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 (1) 4608 (2) 5720 (3) 5719 (4) 4607

### Related Formula textCombinations: ^nC_r = fracn!r!(n-r)!$$\text{Combinations: } {}^nC_r = \frac{n!}{r!(n-r)!}$$ ### Core Logic Total Questions = 20 (A: 8, B: 6, C: 6). Total to attempt = 15. Minimum required from each section = 4. Base attempt gives: 4 (textfrom A) + 4 (textfrom B) + 4 (textfrom C) = 12$4 (\text{from A}) + 4 (\text{from B}) + 4 (\text{from C}) = 12$ questions. We have to distribute the remaining 15 - 12 = 3$15 - 12 = 3$ questions across the sections A, B, and C. Let the additional questions picked be x, y, z$x, y, z$ for sections A, B, C respectively. Then x + y + z = 3$x + y + z = 3$, with constraints based on the maximum questions per section: A max extra = 8 - 4 = 4 Rightarrow x le 4$8 - 4 = 4 \Rightarrow x \le 4$ B max extra = 6 - 4 = 2 Rightarrow y le 2$6 - 4 = 2 \Rightarrow y \le 2$ C max extra = 6 - 4 = 2 Rightarrow z le 2$6 - 4 = 2 \Rightarrow z \le 2$ ### Step 1: Identifying Valid Selection Cases The possible sets of (x, y, z)$(x, y, z)$ are: Case 1: (1, 1, 1)$(1, 1, 1)$ Rightarrow$\Rightarrow$ Total picks: A(5), B(5), C(5) Case 2: (2, 1, 0)$(2, 1, 0)$ and its permutations (respecting constraints). Valid permutations: - A gets 2, B gets 1, C gets 0 Rightarrow$\Rightarrow$ A(6), B(5), C(4) - A gets 2, C gets 1, B gets 0 Rightarrow$\Rightarrow$ A(6), B(4), C(5) - B gets 2, A gets 1, C gets 0 Rightarrow$\Rightarrow$ A(5), B(6), C(4) - C gets 2, A gets 1, B gets 0 Rightarrow$\Rightarrow$ A(5), B(4), C(6) (Note: B(2), C(1) or C(2), B(1) are not allowed if it forces A to take 0, wait, A gets 0 means A(4). A is allowed to have 4.) Let's check permutations of (2, 1, 0)$(2, 1, 0)$: - A(4), B(6), C(5) [x=0, y=2, z=1] - A(4), B(5), C(6) [x=0, y=1, z=2] Case 3: (3, 0, 0)$(3, 0, 0)$ and permutations. Since y le 2$y \le 2$ and z le 2$z \le 2$, only x$x$ can be 3. So, x=3, y=0, z=0$x=3, y=0, z=0$ Rightarrow$\Rightarrow$ A(7), B(4), C(4). ### Step 2: Calculating Combinations per Case Let's list all valid final section breakdowns (A, B, C)$(A, B, C)$: 1) (5, 5, 5) Rightarrow ^8C_5 cdot ^6C_5 cdot ^6C_5 = 56 cdot 6 cdot 6 = 2016$(5, 5, 5) \Rightarrow {}^8C_5 \cdot {}^6C_5 \cdot {}^6C_5 = 56 \cdot 6 \cdot 6 = 2016$ 2) (6, 5, 4) Rightarrow ^8C_6 cdot ^6C_5 cdot ^6C_4 = 28 cdot 6 cdot 15 = 2520$(6, 5, 4) \Rightarrow {}^8C_6 \cdot {}^6C_5 \cdot {}^6C_4 = 28 \cdot 6 \cdot 15 = 2520$ 3) (6, 4, 5) Rightarrow ^8C_6 cdot ^6C_4 cdot ^6C_5 = 28 cdot 15 cdot 6 = 2520$(6, 4, 5) \Rightarrow {}^8C_6 \cdot {}^6C_4 \cdot {}^6C_5 = 28 \cdot 15 \cdot 6 = 2520$ 4) (5, 6, 4) Rightarrow ^8C_5 cdot ^6C_6 cdot ^6C_4 = 56 cdot 1 cdot 15 = 840$(5, 6, 4) \Rightarrow {}^8C_5 \cdot {}^6C_6 \cdot {}^6C_4 = 56 \cdot 1 \cdot 15 = 840$ 5) (5, 4, 6) Rightarrow ^8C_5 cdot ^6C_4 cdot ^6C_6 = 56 cdot 15 cdot 1 = 840$(5, 4, 6) \Rightarrow {}^8C_5 \cdot {}^6C_4 \cdot {}^6C_6 = 56 \cdot 15 \cdot 1 = 840$ 6) (4, 6, 5) Rightarrow ^8C_4 cdot ^6C_6 cdot ^6C_5 = 70 cdot 1 cdot 6 = 420$(4, 6, 5) \Rightarrow {}^8C_4 \cdot {}^6C_6 \cdot {}^6C_5 = 70 \cdot 1 \cdot 6 = 420$ 7) (4, 5, 6) Rightarrow ^8C_4 cdot ^6C_5 cdot ^6C_6 = 70 cdot 6 cdot 1 = 420$(4, 5, 6) \Rightarrow {}^8C_4 \cdot {}^6C_5 \cdot {}^6C_6 = 70 \cdot 6 \cdot 1 = 420$ 8) (7, 4, 4) Rightarrow ^8C_7 cdot ^6C_4 cdot ^6C_4 = 8 cdot 15 cdot 15 = 1800$(7, 4, 4) \Rightarrow {}^8C_7 \cdot {}^6C_4 \cdot {}^6C_4 = 8 \cdot 15 \cdot 15 = 1800$ ### Step 3: Summing the Total Ways Total ways = 2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$2016 + 2520 + 2520 + 840 + 840 + 420 + 420 + 1800$ Total ways = 2016 + 5040 + 1680 + 840 + 1800 = 11376$2016 + 5040 + 1680 + 840 + 1800 = 11376$ ### Pattern Recognition Combinatorial distribution with rigid lower bounds is solved by shifting the baseline. Allocate the minimums immediately (4+4+4=12$4+4+4=12$), then distribute the remaining items via casework ensuring upper capacities aren't breached. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

### Related Formula textWays to distribute n text identical objects among r text persons = ^n+r-1C_r-1$$\text{Ways to distribute } n \text{ identical objects among } r \text{ persons } = ^{n+r-1}C_{r-1}$$ ### Core Logic First, distribute 2 apples to each of the 3 children to satisfy the minimum requirement. Remaining apples = 21 - 3 times 2 = 15$21 - 3 \times 2 = 15$. Now distribute the remaining 15 identical apples among the 3 children without restrictions. textNumber of ways = ^15+3-1C_3-1 = ^17C_2$$\text{Number of ways} = ^{15+3-1}C_{3-1} = ^{17}C_2$$ = frac17 times 162 = 136$$= \frac{17 \times 16}{2} = 136$$ ### Pattern Recognition Beggar's Method: For x_1+x_2+dots+x_r = n$x_1+x_2+\dots+x_r = n$ with x_i ge k$x_i \ge k$, pre-allocate k$k$ to each and apply standard distribution formula on remainder. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

### Related Formula ^nC_r + ^nC_r-1 = ^n+1C_r$$^nC_r + ^nC_{r-1} = ^{n+1}C_r$$ ### Core Logic Simplify the binomial combination: ^6C_m + 2(^6C_m+1) + ^6C_m+2 = (^6C_m + ^6C_m+1) + (^6C_m+1 + ^6C_m+2)$$^6C_m + 2(^6C_{m+1}) + ^6C_{m+2} = (^6C_m + ^6C_{m+1}) + (^6C_{m+1} + ^6C_{m+2})$$ Using Pascal's rule, this becomes: ^7C_m+1 + ^7C_m+2 = ^8C_m+2$$^7C_{m+1} + ^7C_{m+2} = ^8C_{m+2}$$ Given condition: ^8C_m+2 > ^8C_3 = 56$^8C_{m+2} > ^8C_3 = 56$. For N=8$N=8$, the central combinations yield the maximum value: ^8C_4 = 70$^8C_4 = 70$. Others like ^8C_5 = 56$^8C_5 = 56$, which is not strictly greater than 56. So m + 2 = 4 implies m = 2$m + 2 = 4 \implies m = 2$. Solve the permutations ratio: frac^n-1P_3^nP_4 = frac18$$\frac{^{n-1}P_3}{^nP_4} = \frac{1}{8}$$ frac(n-1)(n-2)(n-3)n(n-1)(n-2)(n-3) = frac18 implies frac1n = frac18 implies n = 8$$\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)} = \frac{1}{8} \implies \frac{1}{n} = \frac{1}{8} \implies n = 8$$ Calculate the target expression: ^nP_m+1 + ^n+1C_m = ^8P_3 + ^9C_2$$^nP_{m+1} + ^{n+1}C_m = ^8P_3 + ^9C_2$$ = (8 times 7 times 6) + frac9 times 82$$= (8 \times 7 \times 6) + \frac{9 \times 8}{2}$$ = 336 + 36 = 372$$= 336 + 36 = 372$$ ### Pattern Recognition Binomial coefficient reduction using Pascal's triangle quickly collapses expanded nCr$nCr$ sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

### Core Logic Letters in 'DISTRIBUTION': I(3), T(2), D, S, R, B, U, O, N. There are 9 distinct letters. ### Step 1: Case Analysis **Case 1: 3 alike, 1 distinct** Selection: Choose the letter 'I' (^1C_1$^1C_1$) and 1 from the remaining 8 distinct letters (^8C_1$^8C_1$). Arrangement: ^8C_1 times frac4!3! = 8 times 4 = 32$^8C_1 \times \frac{4!}{3!} = 8 \times 4 = 32$. **Case 2: 2 alike of one kind, 2 alike of another kind** Since only 'I' and 'T' appear at least twice, we must choose both. Arrangement: ^2C_2 times frac4!2!2! = 1 times 6 = 6$^2C_2 \times \frac{4!}{2!2!} = 1 \times 6 = 6$. ### Step 2: Further Cases **Case 3: 2 alike, 2 distinct** Selection: Choose 1 from the 2 repeated sets (^2C_1$^2C_1$) and 2 from the remaining 8 distinct letters (^8C_2$^8C_2$). Arrangement: ^2C_1 times ^8C_2 times frac4!2! = 2 times 28 times 12 = 672$^2C_1 \times ^8C_2 \times \frac{4!}{2!} = 2 \times 28 \times 12 = 672$. **Case 4: All 4 distinct** Selection: Choose 4 from the 9 distinct letters (^9C_4$^9C_4$). Arrangement: ^9C_4 times 4! = 126 times 24 = 3024$^9C_4 \times 4! = 126 \times 24 = 3024$. ### Step 3: Total Words Total = 3024 + 672 + 6 + 32 = 3734$3024 + 672 + 6 + 32 = 3734$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations