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

where [·]$[\cdot]$ represents the greatest integer function.

Core Logic

To find the highest power of 40$40$ dividing 60!$60!$, we first prime factorize 40$40$.

Thus, 40ⁿ = 2³ⁿ × 5ⁿ$40^n = 2^{3n} \times 5^n$. We need to find the exponents of 2$2$ and 5$5$ in 60!$60!$ and restrict n$n$ to satisfy both components simultaneously.

Step 1: Exponent of 2 in 60!

Thus, 60!$60!$ contains 2⁵⁶$2^{56}$.

Step 2: Exponent of 5 in 60!

= 12 + 2 = 14$= 12 + 2 = 14$

Thus, 60!$60!$ contains 5¹⁴$5^{14}$.

Step 3: Calculating Limiting Factor

From the factors, 60!$60!$ can be written as 2⁵⁶ × 5¹⁴ × K$2^{56} \times 5^{14} \times K$.

We need to construct factors of 40 = 2³ × 5$40 = 2^3 \times 5$. From 2⁵⁶$2^{56}$, we can form (2³)¹⁸$(2^3)^{18}$ with a remainder, so 2³$2^3$ limits at 18$18$. From 5¹⁴$5^{14}$, we can form 5¹⁴$5^{14}$, so 5$5$ limits at 14$14$.

The limiting factor is the exponent of 5$5$, which is 14$14$.

Therefore, the maximum value of n$n$ is 14$14$.

Pattern Recognition

For a composite base C = p₁a₁ p₂a₂$C = p_1^{a_1} p_2^{a_2}$, the highest power is ( [ Ep₁(n!)a₁ ], [ Ep₂(n!)a₂ ] )$\min \left( \left[ \frac{E_{p_1}(n!)}{a_1} \right], \left[ \frac{E_{p_2}(n!)}{a_2} \right] \right)$. Generally, the larger prime (here 5$5$) dictates the bottleneck.

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Class 11 Maths: Permutations and Combinations Class 11 Maths: Number Theory

Core Logic

S = 1, 2, 3, , 9$S = \{1, 2, 3, \dots, 9\}$ For x$x$: 9-digit number where ONLY ONE digit is repeated exactly twice. This means the number contains 8 distinct digits from S$S$. One digit appears twice, and 7 digits appear once. Total length = 2 + 7 = 9$2 + 7 = 9$. Calculation for x$x$:

• Select 1 digit to be repeated twice: ⁹C₁$^{9}C_1$
• Select 7 digits out of the remaining 8: ⁸C₇$^{8}C_7$
• Arrange these 9 digits (where 2 are identical): (9!)/(2!)$\frac{9!}{2!}$

Step 1: Determine y

For y$y$: 9-digit number where EXACTLY TWO digits are repeated twice each. This means we select 2 digits to appear twice, and we need 9 - 4 = 5$9 - 4 = 5$ more digits (which must all be distinct and appear once). Total length = 2 + 2 + 5 = 9$2 + 2 + 5 = 9$. Calculation for y$y$:

• Select 2 digits to be repeated twice: ⁹C₂$^{9}C_2$
• Select 5 digits out of the remaining 7: ⁷C₅$^{7}C_5$
• Arrange these 9 digits (two pairs of identicals): (9!)/(2! × 2!)$\frac{9!}{2! \times 2!}$

Step 2: Ratio of x and y

Simplifying the fraction:

Let's carefully compute: x = (72)/(2) × 9! = 36 × 9!$x = \frac{72}{2} \times 9! = 36 \times 9!$ y = 36 × 21 × (9!)/(4) = 9 × 21 × 9! = 189 × 9!$y = 36 \times 21 \times \frac{9!}{4} = 9 \times 21 \times 9! = 189 \times 9!$

Step 3: Final Relation

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Class 11 Mathematics: Permutations and Combinations

Core Logic

The lift can go to floors 1 through 10. It does not stop at 1, 2, or 3. The available floors for exit are 4, 5, 6, 7, 8, 9, 10. Number of available floors n = 7$n = 7$. The three persons must exit at three different floors. So we must choose 3 distinct floors from the 7 available, and assign them to the 3 distinct persons.

Execution

Ways to choose 3 floors from 7: ⁷C₃$^{7}C_{3}$. Ways to arrange 3 persons on these 3 floors: 3!$3!$.

Total ways = ⁷C₃ × 3!$= ^{7}C_{3} \times 3!$

Pattern Recognition

Selecting sets for distinct people inherently combines combination selection with factorial permutation (ⁿCᵣ × r!$^nC_r \times r!$ or directly ⁿPᵣ$^nP_r$). Restricting available floors simply reduces n$n$.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Core Logic

This is a permutations problem with row constraints. We compute the total arrangements of placing 5 distinct letters into 8 boxes and then subtract the invalid cases where one or more rows are left completely empty.

Step 1: Compute total unrestricted arrangements

The grid has a total of 8 boxes. We have 5 distinct letters (A, B, C, D, E):

Step 2: Identify and subtract the invalid empty-row cases

Let the rows be R₁$R_1$, R₂$R_2$, and R₃$R_3$, with box counts 3, 2, and 3 respectively. Since we must distribute 5 letters, it is impossible for 2 rows to be empty simultaneously (as the remaining single row would have at most 3 boxes, which cannot fit 5 letters). Thus, we only subtract cases where exactly one row is empty:

• Case 1: Row R₁$R_1$ (3 boxes) is empty. The 5 letters must go to the remaining 5 boxes of R₂$R_2$ and R₃$R_3$:

• Case 2: Row R₃$R_3$ (3 boxes) is empty. Same as Case 1, the 5 letters must go to the remaining 5 boxes of R₁$R_1$ and R₂$R_2$:

• Case 3: Row R₂$R_2$ (2 boxes) is empty. The 5 letters must go to the remaining 6 boxes of R₁$R_1$ and R₃$R_3$:

Step 3: Calculate the final valid arrangements

Subtracting all empty-row cases from the total arrangements:

Pattern Recognition

Inclusion-Exclusion Principle: For distribution problems with simple boundary exclusions, subtracting the complement set (invalid configurations) is mathematically much cleaner than calculating all possible partitions of row assignments.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations

Related Formula

The exponent of a prime p$p$ in N!$N!$ is given by Legendre's formula:

Core Logic

To find the highest power of 3$3$ that divides 50!$50!$, calculate the sum of the greatest integer functions for successive powers of 3$3$ up to 50$50$.

Step 1: Computation

Applying the formula for N = 50$N = 50$ and p = 3$p = 3$:

Pattern Recognition

Quickly divide by powers of 3$3$: 50/3 arrow 16$50/3 \rightarrow 16$; 16/3 arrow 5$16/3 \rightarrow 5$; 5/3 arrow 1$5/3 \rightarrow 1$. Summing them up yields 16 + 5 + 1 = 22$16 + 5 + 1 = 22$ instantly.

Chapter Mix

Class 11 Mathematics: Permutations and Combinations