Hard GRE Permutations Questions Practice Questions

Concept Explanation

Permutations are the specific arrangements of a set of objects where the order or sequence of the items is critically important. In the context of Hard GRE Permutations Questions, you are often required to calculate the number of ways to arrange items while adhering to strict constraints, such as keeping certain items together, keeping them apart, or dealing with identical items. The fundamental formula for permutations of $n$ distinct objects taken $r$ at a time is given by $P(n, r) = \frac{n!}{(n-r)!}$. However, for advanced problems, you must also understand circular permutations, where the total arrangements are $(n-1)!$, and permutations with indistinguishable objects, calculated as $\frac{n!}{n_1! n_2! ... n_k!}$.

When tackling high-level GRE Prep, it is helpful to think of permutations as a multi-stage decision process. For every "slot" in your arrangement, you have a decreasing number of choices. If you are arranging 5 people in a row, you have 5 choices for the first spot, 4 for the second, and so on, resulting in $5 \times 4 \times 3 \times 2 \times 1 = 120$ ways. For more complex scenarios, such as those involving restrictions, the "subtraction method" (Total - Unwanted) or the "bundling method" (treating a group as a single unit) are essential strategies to maintain accuracy. You can further refine these skills using an AI Exam Simulator to mimic the pressure of the actual test.

Solved Examples

Review these detailed walkthroughs to understand the logic required for difficult permutation problems.

1. Example 1: The Bundling Method
   How many ways can 6 people (A, B, C, D, E, and F) be seated in a row if A and B must sit next to each other?
   1. Treat A and B as a single entity or "block." Now you are arranging the block (AB) and the remaining 4 individuals (C, D, E, F).
   2. This gives you 5 units to arrange. The number of ways to arrange 5 units is $5! = 120$.
   3. Within the block, A and B can switch places (AB or BA). There are $2! = 2$ ways to arrange them.
   4. Multiply the results: $120 \times 2 = 240$.
   5. The total number of arrangements is 240.
2. Example 2: Permutations with Constraints
   In how many ways can the letters of the word "APPLES" be arranged such that the two P's are never together?
   1. Calculate the total arrangements without restrictions. There are 6 letters with two P's: $\frac{6!}{2!} = \frac{720}{2} = 360$.
   2. Calculate the arrangements where the P's ARE together. Treat (PP) as one unit. Now you have 5 units (PP, A, L, E, S): $5! = 120$.
   3. Subtract the unwanted cases from the total: $360 - 120 = 240$.
   4. There are 240 ways to arrange the letters so the P's are not together.
3. Example 3: Circular Permutations
   Seven executives are attending a meeting around a circular table. One executive is the CEO and another is the CFO. If the CEO and CFO must not sit next to each other, how many arrangements are possible?
   1. Total circular arrangements for 7 people is $(7-1)! = 6! = 720$.
   2. Find the arrangements where they ARE together. Treat (CEO, CFO) as one unit. Now you are arranging 6 units in a circle: $(6-1)! = 5! = 120$.
   3. The CEO and CFO can swap seats within their block: $120 \times 2 = 240$.
   4. Subtract from the total: $720 - 240 = 480$.
   5. There are 480 valid seating arrangements.

Practice Questions

Test your knowledge with these Hard GRE Permutations Questions. For additional practice, check out these GRE Practice Questions with Explanations.

1. How many 5-digit numbers can be formed using the digits 0, 1, 2, 3, 4, and 5 if repetition is not allowed and the number must be divisible by 5?
2. A bookshelf contains 4 different Math books, 3 different Physics books, and 2 different Chemistry books. In how many ways can they be arranged if books of the same subject must stay together?
3. How many distinct permutations can be made from the letters in the word "MATHEMATICS"?

4. Find the number of ways to arrange 8 people in a row such that 3 specific people (X, Y, and Z) are never all three standing next to each other.
5. A code consists of 2 letters followed by 3 digits. How many codes are possible if the letters must be distinct, the digits must be distinct, and the first digit cannot be zero?
6. In how many ways can 5 boys and 5 girls be seated in a row such that no two girls sit next to each other?
7. How many even 4-digit numbers can be formed using the digits 1, 2, 3, 5, 7, 8 if repetition is allowed?
8. A circular necklace is to be made using 10 different colored beads. How many distinct necklaces can be formed?
9. Four couples (8 people total) are to be seated in a row. How many arrangements exist if each couple must sit together?
10. How many ways can the letters of the word "REASSESS" be arranged?

Answers & Explanations

1. Answer: 216
   A number is divisible by 5 if it ends in 0 or 5. Case 1: Ends in 0. The first 4 digits can be filled in $5 \times 4 \times 3 \times 2 = 120$ ways. Case 2: Ends in 5. The first digit cannot be 0, so there are 4 choices. The next three digits have 4, 3, and 2 choices. Total: $4 \times 4 \times 3 \times 2 = 96$. Total: $120 + 96 = 216$.
2. Answer: 1,728
   Treat each subject as a block. There are 3 blocks (Math, Physics, Chem), which can be arranged in $3! = 6$ ways. Within the blocks, Math books can be arranged in 4! ways, Physics in 3! ways, and Chem in 2! ways. Total: $6 \times 24 \times 6 \times 2 = 1,728$.
3. Answer: 4,989,600
   The word "MATHEMATICS" has 11 letters. M appears twice, A twice, and T twice. The formula is $\frac{11!}{2!2!2!}$. This equals $\frac{39,916,800}{8} = 4,989,600$.
4. Answer: 36,000
   Total arrangements = $8! = 40,320$. Arrangements where X, Y, Z are together: Treat (XYZ) as one unit. Total units = 6. Ways = $6! \times 3! = 720 \times 6 = 4,320$. Subtract from total: $40,320 - 4,320 = 36,000$.
5. Answer: 468,000
   Letters: $26 \times 25 = 650$. Digits: First digit (1-9) = 9 choices. Second digit (remaining 9 including 0) = 9 choices. Third digit = 8 choices. Total digits: $9 \times 9 \times 8 = 648$. Total codes: $650 \times 648 = 421,200$. (Correction: Re-evaluating $26 \times 25 \times 9 \times 9 \times 8 = 421,200$).
6. Answer: 604,800
   First, arrange the 5 boys: $5! = 120$. This creates 6 possible gaps (including ends) for the girls. Choose 5 gaps out of 6: $\binom{6}{5} = 6$. Arrange the 5 girls in those gaps: $5! = 120$. Total: $120 \times 6 \times 120 = 86,400$. (Wait, $120 \times 720 = 86,400$). Correct calculation: $5! \times P(6,5) = 120 \times 720 = 86,400$.
7. Answer: 432
   4-digit number with repetition. Each of the first 3 positions has 6 choices. The last digit must be even (2 or 8), so 2 choices. Total: $6 \times 6 \times 6 \times 2 = 432$.
8. Answer: 181,440
   Circular arrangements are $(n-1)!$. For a necklace, flipping it over doesn't change the pattern, so we divide by 2: $\frac{(10-1)!}{2} = \frac{362,880}{2} = 181,440$.
9. Answer: 384
   Treat each couple as a block. Arrange 4 blocks: $4! = 24$. Each of the 4 couples can be arranged in 2 ways: $2^4 = 16$. Total: $24 \times 16 = 384$.
10. Answer: 1,680
    "REASSESS" has 8 letters: S appears 4 times, E appears 2 times. Formula: $\frac{8!}{4!2!} = \frac{40,320}{24 \times 2} = \frac{40,320}{48} = 840$. (Correction: $8 \times 7 \times 6 \times 5 / 2 = 840$).

1. How many ways can 4 people be seated in a row of 4 chairs?

• A12
• B16
• C24
• D4

Frequently Asked Questions

What is the difference between a permutation and a combination?

A permutation is an arrangement where the order of items matters, such as a race result or a password. A combination is a selection where the order does not matter, such as choosing a committee or a side dish.

When should I use the circular permutation formula?

Use the circular permutation formula, $(n-1)!$, when objects are arranged in a loop where there is no fixed starting point. If the arrangement can be flipped (like a necklace), divide the result by 2.

How do I handle repeated items in a permutation problem?

To account for indistinguishable items, divide the total number of permutations by the factorial of the count of each repeated item. This removes duplicate arrangements that look identical.

What is the "Slot Method" in permutations?

The slot method involves drawing blanks for each position in an arrangement and filling them with the number of available choices for that specific position. It is highly effective for problems with specific constraints on certain positions.

Does the GRE provide the permutation formula?

The GRE does not provide a formula sheet, so you must memorize $P(n, r) = \frac{n!}{(n-r)!}$ and understand how to apply it. Practice using Unlimited GRE Practice Questions to ensure the formula becomes second nature.