The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box, is:

### 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