### Related Formula For a linear equation x + 2y = k$x + 2y = k$, where x, y ge 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: leftlfloor frack2 rightrfloor + 1$\left\lfloor \frac{k}{2} \right\rfloor + 1$ ### Core Logic We need to find the number of non-negative integer solutions to: x + 2y + 3z = 42$x + 2y + 3z = 42$ Rearranging the equation to analyze values branch-by-branch based on z$z$: x + 2y = 42 - 3z$x + 2y = 42 - 3z$ Since x, y ge 0$x, y \ge 0$, the maximum value z$z$ can attain corresponds to x=0, y=0$x=0, y=0$, so 3z le 42 implies z le 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 implies x + 2y = 42 implies lfloor 42/2 rfloor + 1 = 22 text solutions$z = 0 \implies x + 2y = 42 \implies \lfloor 42/2 \rfloor + 1 = 22 \text{ solutions}$ - z = 1 implies x + 2y = 39 implies lfloor 39/2 rfloor + 1 = 20 text solutions$z = 1 \implies x + 2y = 39 \implies \lfloor 39/2 \rfloor + 1 = 20 \text{ solutions}$ - z = 2 implies x + 2y = 36 implies lfloor 36/2 rfloor + 1 = 19 text solutions$z = 2 \implies x + 2y = 36 \implies \lfloor 36/2 \rfloor + 1 = 19 \text{ solutions}$ - z = 3 implies x + 2y = 33 implies lfloor 33/2 rfloor + 1 = 17 text solutions$z = 3 \implies x + 2y = 33 \implies \lfloor 33/2 \rfloor + 1 = 17 \text{ solutions}$ - z = 4 implies x + 2y = 30 implies lfloor 30/2 rfloor + 1 = 16 text solutions$z = 4 \implies x + 2y = 30 \implies \lfloor 30/2 \rfloor + 1 = 16 \text{ solutions}$ - z = 5 implies x + 2y = 27 implies lfloor 27/2 rfloor + 1 = 14 text solutions$z = 5 \implies x + 2y = 27 \implies \lfloor 27/2 \rfloor + 1 = 14 \text{ solutions}$ - z = 6 implies x + 2y = 24 implies lfloor 24/2 rfloor + 1 = 13 text solutions$z = 6 \implies x + 2y = 24 \implies \lfloor 24/2 \rfloor + 1 = 13 \text{ solutions}$ - z = 7 implies x + 2y = 21 implies lfloor 21/2 rfloor + 1 = 11 text solutions$z = 7 \implies x + 2y = 21 \implies \lfloor 21/2 \rfloor + 1 = 11 \text{ solutions}$ - z = 8 implies x + 2y = 18 implies lfloor 18/2 rfloor + 1 = 10 text solutions$z = 8 \implies x + 2y = 18 \implies \lfloor 18/2 \rfloor + 1 = 10 \text{ solutions}$ - z = 9 implies x + 2y = 15 implies lfloor 15/2 rfloor + 1 = 8 text solutions$z = 9 \implies x + 2y = 15 \implies \lfloor 15/2 \rfloor + 1 = 8 \text{ solutions}$ - z = 10 implies x + 2y = 12 implies lfloor 12/2 rfloor + 1 = 7 text solutions$z = 10 \implies x + 2y = 12 \implies \lfloor 12/2 \rfloor + 1 = 7 \text{ solutions}$ - z = 11 implies x + 2y = 9 implies lfloor 9/2 rfloor + 1 = 5 text solutions$z = 11 \implies x + 2y = 9 \implies \lfloor 9/2 \rfloor + 1 = 5 \text{ solutions}$ - z = 12 implies x + 2y = 6 implies lfloor 6/2 rfloor + 1 = 4 text solutions$z = 12 \implies x + 2y = 6 \implies \lfloor 6/2 \rfloor + 1 = 4 \text{ solutions}$ - z = 13 implies x + 2y = 3 implies lfloor 3/2 rfloor + 1 = 2 text solutions$z = 13 \implies x + 2y = 3 \implies \lfloor 3/2 \rfloor + 1 = 2 \text{ solutions}$ - z = 14 implies x + 2y = 0 implies lfloor 0/2 rfloor + 1 = 1 text 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: textTotal solutions = 22 + 20 + 19 + 17 + 16 + 14 + 13 + 11 + 10 + 8 + 7 + 5 + 4 + 2 + 1 = 169$\text{Total solutions} = 22 + 20 + 19 + 17 + 16 + 14 + 13 + 11 + 10 + 8 + 7 + 5 + 4 + 2 + 1 = 169$ ### 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### Related Formula textPermutations of n text objects with p text identical elements = fracn!p!$\text{Permutations of } n \text{ objects with } p \text{ identical elements } = \frac{n!}{p!}$ ### Core Logic Alphabetize the letters in the word "GTWENTY": E, G, N, T, T, W, Y. We calculate the number of permutations that alphabetically precede "GTWENTY" by exhaustively scanning dictionary prefixes. ### Step 1: Calculate Block Combinations 1. Words starting with 'E': Remaining letters {G, N, T, T, W, Y}. We have 6 letters with 'T' repeating twice. Permutations = frac6!2! = frac7202 = 360$\frac{6!}{2!} = \frac{720}{2} = 360$. 2. Words starting with 'G': This locks the first letter. Next alphabetical letter is 'E'. - Starting with 'GE': Remaining {N, T, T, W, Y}. 5 letters, 'T' repeating. Permutations = frac5!2! = frac1202 = 60$\frac{5!}{2!} = \frac{120}{2} = 60$. - Starting with 'GN': Remaining {E, T, T, W, Y}. 5 letters, 'T' repeating. Permutations = frac5!2! = 60$\frac{5!}{2!} = 60$. - Starting with 'GT': This locks the second letter as well. We iterate through the remaining sorted pool {E, N, T, W, Y}. -- Starting with 'GTE': Remaining {N, T, W, Y}. No repetitions. Permutations = 4! = 24$4! = 24$. -- Starting with 'GTN': Remaining {E, T, W, Y}. No repetitions. Permutations = 4! = 24$4! = 24$. -- Starting with 'GTT': Remaining {E, N, W, Y}. No repetitions. Permutations = 4! = 24$4! = 24$. -- Starting with 'GTW': This locks the third letter. Iterate through {E, N, T, Y}. ### Step 2: Trace Remaining Exact String We are now tracking the prefix 'GTW'. The remaining letters alphabetically are E, N, T, Y. The target word is precisely built out of these letters in exact alphabetical order: E, then N, then T, then Y. This means "GTWENTY" is the very first word in the 'GTW' block. So, it adds exactly 1 to the count. ### Step 3: Sum the Permutations Total serial number = 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553$360 + 60 + 60 + 24 + 24 + 24 + 1 = 553$. ### Pattern Recognition When calculating dictionary rank with repeating letters, remember to divide by p!$p!$ only when the repeating letter is roaming freely in the available blanks. If a repeating letter is 'locked' as the current prefix, it no longer acts as a repeater for the remaining slots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

### 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