The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is

Related Formula

For a strictly increasing function f: A → B$f: A \to B$ where |A| = m$|A| = m$ and |B| = n$|B| = n$, the number of functions without restrictions is nm$\binom{n}{m}$.

Core Logic

We need strictly increasing functions f: 1,2,3,4,5,6 → 1,2, ,9$f: \{1,2,3,4,5,6\} \to \{1,2,\dots,9\}$ subject to f(i) ≠ i$f(i) \neq i$. Since f$f$ is strictly increasing, f(i) ≥ i$f(i) \geq i$ must always hold because the target values are drawn from an equally spaced domain. If f(i) = i$f(i) = i$ for any i$i$, it forces a strict ladder down to 1. But we are given f(i) ≠ i$f(i) \neq i$. Thus, f(i) > i$f(i) > i$ for all 1 ≤ i ≤ 6$1 \leq i \leq 6$. This implies f(1) ≥ 2$f(1) \geq 2$.

Step 1: Case Analysis on f(1)

Since f(1) > 1$f(1) > 1$, we evaluate possible starting points: Case 1: f(1) = 2$f(1) = 2$ Remaining 5 values f(2)$f(2)$ to f(6)$f(6)$ must be strictly increasing and chosen from 3, 4, 5, 6, 7, 8, 9$\{3, 4, 5, 6, 7, 8, 9\}$ (7 available numbers). Since f(1)=2$f(1)=2$, f(i)>i$f(i)>i$ is naturally preserved for subsequent elements (e.g., f(2) ≥ 3 > 2$f(2) \ge 3 > 2$). Number of ways = 75 = 21$= \binom{7}{5} = 21$.

Step 2: Subsequent Cases

Case 2: f(1) = 3$f(1) = 3$ Remaining 5 values chosen from 4, 5, 6, 7, 8, 9$\{4, 5, 6, 7, 8, 9\}$ (6 available numbers). Number of ways = 65 = 6$= \binom{6}{5} = 6$.

Case 3: f(1) = 4$f(1) = 4$ Remaining 5 values chosen from 5, 6, 7, 8, 9$\{5, 6, 7, 8, 9\}$ (5 available numbers). Number of ways = 55 = 1$= \binom{5}{5} = 1$.

Case 4: f(1) = 5$f(1) = 5$ Requires choosing 5 values from 6,7,8,9$\{6,7,8,9\}$, which is impossible.

Step 3: Total Sum

Total number of valid functions = 21 + 6 + 1 = 28$21 + 6 + 1 = 28$.

Pattern Recognition

For f(i) ≠ i$f(i) \neq i$ on strictly increasing integer arrays, f(x) - x > 0$f(x) - x > 0$. Using the substitution g(x) = f(x) - x$g(x) = f(x) - x$, you convert a constrained increasing function into a standard non-decreasing one, or simply pivot on f(1)$f(1)$ and sum the cascading binomials.

Chapter Mix

Class 11 Maths: Permutations and Combinations Class 12 Maths: Functions

Related Formula

Modular arithmetic reductions for power cycles:

Core Logic

Analyze condition p$p$: (6^m + 9ⁿ)$(6^m + 9^n)$ is divisible by 5. 6 ≡ 1 5 ⇒ 6^m ≡ 1^m ≡ 1 5$6 \equiv 1 \pmod 5 \Rightarrow 6^m \equiv 1^m \equiv 1 \pmod 5$. 9 ≡ -1 5 ⇒ 9ⁿ ≡ (-1)ⁿ 5$9 \equiv -1 \pmod 5 \Rightarrow 9^n \equiv (-1)^n \pmod 5$. For the sum to be divisible by 5: 1 + (-1)ⁿ ≡ 0 5 ⇒ (-1)ⁿ = -1$1 + (-1)^n \equiv 0 \pmod 5 \Rightarrow (-1)^n = -1$. This implies n$n$ must be an ODD integer.

Since m in 1, 2, , 50$m \in \{1, 2, \dots, 50\}$, m$m$ can be anything (50 choices). Since n$n$ must be odd in 1, , 50$\{1, \dots, 50\}$, n$n$ has 25 choices. p = 50 × 25 = 1250$p = 50 \times 25 = 1250$.

Step 1: Compute q

Analyze condition q$q$: (m + n)$(m + n)$ is the square of a prime number. Max value of m+n = 50+50 = 100$m+n = 50+50 = 100$. Primes whose squares are ≤ 100$\leq 100$: 2, 3, 5, 7$2, 3, 5, 7$. Their squares are 4, 9, 25, 49$4, 9, 25, 49$. So m+n$m+n$ can be 4, 9, 25, 49$4, 9, 25, 49$.

Match List-I with List-II:

Explanation for counts: If m+n = S$m+n = S$, and m, n ≥ 1$m, n \geq 1$, the number of ways is S-1$S-1$ (since S ≤ 50$S \leq 50$). For S=4$S=4$: 3 ways. For S=9$S=9$: 8 ways. For S=25$S=25$: 24 ways. For S=49$S=49$: 48 ways. q = 3 + 8 + 24 + 48 = 83$q = 3 + 8 + 24 + 48 = 83$.

Step 2: Final Sum

Pattern Recognition

Modular exponentiation immediately shrinks large powers to ± 1$\pm 1$. The sum m+n=S$m+n=S$ where 1 ≤ m,n ≤ N$1 \le m,n \le N$ has exactly S-1$S-1$ solutions if S ≤ N$S \le N$, allowing instant combinatorics tallying without manual counting.

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Core Logic

To find the maximum power n$n$ such that 7ⁿ$7^n$ divides 101!$101!$, we need to find the exponent of the prime 7$7$ in the prime factorization of 101!$101!$.

Step 1: Apply Legendre's Formula

n = 14 + 2 + 0$n = 14 + 2 + 0$ n = 16$n = 16$

Pattern Recognition

For prime p$p$ in N!$N!$, iteratively divide N$N$ by p$p$ taking only integer parts and sum them. Fast mental math: 101 ÷ 7 = 14$101 \div 7 = 14$, 14 ÷ 7 = 2$14 \div 7 = 2$. 14+2=16$14+2=16$.

Chapter Mix

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

Related Formula

Core Logic

A pentagon requires 5 distinct points as vertices. No three points can be collinear to form a proper polygon. Since the available points lie on the three sides of a triangle, picking 3 points from the same side will form a degenerate straight line segment rather than building a strictly convex vertex frame.

Therefore, we must select the 5 points distributed across the three sides (AB with 4 points, BC with 5 points, AC with 4 points) such that a maximum of 2 points is selected from any one side.

Step 1: Case Breakdown

We need to choose 5 points total from the three groups (4, 5, 4) with the condition that no group contributes more than 2 points. The only valid numerical partitions of 5 into 3 parts bounded by 2 are:

• Case 1: 2 points from AB, 2 points from BC, 1 point from AC
• Case 2: 2 points from AB, 1 point from BC, 2 points from AC
• Case 3: 1 point from AB, 2 points from BC, 2 points from AC

Step 2: Calculating Combinations for Each Case

Case 1: (2 from AB, 2 from BC, 1 from AC)

Case 2: (2 from AB, 1 from BC, 2 from AC)

Case 3: (1 from AB, 2 from BC, 2 from AC)

Step 3: Total Pentagons

Total number of pentagons is the sum of all valid cases:

Pattern Recognition

For polygon formation from collinear sets, always frame it as a restricted partition problem. A polygon of k$k$ sides requires selecting k$k$ vertices such that no maximum allowable threshold of collinearity is breached (for a strict polygon, no 3 points on a line).

Chapter Mix

Class 11 Maths: Permutations and Combinations

Related Formula

Complementary counting principle:

Core Logic

Set S = 1, 2, 3, , 11$S = \{1, 2, 3, \dots, 11\}$ contains 11 elements (6 odd: 1, 3, 5, 7, 9, 11$\{1, 3, 5, 7, 9, 11\}$ and 5 even: 2, 4, 6, 8, 10$\{2, 4, 6, 8, 10\}$).

• Total possible subsets of S = 2¹¹ = 2048$S = 2^{11} = 2048$.
• Subsets where product is odd consist entirely of odd numbers: 2⁶ = 64$2^6 = 64$.
• Singletons with even product: 5$5$ (the even numbers themselves).

Step 1: Complementary Subtraction

Excluded subsets:

• Empty set (size 0): 1$1$
• Singletons with odd elements: 6$6$
• Singletons with even elements: 5$5$
• Subsets of size ≥ 2$\ge 2$ with only odd elements: 2⁶ - 1 - 6 = 57$2^6 - 1 - 6 = 57$

Required count:

Pattern Recognition

Complement method: Total subsets minus subsets containing only odd elements minus even singletons.

Chapter Mix

Class 11 Maths: Permutations and Combinations