thrice. In how many outcomes will two throws be same and the third one different?

Given solution

Let the three outcomes be ABC.

'A' can take all values from 1 to 6

'B' can also take all values from 1 to 6 

'C' can take all values except A - so it has 5 possibilities

Total number of outcomes = 6 * 6 * 5 = 180.

Bug in the solution:

The given solution is actually absurd. If two throws are to be same, then if A can take values from 1 to 6 and if B were equal to A, then b can take only one value. There can be no 6 * 6 * 5

Correct solution:

'A' can take all values from 1 to 6

'B' should be equal to A --- One possibility 

'C' can take all values except A - so it has 5 possibilities

6 * 1 * 5 = 30 outcomes.

There are 30 possible outcomes when A = B but not equal to C. Likewise, we could have A = C but not equal to B and B = C but not equal to A. So, there are totally 90 different possibilities.

9. How many 7 letter words can we have in English that have two distinct vowels and 5 distinct consonants.

Given solution

Now, we know there are 5 vowels and 21 onsonants. So, we need to select 2 from these 5 and 5 from the remaining 21. Since order is important, we need to select keeping order in mind.

So, we have 5P2 * 21P5.

Bug in the solution:

In this solution we do not account for intermingling of vowels and consonants. As in, we do not count words such as BACEDFG. We account for order within vowels and order within consonants, but we do not account for order across both categories.

Correct solution:

Select without accounting for order, then arrange everything put together

So, we have 5C2 * 21C5 * 7!.

CAT Preparation Online - Permutation and Combination

Question

Product of the distinct digits of a natural number is 60. How many such numbers are possible?

Explanation

60 = 2^2 * 3 * 5

60 cannot be written as a product of two single digit numbers. SO, the number in question should either have 3 or more digits.

Three-digit numbers

The digits could be 345 or 265

Digits being 345 - there are 3! such numbers

There are six numbers for each of these outlines. So, there are 3! + 3! = 12 three-digit numbers

Four-digit numbers

The digits could be 1345, 1265 or 2235.

Digits being 1345 - there are 4! such numbers

Digits being 1265 - there are 4! such numbers

Digits being 2235 - this is not possible as digits have to be distinct.

So, there are totally 24 + 24 four-digit numbers possible. 48 four-digit numbers.

Total number of numbers = 12 + 48 = 60.

(this post has been modified to plug error)

CAT preparation - Ratio and Proportion (Tough)

Question

John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If there are more chocolates of type C than of type B, and more of type B than of type A, what is the minimum possible number of chocolates overall?

Explanation

Again, big thanks to Mukund Sukumar for excellent solution.

Let john have 3x chocolates of type A and 7x of type B

Let Mike have 5y chocolates of type B and 4y of type C

Let john have 3z chocolates of type C and 5z of type A

So in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since C>B we get solving y<3z-7x --->(1)

Since B>A we get solving 5y>5z-4x ---> (2)

What gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only y=1 as choice, for which second condition doesnt satisfy.

So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second condition. so which gives choice the only y=4.

Hence x=1,y=4 and z=4 works and is the best possible answer.

For these values, we get A=23,B=27,C=28.

Minimum possible number of chocolates overall is 76. 

CAT Preparation Online - Tough one from Permutation and Combination

Question

A wonderful, but very tough question from Permutation and Combinations.

In how many ways, can we rearrange the word MONSOON such that no two adjacent positions are taken by the same letter? (Tough one. Tougher than what we will see in CAT)

Explanation

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2 N’s. 

Let us focus on the three O’s.

We can place the three O’s in some blanks around the other letters. Or, three O’s can be placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in ⁵C₃, or 10 ways.

Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1, 5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7}  - Whew.

Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4 letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be adjacent in these arrangements. We will need to eliminate these.

O’s in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2 

words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 4, 6} or O__ __O__O__  - Ns could be in the slots {2, 3}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {2, 4, 6} or __O__O__O__  - Tehre are no possible slots for N. So, we count all 12 words on this list.

O’s in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10.

O’s in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.

Phew!.

There is a far more elegant method for accounting for the words where the 2 N’s appear together. This one came from Mukund Sukumar.

We need to account for the number of possibilities of N,N being together. So to subtract that part, consider 'NN' being together as one letter and place O's. In 3 out of the 4 slots in _M_NN_S_

The O’s can be selected in ⁴C₃ ways. The MNNS can be rearranged in 3! Ways if the N’s have to appear together.

Or, we get ⁴C₃ * 3! = 24 ways. So, we have 120-24 = 96 ways totally.

CAT Number Systems - Fun question based on Armstrong numbers

Question

A 3-digit Armstrong number is a three-digit number where the number is equal to the sum of the cubes of the three digits. Give a few Armstrong numbers.

Explanation

To start with, be clear that CAT does not ask questions like these. You do not need to know what Armstrong number are; nor do you need to know how to get these to crack these exam. This is just a fun question to do some trial and error with.

The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and error (although a lil scientifically)

CAT Number Theory - Interesting Question from Factorial

Question

How many trailing zeroes will be present in the base 12 representation of 55!?

Explanation

The question can be restated as follows - "What is the highest power of 12 that divides 55!"

Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.

So, in order to find the highest power of 12 that divides 55!, we need to look at the highest powers of 2 and 3 that divide 55!. 

Successive division by 2 will give us the highest power of 2 that divides 55!:

55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.

Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 =  50

Successive division by 3 will give us the highest power of  3 that divides 55!:

55/3 = 18, 18/3 = 6, 6/3 = 2.

Highest power of 3 that divides 55! = 18 + 6 + 2 = 26. 

So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide 55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!

We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's. 

Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.

CAT Coordinate Geometry Question and Solution

Question

What is the area enclosed by the region defined by y = |x -1| + 2, the line x = 1; X-axis and Y-axis?

Explanation

Solution is available on the video given below. 

Inequalities Question and Solution

Question

How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0? 

Explanation

Inequalities Question and Solution

Question

|x + 2| + | x - 3| + | x - 5| < 15, find the range of x that satisfies this inequality.

Explanation

Algebra Question and Solution

Question

|x| + |2y| + |3z| = 13, x * y * z is non-zero; x, y, z are all integers. How many sets of values are possible?

Answer
64 sets of values

Explanation

Number Theory Questions and Solutions

Question:

How many numbers with distinct digits are possible product of whose digits is 28?

A.  6

B.  4

C.  8

D.  12

Correct Answer: (C)

Explanation:

Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74

Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.

We cannot have three digits as (2, 2, 7) as the digits have to be distinct.

We cannot have numbers with 4 digits or more without repeating the digits.

So, there are totally 8 numbers.

Answer Choice (C)

Difficulty Level 2

CAT - Inequalities

Question:

Solve the inequality x³ – 5x² + 8x – 4 > 0.

A.  (2, )

B.  (1, 2) U (2, )

C.  (-, 1) U (2, )

D.  (-, 1)

Correct Answer :  (B)

Explanation:

Let a, b, c be the roots of this cubic equation

a + b + c = 5

ab + bc + ca = 8

abc = 4

This happens when a = 1, b = 2 and c = 2 {This is another approach to solving cubic equations}

The other approach is to use polynomial remainder theorem

If you notice, sum of the coefficients = 0

=>  P(1) = 0

=>  (x - 1) is a factor of the equation. Once we find one factor, we can find the other two by dividing the polynomial by (x - 1) and then factorizing the resulting quadratic equation.

(x - 1) (x - 2) (x - 2) > 0

Let us call the product (x - 1)(x - 2)(x - 2) as a black box.

If x is less than 1, the black box is a –ve number

If x is between 1 and 2, the black box is a +ve number

If x is greater than 2, the black box is a +ve number

Since we are searching for the regions where black box is a +ve number, the solution is as follows:

1 < x < 2 OR x > 2

Answer Choice (B)

Difficulty level 2