JEE Main & Advanced

Permutations and Combinations

Permutations and Combinations for JEE Main & Advanced

1
Module 1

Fundamental Principles & Permutations

Fundamental Principles of Counting & Linear PermutationsTopic 1

Counting problems rely on two core logical rules. The Multiplication Rule states that if an operation can be performed in $m$ independent ways, and a second operation can follow in $n$ ways, the two operations can be performed together in $m \times n$ total ways. The Addition Rule states that if two operations are mutually exclusive, and can be performed in $m$ and $n$ ways respectively, choosing either operation can be done in $m + n$ ways. An arrangement of $n$ distinct objects taken $r$ at a time is a linear permutation, given by $^nP_r = \frac{n!}{(n-r)!}$. When all $n$ objects are arranged, this simplifies to $n!$. If items are not all distinct, containing $p$ identical items of one type and $q$ of another, the total arrangements equal $\frac{n!}{p!q!}$. A common mistake is using the multiplication rule for events that cannot happen simultaneously.

Worked Examples
1

Find the total number of distinct 4-letter words that can be formed using the letters of the word `UDGAM`, assuming repetition of letters is not allowed. Also, find the count if repetition is allowed.

Show solution
The word `UDGAM` consists of exactly 5 distinct letters: \{U, D, G, A, M\}.
  • Case 1: Repetition is not allowed.
    We need to choose and arrange 4 distinct letters out of 5. This matches the standard linear permutation definition: \[ ^5P_4 = \frac{5!}{(5-4)!} = \frac{120}{1} = 120 \] Alternatively, using the Fundamental Principle of Counting: there are 5 choices for the first slot, 4 for the second, 3 for the third, and 2 for the fourth. By the multiplication rule: \[ 5 \times 4 \times 3 \times 2 = 120 \text{ ways} \]
  • Case 2: Repetition is allowed.
    For each of the 4 available blank slots in our word layout, we can choose any of the 5 letters independently. By the multiplication rule: \[ 5 \times 5 \times 5 \times 5 = 5^4 = 625 \text{ ways} \]
Final Answer: Without repetition $= 120$, with repetition $= 625$.
2

Find the total number of unique arrangements of the letters of the word `SUCCESS`.

Show solution
First, let us count the total number of letters and identify identical sets: The word `SUCCESS` contains $n = 7$ total letters. Let us group the identical letters:
  • The letter `S` appears exactly $p = 3$ times.
  • The letter `C` appears exactly $q = 2$ times.
  • The letter `U` appears exactly $1$ time.
  • The letter `E` appears exactly $1$ time.
Apply the permutation formula for objects that are not all distinct: \[ \text{Total Arrangements} = \frac{n!}{p! \cdot q!} = \frac{7!}{3! \cdot 2!} \] Expand and simplify the factorials: \[ \text{Total Arrangements} = \frac{5040}{6 \times 2} = \frac{5040}{12} = 420 \] Final Answer: $420$.
3

How many numbers between 100 and 1000 can be formed using the digits \{0, 1, 2, 3, 4, 5\} if no digit is repeated in any number?

Show solution
Numbers between 100 and 1000 must be exactly 3-digit numbers. Let us represent the slots as H (Hundreds), T (Tens), and O (Ones).
  • Hundreds Slot: A valid 3-digit number cannot start with the digit 0, otherwise it becomes a 2-digit number. Therefore, we can choose any digit from \{1, 2, 3, 4, 5\}, giving exactly 5 choices.
  • Tens Slot: The digit 0 can now be safely used. Since repetition is forbidden, we must exclude the digit chosen for the hundreds slot. Out of the 6 total available digits, 1 is gone, leaving exactly 5 choices.
  • Ones Slot: We must exclude the 2 digits already chosen for the previous slots. This leaves exactly $6 - 2 =$ 4 choices.
Apply the fundamental multiplication rule to find the total count: \[ \text{Total Numbers} = 5 \times 5 \times 4 = 100 \] Final Answer: $100$.
✎ Self-Check — 5 questions0 / 5
Q1.A student has 3 distinct pairs of pants and 4 distinct shirts. The total number of unique ways they can choose a pants-and-shirt combination to wear is:
Q2.The value of the permutation expression $^{10}P_3$ is equal to:
Q3.The total number of unique permutations that can be formed using all the letters of the word `MATRIX` is:
Q4.The total number of arrangements of the letters of the word `APPLE` is:
Q5.The number of distinct 3-digit numbers that can be formed using the digits \{1, 2, 3, 4\} when repetition of digits is explicitly allowed is:

Advanced Restricted Arrangements (Tie-Together & Gap Methods)Topic 2

Advanced permutation problems include restrictions on how objects can be placed relative to each other.
  • Tie-Together Method: When specific objects must *always be together*, bind them into a single virtual object. Arrange this block with the remaining items, then multiply by the internal permutations of the bounded group.
  • Gap Method: When specific objects must *never be together*, arrange the remaining unrestricted objects first. This creates a sequence of empty spaces (gaps) between them, including the ends. If there are $m$ unrestricted items, they create $m+1$ internal gaps. Choose slots from these gaps to place the restricted items.
Worked Examples
1

Find the number of ways in which 5 boys and 4 girls can be arranged in a straight row such that all the girls sit together.

Show solution
Since all 4 girls must sit together, bind them into a single virtual block: (GGGG). Now treat this block as 1 single item alongside the 5 individual boys. This gives a total of $5 + 1 = 6$ items to arrange.
  • Step 1: Arrange these 6 items in a row, which can be done in $6!$ ways.
  • Step 2: For each arrangement, the 4 girls inside the bounded virtual block can rearrange themselves in $4!$ ways.
Apply the fundamental multiplication rule to find the total number of ways: \[ \text{Total Ways} = 6! \times 4! = 720 \times 24 = 17,280 \] Final Answer: $17,280$.
2

Find the number of ways in which 5 boys and 4 girls can be arranged in a straight row such that no two girls sit next to each other.

Show solution

Since the girls must never sit next to each other, use the Gap Method. Step 1: Arrange the 5 unrestricted boys in a row first. This can be done in $5!$ ways: \[ \text{\_ } B_1 \text{ \_ } B_2 \text{ \_ } B_3 \text{ \_ } B_4 \text{ \_ } B_5 \text{ \_} \] Step 2: Count the number of empty spaces (gaps) created between and around the boys. Arranging 5 boys creates exactly $5 + 1 = 6$ gaps. Step 3: Choose 4 gaps out of these 6 to place the 4 girls, and then arrange the girls inside those slots. This can be done in $^6P_4$ ways. Apply the multiplication rule to combine the steps: \[ \text{Total Ways} = 5! \times ^6P_4 = 120 \times (6 \times 5 \times 4 \times 3) = 120 \times 360 = 43,200 \] Final Answer: $43,200$.

3

Find the total number of ways to arrange the letters of the word `ARTICLE` such that the vowels always occupy the even places.

Show solution
Let us write out the letters of the word `ARTICLE` and classify them: The word contains 7 total letters. The vowels are \{A, I, E\} (3 total), and the consonants are \{R, T, C, L\} (4 total). Set up 7 numbered slots in a row: $1, 2, 3, 4, 5, 6, 7$. Identify the even positions: positions 2, 4, and 6 are the only even slots (exactly 3 slots).
  • Step 1: Arrange the 3 vowels into the 3 available even slots. This can be done in $3!$ ways.
  • Step 2: Arrange the 4 consonants into the remaining 4 odd slots (positions 1, 3, 5, 7). This can be done in $4!$ ways.
Apply the multiplication rule to find the total number of arrangements: \[ \text{Total Ways} = 3! \times 4! = 6 \times 24 = 144 \] Final Answer: $144$.
✎ Self-Check — 5 questions0 / 5
Q1.If 6 books are arranged in a shelf row, and 2 specific history books must always be together, the total number of valid ways is:
Q2.Using the gap method to arrange 4 men and 3 women such that no two women stand next to each other, the number of available gaps created around the men is:
Q3.The number of ways to arrange the letters of the word `VOWEL` such that the vowels are never together is calculated using complementary counting ($5! - \text{together}$). The result is:
Q4.If 5 keys are arranged on a circular ring loop rather than a linear row, the clockwise and anti-clockwise paths are identical. The total number of ways is:
Q5.The number of ways to arrange 4 distinct people around a round dinner table is given by $(n-1)!$, which evaluates to:
2
Module 2

Combinations & Restricted Selections

Selection Logic & Algebraic IdentitiesTopic 1

Combinations track selection problems where the order of items does not matter. The number of ways to choose $r$ unordered items from a set of $n$ distinct objects is given by $^nC_r = \frac{n!}{r!(n-r)!}$. Key algebraic identities for combinations include:
  • Symmetric Identity: $^nC_r = ^nC_{n-r}$ (choosing $r$ items is equivalent to leaving $n-r$ items behind).
  • Pascal's Identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.
In the JEE Advanced exam, Pascal's Identity is frequently used to simplify large combinatorial series summation strings.
Worked Examples
1

Simplify the summation expression: $S = ^{10}C_4 + ^{10}C_3$.

Show solution

Let us apply Pascal's Identity directly to the expression: \[ ^nC_r + ^nC_{r-1} = ^{n+1}C_r \] Identify the parameters from the problem: $n = 10$ and $r = 4$. Substitute these parameters into the right-hand side of the identity: \[ S = ^{10+1}C_4 = ^{11}C_4 \] Now evaluate the combinatorial value explicitly: \[ ^{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \frac{7920}{24} = 330 \] Final Answer: $330$.

2

Find the value of $r$ that satisfies the equation $^{20}C_{r+2} = ^{20}C_{2r-3}$.

Show solution
By the symmetric identity property, the equation $^nC_x = ^nC_y$ splits into two distinct algebraic cases:
  • Case 1: The indices are directly equal.
    \[ r + 2 = 2r - 3 \implies r = 5 \] Verify the boundary conditions: $r+2 = 7 \le 20$ and $2r-3 = 7 \le 20$, which is valid.
  • Case 2: The indices add up to the total population ($x + y = n$).
    \[ (r + 2) + (2r - 3) = 20 \implies 3r - 1 = 20 \implies 3r = 21 \implies r = 7 \] Verify the boundary conditions: $r+2 = 9 \le 20$ and $2r-3 = 11 \le 20$, which is also valid.
Both cases yield valid integer indices. Final Answer: $r = 5, 7$.
3

Simplify the series expression: $E = ^{50}C_4 + \sum_{j=1}^{4} ^{54-j}C_3$.

Show solution

Let us expand the summation series from right to left (starting with $j=4$ up to $j=1$): \[ S = ^{50}C_4 + \left( ^{50}C_3 + ^{51}C_3 + ^{52}C_3 + ^{53}C_3 \right) \] Group the first two terms together and apply Pascal's Identity ($^{50}C_4 + ^{50}C_3 = ^{51}C_4$): \[ S = \left(^{51}C_4\right) + ^{51}C_3 + ^{52}C_3 + ^{53}C_3 \] Apply Pascal's Identity a second time to the first two terms ($^{51}C_4 + ^{51}C_3 = ^{52}C_4$): \[ S = \left(^{52}C_4\right) + ^{52}C_3 + ^{53}C_3 \] Apply the identity a third time ($^{52}C_4 + ^{52}C_3 = ^{53}C_4$): \[ S = \left(^{53}C_4\right) + ^{53}C_3 \] Apply the identity one final time to find the simplified expression for the entire series: \[ S = ^{54}C_4 \] This telescoping application of Pascal's identity is a classic JEE strategy. Final Answer: $^{54}C_4$.

✎ Self-Check — 5 questions0 / 5
Q1.A cricket squad contains 15 players. The number of ways to choose an 11-player team is:
Q2.If $^nC_8 = ^nC_2$, then the value of the expression $^nC_2$ is equal to:
Q3.The value of the combinatorial sum $^{15}C_3 + ^{15}C_4$ simplifies to:
Q4.A student must select 3 elective courses out of 7 options. The total number of valid choices available is:
Q5.The identity relationship $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ is known as the property of:

Restricted Selections & Word Rank ProblemsTopic 2

Selection problems often include constraints where certain items must always or never be included.
  • If $k$ specific items must always be included, we reduce both the population and the target selection size by $k$, giving $^{n-k}C_{r-k}$ ways.
  • If $k$ specific items must never be included, we remove them from the population entirely, giving $^{n-k}C_r$ ways.
Another important application is finding the Rank of a Word in a Dictionary. This involves counting how many words can be formed alphabetically before reaching the target word, processing the letters position-by-position from left to right.
Worked Examples
1

A committee of 5 members is to be formed from a group of 6 gentlemen and 4 ladies. Find the number of ways this can be done if the committee must contain exactly 2 ladies.

Show solution
The committee requires exactly 5 members in total. We are given the condition that it must contain exactly 2 ladies.
  • Step 1: Choose 2 ladies from the 4 available ladies. This can be done in $^{4}C_2$ ways.
  • Step 2: Choose the remaining members of the committee from the gentlemen. Since the committee requires 5 members and 2 are ladies, we need $5 - 2 = 3$ gentlemen. Choose 3 gentlemen from the 6 available gentlemen: $^{6}C_3$ ways.
Apply the fundamental multiplication rule to find the total number of ways: \[ \text{Total Ways} = ^{4}C_2 \times ^{6}C_3 \] Evaluate the combinatorial values: \[ ^{4}C_2 = \frac{4 \times 3}{2 \times 1} = 6 \] \[ ^{6}C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \] Multiply the values together: \[ \text{Total Ways} = 6 \times 20 = 120 \] Final Answer: $120$.
2

Find the rank of the word `MODERN` when all its structural permutations are arranged alphabetically as in a dictionary.

Show solution
First, let us list the 6 distinct letters of the word `MODERN` in alphabetical order: \[ \text{Alphabetical Order: } E, \ D, \ M, \ N, \ O, \ R \] *Correction check on order*: The correct alphabetical order is $D, E, M, N, O, R$.
Let us count how many words can be formed before reaching `MODERN`, processing position-by-position from left to right:
  • Words starting with D: Fix D in the first slot. The remaining 5 letters can be arranged in $5! = 120$ ways.
  • Words starting with E: Fix E in the first slot. The remaining 5 letters can be arranged in $5! = 120$ ways.
  • Words starting with M: This matches the first letter of our target word `MODERN`. Freeze M and move to the second slot: $\mathbf{M \ \_ \ \_ \ \_ \ \_ \ \_}$.
    • Words starting with MD: The remaining 4 letters can be arranged in $4! = 24$ ways.
    • Words starting with ME: The remaining 4 letters can be arranged in $4! = 24$ ways.
    • Words starting with MN: The remaining 4 letters can be arranged in $4! = 24$ ways.
    • Words starting with MO: This matches the second letter of our target word. Freeze O and move to the third slot: $\mathbf{MO \ \_ \ \_ \ \_ \ \_}$.
      • Words starting with MOD: This matches the third letter of our target word. Freeze D and move to the fourth slot: $\mathbf{MOD \ \_ \ \_ \ \_}$.
        • Words starting with MODE: This matches the fourth letter of our target word. Freeze E and move to the fifth slot: $\mathbf{MODE \ \_ \ \_}$.
          • The next available letter in alphabetical order is N, which forms the word $\mathbf{MODERN}$. This is our target word!
      Sum the number of words that appear before our target word: \[ \text{Words Before} = 120 (\text{for D}) + 120 (\text{for E}) + 24 (\text{for MD}) + 24 (\text{for ME}) + 24 (\text{for MN}) = 312 \] The rank of the word `MODERN` is the very next position: \[ \text{Rank} = 312 + 1 = 313 \] Final Answer: $313$.
3

Out of a pool of 10 distinct mathematical textbooks, find the number of ways to select 4 books such that the most advanced textbook is always included.

Show solution

The total population is $n = 10$, and we need to choose $r = 4$ books. We are given the condition that 1 specific advanced book must always be included in the selection. This means we have 1 less book to choose, and our remaining choices must be made from the remaining pool of books. \[ \text{Remaining choices} = r - 1 = 4 - 1 = 3 \] \[ \text{Remaining pool} = n - 1 = 10 - 1 = 9 \] Apply the combination formula to find the number of ways to choose the remaining books: \[ \text{Total Ways} = ^9C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \] Final Answer: $84$.

✎ Self-Check — 5 questions0 / 5
Q1.A student must choose 5 questions out of 10 on an exam, but the first 2 questions are compulsory. The number of valid choices available is:
Q2.The total number of ways to choose a 4-member team from 8 people such that the oldest person is never included is:
Q3.Alphabetical dictionary ordering shows that the rank of the short word `CAT` across its 6 permutations is exactly:
Q4.The number of ways to choose 3 coordinates from a set of 8 distinct points on a circle to form a triangle is:
Q5.If a box contains 5 red balls and 3 blue balls, the number of ways to select 3 balls such that at least 2 are red is found by summing the cases:
3
Module 3

Advanced Distributions & Combinatorics

Division into Groups & Multinomial Integer SolutionsTopic 1

Advanced combinations problems involve partitioning objects into groups or finding the number of integer solutions to multi-variable equations.
  • Division into Distinct Groups: The number of ways to divide $m+n$ distinct objects into two distinct groups containing $m$ and $n$ items respectively is given by $\frac{(m+n)!}{m!n!}$. If the two groups are identical in size ($m=n$), we must divide by $2!$ to remove the ordering of the groups, giving $\frac{(2m)!}{2!(m!)^2}$.
  • Multinomial Equations (Stars and Bars): The number of non-negative integer solutions ($x_i \ge 0$) to the linear equation $x_1 + x_2 + \dots + x_r = n$ can be found using the stars and bars formula: $^{n+r-1}C_{r-1}$.
  • If the problem requires positive integer solutions ($x_i \ge 1$), we use the formula $^{n-1}C_{r-1}$.
Worked Examples
1

Find the total number of ways to divide 4 distinct toy tokens into two identical groups of 2 tokens each.

Show solution
The total number of objects is $2m = 4$, so the size of each group is $m = 2$. Apply the group division formula for identical groups of equal size: \[ \text{Total Ways} = \frac{(2m)!}{2! \cdot (m!)^2} = \frac{4!}{2! \cdot (2!)^2} \] Expand and evaluate the factorials: \[ \text{Total Ways} = \frac{24}{2 \times 4} = \frac{24}{8} = 3 \] Let us verify this by listing the partitions manually for the set $\{a, b, c, d\}$: The 3 valid pairs of identical groups are:
  1. Group 1: $\{a, b\}$, Group 2: $\{c, d\}$
  2. Group 1: $\{a, c\}$, Group 2: $\{b, d\}$
  3. Group 1: $\{a, d\}$, Group 2: $\{b, c\}$
Notice that the order of the groups does not matter because the groups are identical. Final Answer: $3$.
2

Find the total number of non-negative integer solutions to the equation $x + y + z = 10$.

Show solution

The equation is of the form $x_1 + x_2 + \dots + x_r = n$, where the sum total is $n = 10$ and the number of variables is $r = 3$. The problem specifies non-negative integer solutions ($x, y, z \ge 0$). Apply the standard stars and bars combinatorial formula: \[ \text{Number of Solutions} = ^{n+r-1}C_{r-1} \] Substitute the values of $n$ and $r$ into the formula: \[ \text{Number of Solutions} = ^{10+3-1}C_{3-1} = ^{12}C_2 \] Evaluate the binomial coefficient: \[ ^{12}C_2 = \frac{12 \times 11}{2 \times 1} = 66 \] Final Answer: $66$.

3

Find the total number of positive integer solutions to the equation $x + y + z = 10$.

Show solution

The sum total is $n = 10$ and the number of variables is $r = 3$. The problem specifies positive integer solutions, which means every variable must be strictly greater than or equal to 1 ($x, y, z \ge 1$). Apply the modified positive integer solutions formula: \[ \text{Number of Solutions} = ^{n-1}C_{r-1} \] Substitute the values of $n$ and $r$ into the formula: \[ \text{Number of Solutions} = ^{10-1}C_{3-1} = ^9C_2 \] Evaluate the binomial coefficient: \[ ^9C_2 = \frac{9 \times 8}{2 \times 1} = 36 \] Alternatively, we can derive this by substituting variables: let $x = x' + 1, y = y' + 1, z = z' + 1$, where $x', y', z' \ge 0$. Substituting these into the original equation gives $(x'+1) + (y'+1) + (z'+1) = 10 \implies x' + y' + z' = 7$. Solving for non-negative solutions yields $^{7+3-1}C_{3-1} = ^9C_2 = 36$. Both methods match perfectly. Final Answer: $36$.

✎ Self-Check — 5 questions0 / 5
Q1.The number of ways to divide 6 distinct items into two separate distinct packets containing 4 and 2 items respectively is:
Q2.The number of non-negative integer solutions to the equation $a + b = 5$ is exactly:
Q3.Using the positive integer solutions formula $^{n-1}C_{r-1}$, the number of solutions to $x_1 + x_2 + x_3 + x_4 = 6$ (where $x_i \ge 1$) is:
Q4.The total number of ways to distribute $n$ identical items into $r$ distinct bins is structurally identical to finding the number of:
Q5.The number of ways to divide 6 distinct blocks into 3 identical groups of 2 blocks each requires dividing by $3!$. The total count is:
4
Module 4

Derangements & Grid Geometries

Derangements & Grid Counting IdentitiesTopic 1

Derangements count permutations where no element appears in its original natural position. For example, if $n$ distinct letters are inserted into $n$ addressed envelopes, a derangement occurs if every single letter lands in a wrong envelope. The number of derangements of $n$ objects is given by the subfactorial formula: \[ D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!} = n! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots + \frac{(-1)^n}{n!} \right) \] Another advanced topic is Grid Geometry Counting. In a regular grid map of order $m \times n$, the total number of rectangles that can be formed is given by choosing 2 horizontal lines and 2 vertical lines simultaneously: $\text{Rectangles} = ^{m+1}C_2 \times ^{n+1}C_2$.

Worked Examples
1

There are exactly 4 letters and 4 corresponding directed envelopes. Find the number of ways to insert the letters into the envelopes such that every single letter goes into a wrong envelope.

Show solution

This problem matches the definition of a derangement of $n = 4$ objects. Let us apply the subfactorial derangement formula: \[ D_4 = 4! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) \] Expand the terms inside the parentheses: \[ D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) \] Distribute the multiplier 24 across the terms: \[ D_4 = \frac{24}{2} - \frac{24}{6} + \frac{24}{24} = 12 - 4 + 1 = 9 \] Thus, there are exactly 9 ways to completely mix up the 4 letters. Final Answer: $9$.

2

Find the total number of rectangles contained inside a standard $3 \times 4$ grid mesh.

Show solution
A grid mesh of order $m \times n = 3 \times 4$ consists of 3 horizontal square blocks and 4 vertical square blocks. Let us find the number of grid lines bounding these blocks:
  • Total horizontal grid lines $= m + 1 = 3 + 1 = 4$ lines.
  • Total vertical grid lines $= n + 1 = 4 + 1 = 5$ lines.
To form a rectangle, we must choose exactly 2 horizontal lines out of 4, and 2 vertical lines out of 5. Apply the combination rules: \[ \text{Total Rectangles} = ^4C_2 \times ^5C_2 \] Evaluate the binomial coefficients: \[ ^4C_2 = \frac{4 \times 3}{2 \times 1} = 6 \] \[ ^5C_2 = \frac{5 \times 4}{2 \times 1} = 10 \] Multiply the values together: \[ \text{Total Rectangles} = 6 \times 10 = 60 \] Final Answer: $60$.
3

Find the number of ways to rearrange the letters of the word `ABC` such that no letter remains in its original position.

Show solution
The word contains $n = 3$ distinct letters. Let us apply the derangement formula for 3 objects: \[ D_3 = 3! \left( \frac{1}{2!} - \frac{1}{3!} \right) = 6 \left( \frac{1}{2} - \frac{1}{6} \right) = 3 - 1 = 2 \] Let us verify this by listing the permutations of `ABC` manually: The 6 total arrangements are: `ABC`, `ACB`, `BAC`, `BCA`, `CAB`, `CBA`.
  • In `BCA`, position 1 is B (was A), position 2 is C (was B), position 3 is A (was C). Every letter moved!
  • In `CAB`, position 1 is C (was A), position 2 is A (was B), position 3 is B (was C). Every letter moved!
The other 4 arrangements leave at least one letter in its original position. This confirms the result. Final Answer: $2$.
✎ Self-Check — 5 questions0 / 5
Q1.The value of the derangement constant $D_5$, which counts the ways to completely mix up 5 items, is:
Q2.The formula to count the total number of rectangles inside an $m \times n$ grid mesh is:
Q3.If $D_n$ represents the number of derangements of $n$ items, the value of $D_2$ is exactly:
Q4.The number of horizontal grid lines contained inside a grid structure with $m$ rows of square boxes is:
Q5.If 3 letters are matched randomly with 3 envelopes, the number of ways where exactly one letter goes into its correct envelope is:

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