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Permutation and Combination is a classic topic where getting something wrong is often more instructive than getting it right the first time, provided we have another look at what exactly went wrong with the incorrect solution.

To reinforce this idea of “debugging” a wrong solution in order to understand this topic better, I have given a bunch of questions with incorrect solutions that need to be fixed. So, have a go at this. Before you try, let me give a few ground rules.

  1. You should try to figure out what exactly is wrong with this solution. It is not enough to say 240 is wrong as 120 is the correct answer. Try to articulate which step of the given solution is wrong.
  1. There might be 1-2 questions where the answer + solution are indeed correct. Otherwise, there is no thrill in this.

Now, on to the questions and solutions.

 

  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?

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.

This solution has been debugged here

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

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.

This solution has been debugged here

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

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.

This solution has been debugged here

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

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.

This solution has been debugged here

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

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.
This solution has been debugged here

  1. 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

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.

This solution has been debugged here

  1. 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?

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 [ack = 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

This solution has been debugged here

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

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.

This solution has been debugged here

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

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.

This solution has been debugged here

  1. This is an interesting question – Of the 10 questions given here, how many answers are wrong?

Solution: Can the 10th question have two possible answers?