A. किसी वस्तु या घटना की स्थिति/Position of an object or phenomenon
Step 1
Concept
Location is basic to geography because every event happens somewhere. Treat location as a core element in exams.
Step 2
Why this answer is correct
The correct answer is A. किसी वस्तु या घटना की स्थिति / Position of an object or phenomenon. Location is basic to geography because every event happens somewhere. Treat location as a core element in exams.
Step 3
Exam Tip
स्थान भूगोल का आधार है क्योंकि हर घटना किसी जगह पर होती है। परीक्षा में स्थान को भूगोल का मूल तत्व मानें।
C. किसी वस्तु या घटना की स्थान से जुड़ी विशेषताएँ/Location related features of an object or phenomenon
Step 1
Concept
Spatial attributes tell where something is and how it is spread. In exams location is the main clue.
Step 2
Why this answer is correct
The correct answer is C. किसी वस्तु या घटना की स्थान से जुड़ी विशेषताएँ / Location related features of an object or phenomenon. Spatial attributes tell where something is and how it is spread. In exams location is the main clue.
Step 3
Exam Tip
स्थानिक गुण बताते हैं कि कोई वस्तु कहाँ है और उसका फैलाव कैसा है। परीक्षा में स्थान को मुख्य संकेत मानें।
B. क्योंकि यह प्राकृतिक और मानवीय तत्वों को जोड़कर अध्ययन करता है/Because it studies natural and human elements together
Step 1
Concept
Geography links nature and human activities on the earth. Remember it as an integrating subject for exams.
Step 2
Why this answer is correct
The correct answer is B. क्योंकि यह प्राकृतिक और मानवीय तत्वों को जोड़कर अध्ययन करता है / Because it studies natural and human elements together. Geography links nature and human activities on the earth. Remember it as an integrating subject for exams.
Step 3
Exam Tip
भूगोल पृथ्वी पर प्रकृति और मानव गतिविधियों के संबंधों को जोड़ता है। परीक्षा में इसे समाकलनकारी विषय के रूप में याद रखें।
Since \( \binom{10}{2}=45 \), (n=10), so \( \binom{10}{4}=210 \). First find (n), then evaluate the required combination.
Step 2
Why this answer is correct
The correct answer is C. (210). Since \( \binom{10}{2}=45 \), (n=10), so \( \binom{10}{4}=210 \). First find (n), then evaluate the required combination.
Step 3
Exam Tip
\( \binom{10}{2}=45 \) से (n=10), इसलिए \( \binom{10}{4}=210 \)। पहले (n) निकालें, फिर माँगा गया संयोजन रखें।
By Vandermonde's identity, the sum is \( \binom{14}{4}=1001 \). In such sums, view it as total selection after adding upper counts.
Step 2
Why this answer is correct
The correct answer is C. (1001). By Vandermonde's identity, the sum is \( \binom{14}{4}=1001 \). In such sums, view it as total selection after adding upper counts.
Step 3
Exam Tip
वैंडरमोंड सर्वसमिका से योग \( \binom{14}{4}=1001 \) है। ऐसे योग में ऊपर की संख्याएँ जोड़कर कुल चयन देखें।
The product is ( (1+x)^{12} ), so the coefficient is \( \binom{12}{8}=495 \). In binomial products, combine powers first.
Step 2
Why this answer is correct
The correct answer is D. (495). The product is ( (1+x)^{12} ), so the coefficient is \( \binom{12}{8}=495 \). In binomial products, combine powers first.
Step 3
Exam Tip
गुणनफल ( (1+x)^{12} ) है, इसलिए गुणांक \( \binom{12}{8}=495 \)। द्विपद गुणन में घातें जोड़कर सरल करें।
The valid selections are \(A^3N^2\), \(BA^2N^2\), and \(BA^3N\), so the total is (3). In small repetition questions, list valid distributions directly.
Step 2
Why this answer is correct
The correct answer is B. (3). The valid selections are \(A^3N^2\), \(BA^2N^2\), and \(BA^3N\), so the total is (3). In small repetition questions, list valid distributions directly.
Step 3
Exam Tip
वैध चयन \(A^3N^2\), \(BA^2N^2\), और \(BA^3N\) हैं, इसलिए कुल (3) हैं। छोटे बहुलता प्रश्नों में वैध वितरण सीधे लिखें।
Using the limits \(M\le1\), \(I\le4\), \(S\le4\), and \(P\le2\), the total selections are (21). For repeated letters, count bounded integer solutions.
Step 2
Why this answer is correct
The correct answer is C. (21). Using the limits \(M\le1\), \(I\le4\), \(S\le4\), and \(P\le2\), the total selections are (21). For repeated letters, count bounded integer solutions.
Step 3
Exam Tip
अक्षर-सीमाएँ \(M\le1\), \(I\le4\), \(S\le4\), \(P\le2\) रखकर कुल (21) चयन मिलते हैं। बहुलता वाले अक्षरों में सीमित पूर्णांक हल गिनें।
A quadrilateral needs (2) points from each line, so \( \binom{7}{2}\binom{5}{2}=210 \). Taking (3) points from one line cannot form a quadrilateral.
Step 2
Why this answer is correct
The correct answer is C. (210). A quadrilateral needs (2) points from each line, so \( \binom{7}{2}\binom{5}{2}=210 \). Taking (3) points from one line cannot form a quadrilateral.
Step 3
Exam Tip
चतुर्भुज के लिए हर रेखा से (2) बिंदु चाहिए, इसलिए \( \binom{7}{2}\binom{5}{2}=210 \)। एक ही रेखा से (3) बिंदु लेने पर चतुर्भुज नहीं बनेगा।
There are (2) ways to choose the special person and then (4) from the other (8), so \(2\binom{8}{4}=140\). Exactly one means two symmetric cases.
Step 2
Why this answer is correct
The correct answer is A. (140). There are (2) ways to choose the special person and then (4) from the other (8), so \(2\binom{8}{4}=140\). Exactly one means two symmetric cases.
Step 3
Exam Tip
विशेष व्यक्ति चुनने के (2) तरीके और बाकी (4) व्यक्ति (8) में से, इसलिए \(2\binom{8}{4}=140\)। ठीक एक का अर्थ है दो समान मामले।
If both are selected, count \( \binom{11}{4} \); if neither is selected, count \( \binom{11}{6} \), giving (792). Split either both or neither into two cases.
Step 2
Why this answer is correct
The correct answer is C. (792). If both are selected, count \( \binom{11}{4} \); if neither is selected, count \( \binom{11}{6} \), giving (792). Split either both or neither into two cases.
Step 3
Exam Tip
दोनों चुने जाने पर \( \binom{11}{4} \) और दोनों न चुने जाने पर \( \binom{11}{6} \), कुल (792) है। या तो दोनों या कोई नहीं को दो अलग मामलों में बाँटें।
The first committee can be formed in \( \binom{10}{3} \) ways and the second in \( \binom{7}{4} \) ways, giving (4200). For named committees, the stages matter.
Step 2
Why this answer is correct
The correct answer is C. (4200). The first committee can be formed in \( \binom{10}{3} \) ways and the second in \( \binom{7}{4} \) ways, giving (4200). For named committees, the stages matter.
Step 3
Exam Tip
पहली समिति \( \binom{10}{3} \) और दूसरी \( \binom{7}{4} \) तरीकों से बनेगी, कुल (4200) है। नामित समितियों में चरणों का क्रम महत्त्व रखता है।
For an even sum, the number of odd selected numbers can be (0,2,4), giving (255). Focus on how many odd elements are selected, not only on total selection size.
Step 2
Why this answer is correct
The correct answer is C. (255). For an even sum, the number of odd selected numbers can be (0,2,4), giving (255). Focus on how many odd elements are selected, not only on total selection size.
Step 3
Exam Tip
सम योग के लिए विषम संख्याओं की संख्या (0,2,4) हो सकती है, कुल (255) है। चयन की संख्या पर नहीं, विषम चुने गए तत्वों की संख्या पर ध्यान दें।
For an odd sum, choose (1) odd and (2) even or (3) odd numbers, giving (570). Making a parity case table reduces errors.
Step 2
Why this answer is correct
The correct answer is B. (570). For an odd sum, choose (1) odd and (2) even or (3) odd numbers, giving (570). Making a parity case table reduces errors.
Step 3
Exam Tip
योग विषम के लिए (1) विषम और (2) सम या (3) विषम चुनें, कुल (570) है। सम-विषम संयोजन तालिका बनाकर गलती घटती है।
The product is odd only when all four numbers are odd, so \( \binom{8}{4}=70 \). For product parity, check every factor.
Step 2
Why this answer is correct
The correct answer is C. (70). The product is odd only when all four numbers are odd, so \( \binom{8}{4}=70 \). For product parity, check every factor.
Step 3
Exam Tip
गुणनफल विषम तभी होगा जब चारों संख्याएँ विषम हों, इसलिए \( \binom{8}{4}=70 \)। गुणनफल की सम-विषम प्रकृति में सभी कारकों पर ध्यान दें।
(a) is fixed as selected and (b) is fixed as not selected, so the remaining (8) elements are free, giving \(2^8=256\). Fix compulsory states first.
Step 2
Why this answer is correct
The correct answer is B. (256). (a) is fixed as selected and (b) is fixed as not selected, so the remaining (8) elements are free, giving \(2^8=256\). Fix compulsory states first.
Step 3
Exam Tip
(a) निश्चित रूप से चुना और (b) निश्चित रूप से छोड़ा गया, बाकी (8) अवयव स्वतंत्र हैं, इसलिए \(2^8=256\)। निश्चित अवस्थाओं को पहले स्थिर करें।
The selection is \( \binom{3}{2}\binom{6}{3}=60 \). With exactly (2) special elements, take the remaining (3) from ordinary elements.
Step 2
Why this answer is correct
The correct answer is C. (60). The selection is \( \binom{3}{2}\binom{6}{3}=60 \). With exactly (2) special elements, take the remaining (3) from ordinary elements.
Step 3
Exam Tip
चयन \( \binom{3}{2}\binom{6}{3}=60 \) है। ठीक (2) विशेष होने पर बाकी (3) सामान्य अवयवों से लें।
Subtract \( \binom{6}{2} \) subsets containing all three special elements from total \( \binom{9}{5} \), giving (111). In not-type questions, subtract forbidden selections.
Step 2
Why this answer is correct
The correct answer is A. (111). Subtract \( \binom{6}{2} \) subsets containing all three special elements from total \( \binom{9}{5} \), giving (111). In not-type questions, subtract forbidden selections.
Step 3
Exam Tip
कुल \( \binom{9}{5} \) से तीनों विशेष अवयवों वाले \( \binom{6}{2} \) घटाएँ, उत्तर (111) है। नहीं वाले प्रश्न में निषिद्ध चयन घटाएँ।
Subtract \( \binom{15}{3} \) cases with \(x_1\ge6\) from total \( \binom{21}{3} \), giving (875). For upper limits, subtract forbidden solutions.
Step 2
Why this answer is correct
The correct answer is C. (875). Subtract \( \binom{15}{3} \) cases with \(x_1\ge6\) from total \( \binom{21}{3} \), giving (875). For upper limits, subtract forbidden solutions.
Step 3
Exam Tip
कुल \( \binom{21}{3} \) से \(x_1\ge6\) वाले \( \binom{15}{3} \) घटते हैं, उत्तर (875) है। ऊपरी सीमा में प्रतिबंधित हल घटाना आसान है।
After subtracting minimums (5,1,1,1), (12) remains, so \( \binom{15}{3}=455 \). Subtract unequal minimum conditions first.
Step 2
Why this answer is correct
The correct answer is D. (455). After subtracting minimums (5,1,1,1), (12) remains, so \( \binom{15}{3}=455 \). Subtract unequal minimum conditions first.
Step 3
Exam Tip
न्यूनतम (5,1,1,1) घटाने पर (12) बचता है, इसलिए \( \binom{15}{3}=455 \)। असमान न्यूनतम शर्तों को पहले घटाएँ।
After giving (2) pens to each student, (6) pens remain, so \( \binom{8}{2}=28 \). Subtract the minimum condition and then count nonnegative solutions.
Step 2
Why this answer is correct
The correct answer is C. (28). After giving (2) pens to each student, (6) pens remain, so \( \binom{8}{2}=28 \). Subtract the minimum condition and then count nonnegative solutions.
Step 3
Exam Tip
हर विद्यार्थी को (2) देने के बाद (6) कलमें बचती हैं, इसलिए \( \binom{8}{2}=28 \)। न्यूनतम शर्त घटाकर फिर शून्य सहित हल गिनें।
This is the number of nonnegative solutions of \(x_1+x_2+x_3+x_4=6\), so \( \binom{9}{3}=84 \). For unlimited identical types, use stars and bars.
Step 2
Why this answer is correct
The correct answer is B. (84). This is the number of nonnegative solutions of \(x_1+x_2+x_3+x_4=6\), so \( \binom{9}{3}=84 \). For unlimited identical types, use stars and bars.
Step 3
Exam Tip
यह \(x_1+x_2+x_3+x_4=6\) के अशून्येतर नहीं बल्कि शून्य सहित हल हैं, इसलिए \( \binom{9}{3}=84 \)। असीमित समान प्रकारों में सितारे और पट्टियाँ लगाएँ।
Choose the couple \( \binom{6}{1} \), then (3) other couples \( \binom{5}{3} \), and one person from each \(2^3\), giving (480). After fixing exactly one couple, choose only one person from each remaining couple.
Step 2
Why this answer is correct
The correct answer is C. (480). Choose the couple \( \binom{6}{1} \), then (3) other couples \( \binom{5}{3} \), and one person from each \(2^3\), giving (480). After fixing exactly one couple, choose only one person from each remaining couple.
Step 3
Exam Tip
जोड़ा \( \binom{6}{1} \), बाकी (3) दंपति \( \binom{5}{3} \), और उनसे एक-एक व्यक्ति \(2^3\), कुल (480) है। ठीक एक जोड़ा रखने के बाद बाकी जोड़ों से केवल एक व्यक्ति चुनें।
First choose (4) couples and then (1) person from each, giving \( \binom{8}{4}2^4=1120 \). Choosing only one from a couple gives (2) choices.
Step 2
Why this answer is correct
The correct answer is D. (1120). First choose (4) couples and then (1) person from each, giving \( \binom{8}{4}2^4=1120 \). Choosing only one from a couple gives (2) choices.
Step 3
Exam Tip
पहले (4) दंपति चुनें और हर दंपति से (1) व्यक्ति लें, \( \binom{8}{4}2^4=1120 \)। जोड़े से केवल एक चुनने पर (2) विकल्प बनते हैं।
The selection is \( \binom{5}{3}\binom{9}{4}=1260 \). Exactly (3) means the remaining (4) come from outside the fixed group.
Step 2
Why this answer is correct
The correct answer is A. (1260). The selection is \( \binom{5}{3}\binom{9}{4}=1260 \). Exactly (3) means the remaining (4) come from outside the fixed group.
Step 3
Exam Tip
चयन \( \binom{5}{3}\binom{9}{4}=1260 \) है। ठीक (3) का अर्थ है बाकी (4) बाहर वाले समूह से आएँगे।
From one suit there are \( \binom{13}{5} \) selections and there are (4) suits, so the count is (5148). Choose the suit and then the cards.
Step 2
Why this answer is correct
The correct answer is D. (5148). From one suit there are \( \binom{13}{5} \) selections and there are (4) suits, so the count is (5148). Choose the suit and then the cards.
Step 3
Exam Tip
किसी एक सूट से \( \binom{13}{5} \) चयन और (4) सूट हैं, इसलिए (5148) है। सूट चुनना और फिर पत्ते चुनना दो चरण हैं।
