Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_1{}^{15}C_2\). Choose the zero variable and split positive sum (16) among the remaining (3) variables. In exams convert exactly-zero cases into positive distribution.
Step 3
Exam Tip
Zero variable चुनें और बाकी (3) variables में positive sum (16) बांटें। परीक्षा में exactly zero cases को positive distribution में बदलें।
After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.
Step 2
Why this answer is correct
The correct answer is B. \(^{15}C_3\). After removing the minimum sum (10), (12) remains and is distributed among (4) variables. In exams subtract lower bounds and use stars and bars.
Step 3
Exam Tip
Minimum sum (10) हटाने पर (12) बचता है और (4) variables में distribute होता है। परीक्षा में lower bounds subtract करके stars and bars लगाएं।
Choose the (3) positive variables first, then split (12) into (3) positive parts. In exams use choose variables plus positive stars and bars for exactly positive variables.
Step 2
Why this answer is correct
The correct answer is A. \(^{5}C_3{}^{11}C_2\). Choose the (3) positive variables first, then split (12) into (3) positive parts. In exams use choose variables plus positive stars and bars for exactly positive variables.
Step 3
Exam Tip
पहले (3) positive variables चुनें, फिर (12) को (3) positive parts में बांटें। परीक्षा में exactly positive variables में choose variables plus positive stars-bars करें।
After removing the minimum sum (14), (16) remains, so \({}^{16+4-1}C_{3}\) is obtained. In exams subtract unequal lower bounds first.
Step 2
Why this answer is correct
The correct answer is A. \(^{19}C_3\). After removing the minimum sum (14), (16) remains, so \({}^{16+4-1}C_{3}\) is obtained. In exams subtract unequal lower bounds first.
Step 3
Exam Tip
Minimum sum (14) हटाने पर (16) बचता है, इसलिए \({}^{16+4-1}C_{3}\) मिलता है। परीक्षा में unequal lower bounds पहले subtract करें।
Choose the zero variables first, then the remaining two variables form a positive sum of (18). In exams use positive distribution for exactly-zero cases.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_2{}^{17}C_1\). Choose the zero variables first, then the remaining two variables form a positive sum of (18). In exams use positive distribution for exactly-zero cases.
Step 3
Exam Tip
पहले zero variables चुनें, फिर बाकी दो variables positive sum (18) बनाते हैं। परीक्षा में exactly zero cases में positive distribution लगाएं।
Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.
Step 2
Why this answer is correct
The correct answer is B. \(^{20}C_3\). Give (2) first to the four variables, leaving (17). In exams subtract the lower bound and apply non-negative stars and bars.
Step 3
Exam Tip
चार variables को पहले (2) देने पर (17) बचता है। परीक्षा में lower bound घटाकर non-negative stars and bars लगाएं।
In stars and bars, (18) stars and (3) bars are arranged. In exams use \(^{n+r-1}C_{r-1}\) for non-negative solutions.
Step 2
Why this answer is correct
The correct answer is B. \(^{21}C_3\). In stars and bars, (18) stars and (3) bars are arranged. In exams use \(^{n+r-1}C_{r-1}\) for non-negative solutions.
Step 3
Exam Tip
Stars and bars में (18) stars और (3) bars arrange होते हैं। परीक्षा में अऋणात्मक हलों के लिए \(^{n+r-1}C_{r-1}\) लगाएं।
From total \( \binom{12}{2} \), subtract \(3\binom{6}{2}\) cases where a variable is (6) or more, giving (21). For upper bounds, complementary counting is useful.
Step 2
Why this answer is correct
The correct answer is D. (21). From total \( \binom{12}{2} \), subtract \(3\binom{6}{2}\) cases where a variable is (6) or more, giving (21). For upper bounds, complementary counting is useful.
Step 3
Exam Tip
कुल \( \binom{12}{2} \) से किसी चर के (6) या अधिक होने के \(3\binom{6}{2}\) मामले घटते हैं, उत्तर (21) है। ऊपरी सीमा में पूरक गिनती उपयोगी रहती है।
After subtracting minimum values, (a+b+z=7), so \( \binom{9}{2}=36 \). First convert constraints to zero-based variables.
Step 2
Why this answer is correct
The correct answer is A. (36). After subtracting minimum values, (a+b+z=7), so \( \binom{9}{2}=36 \). First convert constraints to zero-based variables.
Step 3
Exam Tip
न्यूनतम मान घटाने पर (a+b+z=7) मिलता है, इसलिए \( \binom{9}{2}=36 \)। पहले शर्तों को शून्य-आधारित बनाइए।
Convert the restrictions on \(x_1\) and \(x_2\) into non-negative variables and take cases for \(x_3\). For bounded variables, add case counts.
Step 2
Why this answer is correct
The correct answer is C. (103). Convert the restrictions on \(x_1\) and \(x_2\) into non-negative variables and take cases for \(x_3\). For bounded variables, add case counts.
Step 3
Exam Tip
\(x_1\) और \(x_2\) की शर्तों को बदलकर गैर-ऋणात्मक चर में लिखें और \(x_3\) के लिए मामले लें। सीमा वाली शर्त में मामलों का योग करें।
The integers between \( \frac{1}{2} \) and \( \frac{11}{2} \) are from ( 1 ) to ( 5 ). For integer solutions, mark only separate points.
Step 2
Why this answer is correct
The correct answer is B. ( 1,2,3,4,5 ). The integers between \( \frac{1}{2} \) and \( \frac{11}{2} \) are from ( 1 ) to ( 5 ). For integer solutions, mark only separate points.
Step 3
Exam Tip
\( \frac{1}{2} \) और \( \frac{11}{2} \) के बीच पूर्णांक ( 1 ) से ( 5 ) तक हैं। पूर्णांक समाधान में केवल अलग-अलग बिंदु चिन्हित करें।
( -3 ) is included and ( 4 ) is excluded. For integer solutions, mark separate points instead of a continuous shade.
Step 2
Why this answer is correct
The correct answer is C. ( -3,-2,-1,0,1,2,3 ). ( -3 ) is included and ( 4 ) is excluded. For integer solutions, mark separate points instead of a continuous shade.
Step 3
Exam Tip
( -3 ) शामिल है और ( 4 ) शामिल नहीं है। पूर्णांक समाधान में लगातार छाया के बजाय अलग बिंदु चिन्हित करें।
C. सबसे छोटा (13), सबसे बड़ा (26)/Smallest (13), greatest (26)
Step 1
Concept
\(12\le x-1<27\) gives \(13\le x<28\), so integers go from (13) to (27). In exams, take the integer before the strict upper boundary.
Step 2
Why this answer is correct
The correct answer is C. सबसे छोटा (13), सबसे बड़ा (26) / Smallest (13), greatest (26). \(12\le x-1<27\) gives \(13\le x<28\), so integers go from (13) to (27). In exams, take the integer before the strict upper boundary.
Step 3
Exam Tip
\(12\le x-1<27\) से \(13\le x<28\), इसलिए पूर्णांक (13) से (27) तक हैं। परीक्षा में upper strict सीमा से पहले वाला integer लें।
Integers greater than \(-\frac{5}{2}\) start at (-2), and the last integer up to \(\frac{13}{3}\) is (4). In exams, choose valid integers at fractional boundaries.
Step 2
Why this answer is correct
The correct answer is C. ({-2,-1,0,1,2,3,4}). Integers greater than \(-\frac{5}{2}\) start at (-2), and the last integer up to \(\frac{13}{3}\) is (4). In exams, choose valid integers at fractional boundaries.
Step 3
Exam Tip
\(-\frac{5}{2}\) से बड़े पूर्णांक (-2) से शुरू होते हैं और \(\frac{13}{3}\) तक (4) अंतिम पूर्णांक है। परीक्षा में fractional boundary पर valid integer चुनें।
A. सबसे छोटा (4), सबसे बड़ा (11)/Smallest (4), greatest (11)
Step 1
Concept
The solution \(6\le x+2<14\) gives \(4\le x<12\), so integers run from (4) to (11). In exams, take the integer just before a strict upper boundary.
Step 2
Why this answer is correct
The correct answer is A. सबसे छोटा (4), सबसे बड़ा (11) / Smallest (4), greatest (11). The solution \(6\le x+2<14\) gives \(4\le x<12\), so integers run from (4) to (11). In exams, take the integer just before a strict upper boundary.
Step 3
Exam Tip
हल \(6\le x+2<14\) से \(4\le x<12\), इसलिए पूर्णांक (4) से (11) तक हैं। परीक्षा में ऊपरी strict सीमा से ठीक पहले वाला पूर्णांक लें।
(-7) is included, and integers less than \(\frac{5}{2}\) go up to (2). In exams, with an integer restriction, mark separate points instead of a continuous line.
Step 2
Why this answer is correct
The correct answer is B. ({-7,-6,-5,-4,-3,-2,-1,0,1,2}). (-7) is included, and integers less than \(\frac{5}{2}\) go up to (2). In exams, with an integer restriction, mark separate points instead of a continuous line.
Step 3
Exam Tip
(-7) शामिल है और \(\frac{5}{2}\) से छोटे पूर्णांक (2) तक हैं। परीक्षा में पूर्णांक प्रतिबंध हो तो रेखा नहीं, अलग बिंदु बनाएं।
Only separate integer points are filled, so \(x\in\mathbb{Z}\). In exams, distinguish a shaded line segment from separate points.
Step 2
Why this answer is correct
The correct answer is A. \({x\in\mathbb{Z}:-3\le x\le 1}\). Only separate integer points are filled, so \(x\in\mathbb{Z}\). In exams, distinguish a shaded line segment from separate points.
Step 3
Exam Tip
केवल अलग-अलग पूर्णांक बिंदु भरे हैं इसलिए \(x\in\mathbb{Z}\) होगा। परीक्षा में रेखा और अलग बिंदुओं में अंतर पहचानें।
The solution is \(-1\le x<\frac{11}{3}\), so the integers are ({-1,0,1,2,3}). In exams write the final answer according to the domain.
Step 2
Why this answer is correct
The correct answer is A. ({-1,0,1,2,3}). The solution is \(-1\le x<\frac{11}{3}\), so the integers are ({-1,0,1,2,3}). In exams write the final answer according to the domain.
Step 3
Exam Tip
हल \(-1\le x<\frac{11}{3}\) है, इसलिए पूर्णांक ({-1,0,1,2,3}) मिलते हैं। परीक्षा में अंतिम उत्तर डोमेन के अनुसार लिखें।
