GMAT Permutation Probability Questions

15 GMAT Quant Questions | Counting Methods Practice Questionbank

You may get two to three questions from Permutation Combination, counting methods and probability in the GMAT quant section - in both variants viz., problem solving and data sufficiency. The concepts tested include selecting one or more objects from a sample space, reordering objects with or without a constraint, questions on number sequences, tossing of coins, rolling a die, picking cards from a pack of cards, conditional probability, probability of exhaustive events, complimentary events, mutually exclusive events and independent events.

You could start navigating through this topic by watching the two embedded videos or jump straight to solving questions. If you know your basics in Permutation and Probability, Jump to Permutation Practice Questions

Watch these 2 videos to learn Permutation Combination basics


GMAT Permutation Probability Practice Questions

  1. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

    1. 3! × 3!
    2. \\frac{4!}{2!})
    3. \\frac{4! × 3!}{2!})
    4. \\frac{3! × 3!}{2!})
    Choice C
    Required number of rearrangements is (4! * 3!) / 2!
    Hint to solve this Permutation Rearrangement Problem

    Condition: The 3 vowels AAU in the word ABACUS appear together
    Step 1: Make the vowels as one unit. Let us call this unit X. So, we have to rearrange BCSX. This can be done in 4! ways.
    Step 2: Rearrange the vowels AAU in the unit. Product of 4! and the answer to Step 2 is the final answer.


  2. How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

    1. 59
    2. \\frac{11!}{2! × 2! × 2!})
    3. 56
    4. 23
    5. \\frac{11!}{3!×2!×2!×2!})
    Choice A
    The number of reorderings in which the first letter is E and the last one is R is 59.
    Hint to solve this Permutation Rearrangement of Letters Question

    Condition: The first letter is E and the last letter is R. Need to form 4-letter words.
    Step 1: Leaving one E and one R, count the number of times the other letters appear
    Step 2: Because some of the letters appear more than once, the second and third position of the required 4-letter word can be different letters or the same
    Step 3: Compute the number of outcomes if the two letters in the second and third position are different. Repeat the exercise if the letters are same. Add the result of the two computations to arrive at the final answer to this reordering of the letters of a word question.


  3. What is the probability that the position in which the consonants appear remain unchanged when the letters of the word "Math" are re-arranged?

    1. \\frac{1}{4})
    2. \\frac{1}{6})
    3. \\frac{1}{3})
    4. \\frac{1}{24})
    5. \\frac{1}{12})
    Choice A
    \\frac{1}{4})
    Approach to solve this probability question

    Step 1 - Denominator: Compute the number of ways in which the letters of the word MATH can be reordered. This value is the denominator to compute the required probability.
    Step 2- Numerator: Compute the number of ways in which the letters of the word MATH can be reordered if the consonants take the 1st, 3rd, and 4th place. Essentially, A will appear in the second place. We have to compute the number of ways in which M, T, and H will reorder in places 1, 3, and 4.
    Step 3: Now that we have computed the numerator and the denominator, the result will be the required probability.



  4. There are 6 boxes numbered 1, 2, ... 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

    1. 5
    2. 21
    3. 33
    4. 60
    5. 6
    Choice B
    The required number of ways to achieve the stated condition is 21.
    Approach to solve this Permutation problem

    Conditions: At least 1 box contains a green ball. Boxes containing green balls are consecutively numbered.
    Step 1: Anything from one to all 6 boxes could contain a green ball.
    Step 2: Count the number of outcomes if only 1 box contains a green ball; count the number of outcomes if 2 out of 6 boxes contain a green ball; continue for all possibilities up to all 6 boxes containing a green ball each.
    Step 3: Add the values counted in step 2 to arrive at the answer.


  5. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

    1. 1
    2. \\frac{1}{256})
    3. \\frac{81}{256})
    4. \\frac{175}{256})
    5. \\frac{144}{256})
    Choice D
    \\frac{175}{256})
    Approach to solve this Probability of Complementary Events Question

    Step 1: Compute the probability of not hitting a target in a shot.
    Step 2: Compute the probability of not hitting the target in all of the 4 shots.
    Step 3: The complement of not hitting the target in all 4 shots is the event of hitting the target at least once. So, subtract the result in step 2 from 1 to compute the required probability.


  6. In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes?

    1. 5 C 3
    2. 5 P 3
    3. 53
    4. 35
    5. 25
    Choice D
    The number of ways of posting the 5 letters in the 3 boxes is 35.
    Hint to solve this Sampling with Replacement Question

    Each of the 5 letters has 3 possibilities of being posted in a box. The product of these possibilities is the answer to this permutation combination question.


  7. Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

    1. 210
    2. 29
    3. 3 * 28
    4. 3 * 29
    5. None of these
    Choice B
    The number of outcomes in which the 3rd coin will turn up heads is 29.
    Approach to solve this Permutation Combination question

    Condition: 10 coins are tossed simultaneously. The 3rd coin is a head.
    Step 1: The number of ways in which the third coin turns a head is 1.
    Step 2: Each of the remaining coins could turn a head or a tail. Compute the total possibilities to arrive at the answer.


  8. In how many ways can the letters of the word "PROBLEM" be rearranged to make seven letter words such that none of the letters repeat?

    1. 7!
    2. 7C7
    3. 77
    4. 49
    5. None of these
    Choice A
    7!
    Approach to solve this GMAT Reordering practice question

    PROBLEM is a 7-letter word in which all letters are distinct. The number of ways in which r distinct objects can be reordered is r!.


  9. Bag x contains 3 red and 5 black balls and bag y contains 4 red and 4 black balls. One bag is selected at random and from the selected bag one ball is drawn. What is the probability that the ball drawn is red?

    1. \\frac{7}{8})
    2. \\frac{7}{16})
    3. \\frac{3}{16})
    4. \\frac{4}{5})
    5. \\frac{9}{16})
    Choice B
    7/16
    Hint to solve this GMAT Probability Sample Question

    Step 1: Compute the probability of selecting bag x and the probability of drawing a red ball from bag x. The product of these two values is the probability of selecting a red ball from bag x.
    Step 2: Compute the probability of selecting bag y and the probability of drawing a red ball from bag y. The product of these two values is the probability of selecting a red ball from bag y.
    Step 3: The sum of the values arrived at step 2 and 3 is the answer to the question.


  10. Data Sufficiency: Set A contains distinct integers: A = {2, 4, 6, -8, x, y}. When two numbers from this set are selected and multiplied, what is the probability that the product is less than zero?

    1. x * y is not equal to zero.
    2. |x| = |y|
    Choice B
    Approach to solve this GMAT Data Sufficiency problem

    Step 1: Compute the different possibilities for x and y given that we have a set of distinct integers. Both x and y could be positive (5 positive and 1 negative); both x and y could be negative (3 positive and 3 negative); one of x or y is positive and the other is negative (4 positive and 1 negative).
    Step 2: From statement 1, we can deduce that neither x nor y is 0. All 3 possibilities listed above are possible. Therefore, we will not be able compute a unique value.
    Step 3: Determine how many of the 3 possibilities will be valid if statement 2 is true. If we get only one possibility out of the 3, we will have a unique answer. In that scenario, statement 2 will be sufficient. Else, combine the statements and evaluate whether we are able to narrow the possibilities down to one of the three.


  11. There are 4 identical pens and 7 identical books. In how many ways can a person select at least one object from this set?

    1. 12
    2. (2 4 – 1)(2 7 – 1)
    3. 11
    4. 211 - 1
    5. 39
    Choice E
    39 ways.
    Approach to solve this GMAT Permutation Combination Problem

    Concept: Selecting from a set of identical objects
    Step 1: Compute the number of ways of selecting none or up to 4 pens from the set of 4 idential pens.
    Step 2: Compute the number of ways of selecting none or up to 7 books from a set of 7 identical books.
    Step 3: The product of the result of steps 2 and 3 will give the number of ways of selecting none or all of the objects
    Step 4: Subtract the only possibility of selecting none of the objects from the result of step 3 to arrive at the answer to this question.


  12. How many odd 4-digit positive integers that are multiples of 5 can be formed without using the digit 3?

    1. 900
    2. 729
    3. 3240
    4. 648
    5. 1296
    Choice D
    648 integers.
    Hint to solve this Permutation Question on Numbers and Digits

    Conditions: Odd multiple of 5; 4-digit positive integer; does not contain the digit 3
    Step 1: Compute the number of possibilities for the unit digit if the number is an odd multiple of 5
    Step 2: Compute the number of possibilities for the thousands place if it cannot be 3.
    Step 3: Compute the number of possibliities for the hundreds and tens place if those digits cannot include 3.
    Step 4: The product of the results of steps 1 to 3 is the answer to the question.


  13. How many six-digit positive integers comprising only the digits 1 or 2 can be formed such that the number is divisible by 3?

    1. 3
    2. 20
    3. 22
    4. 38
    5. 360
    Choice C
    22
    Hint to solve this GMAT Hard Math Question

    Condition: 6-digit positive integers; only digits to be used 1 or 2; divisible by 3
    Step 1: List down possibilities of the form of 6-digit numbers comprising only 1 or 2 that are divisible by 3. Example 111111.
    Step 2: Count the number of integers for each such possibility after factoring in the reorderings as applicable.
    Step 3: Sum of the counts in step 2 is the answer to the question.


  14. How many five-digit positive integers comprising only the digits 1, 2, 3, and 4, each appearing at least once, exist such that the number is divisible by 4?

    1. 120
    2. 24
    3. 72
    4. 60
    5. 54
    Choice D
    60
    Approach to solve this GMAT Numbers and Digits Permutation Question

    Conditions: 5-digit positive integer; only digits found 1, 2, 3, and 4; digits appearing at least once; divisible by 4.
    Step 1: List down possibilities for the two rightmost digits if the number is divisible by 4.
    Step 2: List down possiblities for the first three digits for each possibility in step 2. Reorder each such possibility for the first 3 digits and count the outcomes.
    Step 3: Sum of the counts in step 2 is the answer to the question.


  15. A fair coin is tossed 'n' times. If the number of outcomes in which two heads will appear is 28, what is the value of 'n'?

    1. 14
    2. 6
    3. 7
    4. 8
    5. 32
    Choice D
    8 terms
    Hint to solve this GMAT Combination Practice Question

    Theory: The number of outcomes in which 'r heads will appear when a coin is tossed 'n' times is nCr.
    Step 1: Number of outcomes in which 2 heads will appear in 'n' tosses = nC2.
    Step 2: Equate nC2 to 28 and find the value of n.


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