GRE® Permutation Combination & Probability

Concepts Tested in Permutation Combination

One can expect two to three questions from permutation combination and counting methods. It is imperative that you understand the basics of permutation and combination well so that you will be able to tackle questions from this topic. The following concepts are tested in permutation and combination in the GRE test.

• Sampling with and without replacement
• Sampling where order matters and where order does not matter
• Difference between permutation and combination
• Number sequence examples in counting methods
• Re ordering letters of a word
• Tossing of coins, rolling dice and pack of cards

Concepts Tested in Probability

If you have understood the basics of permutation and combination well, solving questions from probability becomes easy. In fact, many probability questions are a set of two permutation probability questions with the denominator being the total number of outcomes for an event and the numerator being the number of favorable outcomes. The following concepts are tested.

• Computing probabilities for events that are equally likely
• Computing probabilities for independent events that occur together
• Mutually exclusive events
• Conditional probabilities
• Geometric probabilities

Session 1 | 55 Minutes

Session 2 | 80 Minutes

1. There are 5 doors to a lecture room. In how many ways can a student enter the room through a door and leave the room by a different door?

   1. 10
   2. 9
   3. 20
   4. 625
   5. 1024

   Correct AnswerChoice C
   20 ways

   Explanation Video Solution
   Approach to solve this permutation question

   Step 1: Compute the number of ways of entering the room through a door.
   Step 2: Compute the number of ways of leaving the room through a different door.
   Step 3: The number of ways of entering through one door and leaving through a different door is the permutation of the number of outcomes of steps 1 and 2.
   Essentially, the product of values obtained in step 1 and step 2.

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2. In how many ways can 3 students be selected from a group of 12 students to represent a school in the inter school essay competition

   1. 33
   2. 12!
   3. 1320
   4. 220
   5. 36

   Correct AnswerChoice D
   220 ways

   Explanation Video Solution
   Approach to solve this permutation combination question

   This question is a classic example of selecting r objects from a sample space of n objects without replacement and where the order does not matter
   We need 3 students (so without replacement) and all 3 are going to do the same activity - participate in the essay competition (order does not matter).
   It is a nCr question.
   3 students can be selected from 12 in 12C3 ways = 220 ways.

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3. How many words can be formed by re-arranging the letters of the word PROBLEMS such that P and S occupy the first and last position respectively? (Note: The words thus formed need not be meaningful)

   1. \\frac{8}{2})
   2. 8! - 2!
   3. 6!
   4. 8! - 2*7!
   5. 6! * 2!

   Correct AnswerChoice C
   6! rearrangements

   Explanation Video Solution
   Approach to solve this permutation question

   All 8 letters of the word PROBLEMS are distinct.
   Given Constraint: P occupies the first place and S occupies last place.
   So, we have to rearrange 6 distinct letters (other than P and S) in the six available spaces. Computing the number of ways of reordering 6 different objects will give the answer to the question.

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4. GRE Quantitative Comparison Practice Question

   Quantity A	Quantity B
   The probability that a word selected from the set of all rearrangements of the letters of the word "Math" results in "Math"	The probability that a word selected from the set of all rearrangements of the letters of the word "Good" results in "Good"

   1. Quantity A is greater
   2. Quantity B is greater
   3. The two quantities are equal
   4. The relationship cannot be determined from the information given

   Correct AnswerChoice B
   Quantity B is greater

   Explanation Video Solution
   Approach to solve this GRE Probability QC Question

   Step 1: Quantity A : Compute the number of ways the letters of the word MATH can be rearranged. That will be denominator to compute the probability for quantity A.
   Step 2: The numerator is 1 because in only one of the different rearrangements will we get the same word MATH. Compute the probability using information obtained in steps 1 and 2 for quantity A
   Step 3: Quantity B : Compute the number of ways the letters of the word GOOD can be rearranged. That will be denominator to compute the probability for quantity B.
   Step 4: The numerator is 1 because in only one of the different rearrangements will we get the same word Good. Compute the probability using information obtained in steps 1 and 2 for quantity B
   Step 5: Compare results obtained for both quantities and determine the answer.

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5. In how many rearrangements of the letters of the word SCINTILLATING will no two 'I's appear together?

   1. ¹¹C₃ * 13!
   2. \\frac{10!}{2!*2!*2!})
   3. ¹¹C₃ * 3! * 10!
   4. ¹¹C₃\\frac{10!}{2!*2!*2!})
   5. \\frac{11!}{2!*2!*2!})

   Correct AnswerChoice D

   Explanation Video Solution
   Approach to solve this permutation question

   The condition given is that no two 'I's should appear together.
   The same condition can be reworded as "there should be at least one letter in between any two 'I's
   The way to achieve this condition is to place the remaining 10 letters in such a way that there is exactly one gap between any two of these letters.
   There are a total of 11 places where the 'I's can be placed
   Step 1: Compute the number of ways of selecting 3 places for the 3 'I's from these 11 places.
   Step 2: Compute the number of ways to reorder the remaining 10 letters.
   Step 3: Final answer is the product of the values obtained in Steps 1 and 2.

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6. How many squares are there in a chess board?

   1. 64
   2. 65
   3. 4096
   4. 1296
   5. 204

   Correct AnswerChoice E
   204 squares

   Explanation Video Solution
   Approach to solve this permutation question

   Step 1: Note that the chess board comprises squares other than the obvious 1 × 1 squares.
   Step 2: Count the number of 2 × 2 squares all the way up to 8 × 8 squares.
   Step 3: Add the number of squares of dimensions from 1 × 1 squares to 8 × 8 square to find the answer.

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7. What is the probability that two squares(smallest dimension) selected randomly from a chess board will have only one common corner?

   1. \\frac{7}{288})
   2. \\frac{7}{144})
   3. \\frac{7}{126})
   4. \\frac{7}{72})
   5. \\frac{2}{63})

   Correct AnswerChoice B

   Explanation Video Solution
   Approach to solve this GRE probability question

   Step 1: Find the number of ways of selecting two 1 × 1 square in a chess board. That is the denominator of the probability.
   Step 2: Count the total instances when two small 1 × 1 squares have only one common corner
   Step 3: Compute the number of ways of selecting a pair of such squares. That is the numerator of the probability
   Step 4: Compute required probability

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8. 8 directors, the vice chairman and the chairman are to be seated around a circular table. If the chairman should sit between a director and the vice chairman, in how many ways can they be seated?

   1. 9!
   2. 7! * 2
   3. 9! * 2
   4. 8! * 2
   5. 8!

   Correct AnswerChoice D
   8! * 2 ways

   Explanation Video Solution
   Method to solve this GRE Circular Permutation Problem

   Step 1: Make the chairman and vicechairman as one unit. This will ensure that the chairman has vicechairman on one side. The other side of the chairman will definitely be one of the 8 directors.
   Step 2: Compute number of ways of reordering the unit of chairman and vicechairman and the 8 directors around a circular table.
   Step 3: Compute the number of ways the chairman and the vice chairman can reorder between themselves.
   Step 4: The product of the values computed in steps 2 and 3 is the answer.

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9. Ms Li works at an office where the work timing is from 9:00 AM to 6:00 PM. 25% of a year she goes late to office and 35% of a year she leaves early from office. If P is the probability that she works at office the entire day then

   1. 0.25 ≤ P ≤ 0.35
   2. 0.25 ≤ P ≤ 0.65
   3. 0.4 ≤ P ≤ 0.65
   4. 0.35 ≤ P ≤ 0.4
   5. 0.1 ≤ P ≤ 0.6

   Correct AnswerChoice C

   Explanation Video Solution
   Approach to solve this GRE Probability Problem

   Step 1: The scenario in which the probability that Ms Li does not work the entire day is maximized is the scenario when the required probability is minimized. Compute minimum probability in this scenario.
   Step 2: The scenario in which the probability that Ms Li does not work the entire day is minimized is the scenario when the required probability is maximized. Compute maximum probability in this scenario.

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10. For which of the following events will the number of outcomes exceed 50?
    Indicate all such events.

    1. The number of outcomes in which at least three heads appears in 6 consecutive tosses of a fair coin.
    2. The number of outcomes in which the sum of the digits that appear on the facing side is odd when a fair die rolled thrice.
    3. The number of outcomes in which the two cards drawn from a pack of well shuffled cards are both red and face cards.
    4. The number of outcomes in which the vowels appear together when the letters of the word 'PRIORITY' are reordered.
    5. The number of ways of posting 6 different letters in 2 different post boxes such that at least one letter is posted in each of the boxes.
    6. The number of ways of selecting at least one Indian and at least one American for a debate from a group comprising 3 Indians and 4 Americans and no one else.

    Correct AnswerChoices B, D, E, and F

    Explanation Video Solution
    Approach to solve this permutation combination MTOA question

    Compute painstakingly the number of outcomes for each of the 6 answer options and include those answer options in which the number of outcomes exceeds 50. Make sure that you do not include any event for which the answer is 50.

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