Let ⁿCr - 1 = 28, ⁿCᵣ = 56${}^nC_{r - 1} = 28, {}^nC_r = 56$ and ⁿCr + 1 = 70${}^nC_{r + 1} = 70$. Let A(4cost, 4sint), B(2sint, -2cost) and C(3r - n, r² - n - 1$r^2 - n - 1$) be the vertices of a triangle ABC, where t is a parameter. If (3x - 1)² + (3y)² = α$(3x - 1)^2 + (3y)^2 = \alpha$, is the locus of the centroid of triangle ABC, then α$\alpha$ equals:

Related Formula

Core Logic

Let the 7 digits be formed using 1s, 2s, and 3s. We seek combinations whose sum is 11. Since minimum value for 7 digits using '1' is 7, we evaluate the distribution of surplus elements (11 - 7 = 4$11 - 7 = 4$ remaining to add).

Case 1: Using five 1s and two 3s

Digits: 1, 1, 1, 1, 1, 3, 3$\{1, 1, 1, 1, 1, 3, 3\}$

Case 2: Using four 1s, two 2s, and one 3

Digits: 1, 1, 1, 1, 2, 2, 3$\{1, 1, 1, 1, 2, 2, 3\}$

Case 3: Using three 1s and four 2s

Digits: 1, 1, 1, 2, 2, 2, 2$\{1, 1, 1, 2, 2, 2, 2\}$

Step 1: Summing the Total Cases

Pattern Recognition

Always set a base state (e.g., all 1s) to compute the baseline sum, then distribute the remainder explicitly via integer partitions to verify all distinct permutation paths systematically.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Related Formula

Core Logic

The word MATHS has 5 distinct letters: {M, A, T, H, S}. We need to form 6-letter words such that any chosen letter appears ≥ 2$\geq 2$ \times. We analyze combinations by structural frequency cases.

Case 1: Single letter used 6 \times

Format: a a a a a a$a a a a a a$ Choose 1 letter out of 5: 51 = 5$\binom{5}{1} = 5$ words.

Case 2: Two distinct letters used

Subcase 2a: One letter 4 \times, another 2 \times (aaaa bb$aaaa bb$)

Subcase 2b: Both letters used 3 \times each (aaa bbb$aaa bbb$)

Total for Case 2 = 300 + 200 = 500$= 300 + 200 = 500$ words.

Case 3: Three distinct letters used

Format: Each letter appears exactly 2 \times (aa bb cc$aa bb cc$)

Step 1: Calculate Total Words

Pattern Recognition

When constraints enforce frequencies ≥ 2$\geq 2$, organize calculations strictly by number of distinct letters to cover all possibilities without overcounting.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Related Formula

The number of ways to distribute m$m$ distinct items into k$k$ indistinguishable groups is given by the sum of Stirling numbers of the second kind:

Core Logic

Since the offices are indistinguishable, we look at partitions of the number 5$5$ into at most 4$4$ parts. Let's analyze each case based on distribution patterns:

Step 1: Calculate Case Combinations

• Case 1: 5, 0, 0, 0$5, 0, 0, 0$

All employees in one office: (5!)/(5!) = 1 way$\frac{5!}{5!} = 1 \text{ way}$

• Case 2: 4, 1, 0, 0$4, 1, 0, 0$

Four employees in one office, one in another: (5!)/(4!1!) = 5 ways$\frac{5!}{4!1!} = 5 \text{ ways}$

• Case 3: 3, 2, 0, 0$3, 2, 0, 0$

Three employees in one office, two in another: (5!)/(3!2!) = 10 ways$\frac{5!}{3!2!} = 10 \text{ ways}$

• Case 4: 3, 1, 1, 0$3, 1, 1, 0$

Three in one, and one each in two other offices. Since the two single-person offices are identical:

• Case 5: 2, 2, 1, 0$2, 2, 1, 0$

Two groups of two, and one group of one. Since the two two-person offices are identical:

• Case 6: 2, 1, 1, 1$2, 1, 1, 1$

One group of two, three groups of one. Since the three single-person offices are identical:

Step 2: Total Sum

Summing all these distinct non-overlapping partition groups:

Pattern Recognition

Sees: Distinct items placed into identical/indistinguishable containers. Trap: When dividing items into groups of equal sizes (like two groups of 2$2$ or three groups of 1$1$), you must divide by the factorial of the frequency of those groups (2!$2!$ and 3!$3!$) to avoid overcounting permutations.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Related Formula

For a linear equation x + 2y = k$x + 2y = k$, where x, y ≥ 0$x, y \ge 0$ are non-negative integers, the number of distinct integral pairs (x,y)$(x,y)$ is equal to the number of possible non-negative values that y$y$ can take, which is given by:

Core Logic

We need to find the number of non-negative integer solutions to:

Rearranging the equation to analyze values branch-by-branch based on z$z$:

Since x, y ≥ 0$x, y \ge 0$, the maximum value z$z$ can attain corresponds to x=0, y=0$x=0, y=0$, so 3z ≤ 42 z ≤ 14$3z \le 42 \implies z \le 14$.

Step 1: Enumerate Values across all z configurations

Let's count the possible solutions for each integer value of z$z$ from 0$0$ to 14$14$:

• z = 0 x + 2y = 42 42/2 + 1 = 22 solutions$z = 0 \implies x + 2y = 42 \implies \lfloor 42/2 \rfloor + 1 = 22 \text{ solutions}$
• z = 1 x + 2y = 39 39/2 + 1 = 20 solutions$z = 1 \implies x + 2y = 39 \implies \lfloor 39/2 \rfloor + 1 = 20 \text{ solutions}$
• z = 2 x + 2y = 36 36/2 + 1 = 19 solutions$z = 2 \implies x + 2y = 36 \implies \lfloor 36/2 \rfloor + 1 = 19 \text{ solutions}$
• z = 3 x + 2y = 33 33/2 + 1 = 17 solutions$z = 3 \implies x + 2y = 33 \implies \lfloor 33/2 \rfloor + 1 = 17 \text{ solutions}$
• z = 4 x + 2y = 30 30/2 + 1 = 16 solutions$z = 4 \implies x + 2y = 30 \implies \lfloor 30/2 \rfloor + 1 = 16 \text{ solutions}$
• z = 5 x + 2y = 27 27/2 + 1 = 14 solutions$z = 5 \implies x + 2y = 27 \implies \lfloor 27/2 \rfloor + 1 = 14 \text{ solutions}$
• z = 6 x + 2y = 24 24/2 + 1 = 13 solutions$z = 6 \implies x + 2y = 24 \implies \lfloor 24/2 \rfloor + 1 = 13 \text{ solutions}$
• z = 7 x + 2y = 21 21/2 + 1 = 11 solutions$z = 7 \implies x + 2y = 21 \implies \lfloor 21/2 \rfloor + 1 = 11 \text{ solutions}$
• z = 8 x + 2y = 18 18/2 + 1 = 10 solutions$z = 8 \implies x + 2y = 18 \implies \lfloor 18/2 \rfloor + 1 = 10 \text{ solutions}$
• z = 9 x + 2y = 15 15/2 + 1 = 8 solutions$z = 9 \implies x + 2y = 15 \implies \lfloor 15/2 \rfloor + 1 = 8 \text{ solutions}$
• z = 10 x + 2y = 12 12/2 + 1 = 7 solutions$z = 10 \implies x + 2y = 12 \implies \lfloor 12/2 \rfloor + 1 = 7 \text{ solutions}$
• z = 11 x + 2y = 9 9/2 + 1 = 5 solutions$z = 11 \implies x + 2y = 9 \implies \lfloor 9/2 \rfloor + 1 = 5 \text{ solutions}$
• z = 12 x + 2y = 6 6/2 + 1 = 4 solutions$z = 12 \implies x + 2y = 6 \implies \lfloor 6/2 \rfloor + 1 = 4 \text{ solutions}$
• z = 13 x + 2y = 3 3/2 + 1 = 2 solutions$z = 13 \implies x + 2y = 3 \implies \lfloor 3/2 \rfloor + 1 = 2 \text{ solutions}$
• z = 14 x + 2y = 0 0/2 + 1 = 1 solution$z = 14 \implies x + 2y = 0 \implies \lfloor 0/2 \rfloor + 1 = 1 \text{ solution}$

Step 2: Total Sum Computation

Summing all the calculated distribution counts:

Pattern Recognition

Sees: Linear multi-variable diophantine solution constraints. Shortcut: Grouping into alternating arithmetic sequences can expedite the total calculation step instead of adding each discrete integer row manually.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Related Formula

Since both the objects (books) and the containers (shelves) are identical, this problem is equivalent to finding the number of partitions of the integer 8$8$ into at most 4$4$ parts.

Core Logic

Let us systematically list out all the possible distributions based on the number of empty shelves:

• 3 Shelves empty:
• (8, 0, 0, 0) arrow 1 way$(8, 0, 0, 0) \rightarrow 1\text{ way}$
• 2 Shelves empty:
• (7, 1, 0, 0)$(7, 1, 0, 0)$
• (6, 2, 0, 0)$(6, 2, 0, 0)$
• (5, 3, 0, 0)$(5, 3, 0, 0)$
• (4, 4, 0, 0) arrow 4 ways$(4, 4, 0, 0) \rightarrow 4\text{ ways}$
• 1 Shelf empty:
• (6, 1, 1, 0)$(6, 1, 1, 0)$
• (5, 2, 1, 0)$(5, 2, 1, 0)$
• (4, 3, 1, 0)$(4, 3, 1, 0)$
• (4, 2, 2, 0)$(4, 2, 2, 0)$
• (3, 3, 2, 0) arrow 5 ways$(3, 3, 2, 0) \rightarrow 5\text{ ways}$
• 0 Shelves empty:
• (5, 1, 1, 1)$(5, 1, 1, 1)$
• (4, 2, 1, 1)$(4, 2, 1, 1)$
• (3, 3, 1, 1)$(3, 3, 1, 1)$
• (3, 2, 2, 1)$(3, 2, 2, 1)$
• (2, 2, 2, 2) arrow 5 ways$(2, 2, 2, 2) \rightarrow 5\text{ ways}$

Step 1: Total Computations

Summing all these cases together:

Pattern Recognition

Be very careful to identify if containers/objects are identical or distinct. Identical into identical means simple partition of integers. Listing them in descending order ensures no partition is missed or duplicated.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations