Adding \(360^\circ\) three times to \(-725^\circ\) gives \(355^\circ\). In exams, keep the answer between \(0^\circ\) and \(360^\circ\).
Step 2
Why this answer is correct
The correct answer is D. \(355^\circ\). Adding \(360^\circ\) three times to \(-725^\circ\) gives \(355^\circ\). In exams, keep the answer between \(0^\circ\) and \(360^\circ\).
Step 3
Exam Tip
\(-725^\circ\) में (3) बार \(360^\circ\) जोड़ने पर \(355^\circ\) मिलता है। परीक्षा में उत्तर को \(0^\circ\) से \(360^\circ\) के बीच रखें।
\(112^\circ 30'\) means \(112.5^\circ\), so the value is \(5\pi/8\). In exams, first convert minutes into degrees.
Step 2
Why this answer is correct
The correct answer is B. \(5\pi / 8\). \(112^\circ 30'\) means \(112.5^\circ\), so the value is \(5\pi/8\). In exams, first convert minutes into degrees.
Step 3
Exam Tip
\(112^\circ 30'\) का अर्थ \(112.5^\circ\) है, इसलिए मान \(5\pi/8\) मिलता है। परीक्षा में मिनट को पहले डिग्री में बदलें।
A. recombinant DNA और DNA replication/Recombinant DNA and DNA replication
Step 1
Concept
Ligase joins fragments in recombinant DNA and joins Okazaki fragments in replication. Both exam contexts are important.
Step 2
Why this answer is correct
The correct answer is A. recombinant DNA और DNA replication / Recombinant DNA and DNA replication. Ligase joins fragments in recombinant DNA and joins Okazaki fragments in replication. Both exam contexts are important.
Step 3
Exam Tip
Ligase recombinant DNA में fragments जोड़ता है और replication में Okazaki fragments जोड़ता है। ये दोनों exam contexts महत्वपूर्ण हैं।
A. डीएनए खंडों की joining/Joining of DNA fragments
Step 1
Concept
Ligase joins DNA fragments. In biotechnology it is linked with cloning and recombinant DNA.
Step 2
Why this answer is correct
The correct answer is A. डीएनए खंडों की joining / Joining of DNA fragments. Ligase joins DNA fragments. In biotechnology it is linked with cloning and recombinant DNA.
Step 3
Exam Tip
लाइगेज डीएनए fragments को जोड़ता है। जैव प्रौद्योगिकी में यह cloning और recombinant DNA से जुड़ा है।
A. वे DNA molecules को cut करते हैं/They cut DNA molecules
Step 1
Concept
The name molecular scissors comes from their DNA cutting function. This is a basic term in Class 11 biotechnology.
Step 2
Why this answer is correct
The correct answer is A. वे DNA molecules को cut करते हैं / They cut DNA molecules. The name molecular scissors comes from their DNA cutting function. This is a basic term in Class 11 biotechnology.
Step 3
Exam Tip
Molecular scissors नाम DNA cutting function के कारण है। यह Class 11 biotechnology का basic term है।
A. डीएनए को विशेष स्थलों पर काटने वाले जीवाणु एंजाइम/Bacterial enzymes that cut DNA at specific sites
Step 1
Concept
Restriction enzymes are bacterial-origin DNA-cutting tools. This is the main summary of the topic.
Step 2
Why this answer is correct
The correct answer is A. डीएनए को विशेष स्थलों पर काटने वाले जीवाणु एंजाइम / Bacterial enzymes that cut DNA at specific sites. Restriction enzymes are bacterial-origin DNA-cutting tools. This is the main summary of the topic.
Step 3
Exam Tip
प्रतिबंध एंजाइम जीवाणु स्रोत के डीएनए-काटने वाले उपकरण हैं। यही इस टॉपिक का मुख्य सार है।
A. डीएनए को विशेष स्थानों पर काटना/Cutting DNA at specific sites
Step 1
Concept
Restriction enzymes cut DNA at specific recognition sites. For exams remember them as molecular scissors of biotechnology.
Step 2
Why this answer is correct
The correct answer is A. डीएनए को विशेष स्थानों पर काटना / Cutting DNA at specific sites. Restriction enzymes cut DNA at specific recognition sites. For exams remember them as molecular scissors of biotechnology.
Step 3
Exam Tip
प्रतिबंध एंजाइम डीएनए को विशेष पहचान स्थलों पर काटते हैं। परीक्षा में इन्हें जैव प्रौद्योगिकी की आणविक कैंची याद रखें।
Treat (AB) and (CD) as two blocks; (6) objects can be arranged in (6!) ways and the blocks internally in \(2!\times2!\) ways. Exam tip: with multiple blocks, count each block's internal order.
Step 2
Why this answer is correct
The correct answer is A. (2880). Treat (AB) and (CD) as two blocks; (6) objects can be arranged in (6!) ways and the blocks internally in \(2!\times2!\) ways. Exam tip: with multiple blocks, count each block's internal order.
Step 3
Exam Tip
(AB) और (CD) को दो खंड मानें; कुल (6) वस्तुएं (6!) तरीकों से और दोनों खंडों के अंदर \(2!\times2!\) तरीके हैं। परीक्षा में कई blocks हों तो हर block का internal order गिनें।
Zero can occupy only the hundreds or tens place, giving (2) choices; the other (3) places are filled from (6) nonzero digits in \(^{6}P_{3}\) ways. Exam tip: decide the allowed positions of (0) first.
Step 2
Why this answer is correct
The correct answer is B. (300). Zero can occupy only the hundreds or tens place, giving (2) choices; the other (3) places are filled from (6) nonzero digits in \(^{6}P_{3}\) ways. Exam tip: decide the allowed positions of (0) first.
Step 3
Exam Tip
(0) को केवल सैकड़ों या दहाइयों के स्थान पर रख सकते हैं, यानी (2) तरीके; फिर बाकी (3) स्थान (6) nonzero अंकों से \(^{6}P_{3}\) तरीकों से भरते हैं। परीक्षा में (0) की allowed positions पहले तय करें।
Treat each couple as one block with fixed internal order, so only the (4!) blocks are arranged. Exam tip: do not multiply by (2!) when internal order is fixed.
Step 2
Why this answer is correct
The correct answer is A. (24). Treat each couple as one block with fixed internal order, so only the (4!) blocks are arranged. Exam tip: do not multiply by (2!) when internal order is fixed.
Step 3
Exam Tip
हर दंपति को निश्चित अंदरूनी क्रम वाला एक खंड मानें, इसलिए केवल (4!) खंडों को सजाना है। परीक्षा में internal order fixed हो तो (2!) न लगाएं।
Among total (7!) arrangements, the (3!) relative orders of (A,B,C) are equally likely, so \(\frac{7!}{3!}\) are valid. Exam tip: divide by factorial for fixed relative order.
Step 2
Why this answer is correct
The correct answer is A. (840). Among total (7!) arrangements, the (3!) relative orders of (A,B,C) are equally likely, so \(\frac{7!}{3!}\) are valid. Exam tip: divide by factorial for fixed relative order.
Step 3
Exam Tip
कुल (7!) व्यवस्थाओं में (A,B,C) के (3!) सापेक्ष क्रम समान हैं, इसलिए \(\frac{7!}{3!}\) मान्य हैं। परीक्षा में fixed relative order के लिए factorial से divide करें।
First arrange the (5) white balls around a circle in ((5-1)!) ways, then place (3) black balls in \(^{5}P_{3}\) circular gaps. Exam tip: in circular gaps, arrange the base objects first.
Step 2
Why this answer is correct
The correct answer is A. (2880). First arrange the (5) white balls around a circle in ((5-1)!) ways, then place (3) black balls in \(^{5}P_{3}\) circular gaps. Exam tip: in circular gaps, arrange the base objects first.
Step 3
Exam Tip
पहले (5) सफेद गेंदों की गोल व्यवस्था ((5-1)!) है, फिर (5) gaps में (3) काली गेंदें \(^{5}P_{3}\) तरीकों से रखी जाती हैं। परीक्षा में circular gaps में सफेद वस्तुओं को पहले fix करें।
Subtract adjacent-special-car cases \(9\times2!\times{}^{8}P_{6}\) from total \(^{10}P_{8}\). Exam tip: when parking spots are distinct, start from total arrangements.
Step 2
Why this answer is correct
The correct answer is B. (1451520). Subtract adjacent-special-car cases \(9\times2!\times{}^{8}P_{6}\) from total \(^{10}P_{8}\). Exam tip: when parking spots are distinct, start from total arrangements.
Step 3
Exam Tip
कुल \(^{10}P_{8}\) से special cars adjacent वाली \(9\times2!\times{}^{8}P_{6}\) व्यवस्थाएं घटाएं। परीक्षा में parking spots distinct हों तो पहले कुल arrangements लें।
Choose the end for (P) and (E) in (2) ways; the remaining (5) letters include (E) repeated twice, so there are \(\frac{5!}{2!}\) arrangements. Exam tip: divide for a repeated end-letter left inside.
Step 2
Why this answer is correct
The correct answer is B. (240). Choose the end for (P) and (E) in (2) ways; the remaining (5) letters include (E) repeated twice, so there are \(\frac{5!}{2!}\) arrangements. Exam tip: divide for a repeated end-letter left inside.
Step 3
Exam Tip
एक सिरा (P) के लिए और दूसरा (E) के लिए (2) तरीकों से चुने; बाकी (5) अक्षरों में (E) दो बार है, इसलिए \(\frac{5!}{2!}\) व्यवस्थाएं हैं। परीक्षा में repeated end-letter बच जाने पर उसका factorial divide करें।
There are (4) position pairs at distance (2) and (2!) orders for questions (1) and (2), and the remaining (4) questions arrange in (4!) ways. Exam tip: exactly one between means position difference (2).
Step 2
Why this answer is correct
The correct answer is B. (192). There are (4) position pairs at distance (2) and (2!) orders for questions (1) and (2), and the remaining (4) questions arrange in (4!) ways. Exam tip: exactly one between means position difference (2).
Step 3
Exam Tip
प्रश्न (1) और (2) के लिए दूरी (2) वाले (4) position-pairs और (2!) क्रम हैं, बाकी (4) प्रश्न (4!) तरीकों से लगेंगे। परीक्षा में exactly one between का अर्थ position difference (2) है।
Form an (AB) block among (8) objects, then subtract cases where (C) is adjacent to the block on either outside side; total is (2!\times\(8!-2\times7!\)). Exam tip: compound restrictions may need both block and subtraction.
Step 2
Why this answer is correct
The correct answer is D. (10080). Form an (AB) block among (8) objects, then subtract cases where (C) is adjacent to the block on either outside side; total is (2!\times\(8!-2\times7!\)). Exam tip: compound restrictions may need both block and subtraction.
Step 3
Exam Tip
(AB) खंड बनाकर (8) वस्तुओं की व्यवस्थाएं लें, फिर (C) के खंड से adjacent होने वाले (2) बाहरी स्थानों को घटाएं; कुल (2!\times\(8!-2\times7!\)) है। परीक्षा में compound restriction में block और subtraction दोनों लग सकते हैं।
The thousands digit must be from (5,6,7,8,9), and the units digit must be even; case-wise counting gives (400). Exam tip: count overlapping digit choices case by case.
Step 2
Why this answer is correct
The correct answer is A. (400). The thousands digit must be from (5,6,7,8,9), and the units digit must be even; case-wise counting gives (400). Exam tip: count overlapping digit choices case by case.
Step 3
Exam Tip
हजारों का अंक (5,6,7,8,9) में से और इकाई अंक सम होना चाहिए; leading और ending cases अलग गिनने पर (400) मिलता है। परीक्षा में overlapping digit choices को case-wise गिनें।
Choose the other (3) letters from (6), then subtract adjacent special-pair arrangements \(2!\times4!\) from total (5!). Exam tip: handle inclusion and adjacency restrictions separately.
Step 2
Why this answer is correct
The correct answer is C. (2160). Choose the other (3) letters from (6), then subtract adjacent special-pair arrangements \(2!\times4!\) from total (5!). Exam tip: handle inclusion and adjacency restrictions separately.
Step 3
Exam Tip
बाकी (3) अक्षर (6) में से चुनें, फिर (5!) कुल व्यवस्थाओं से special pair adjacent वाली \(2!\times4!\) व्यवस्थाएं घटाएं। परीक्षा में शामिल और adjacent restriction को अलग-अलग संभालें।
Choose the (2) correctly assigned prizes in \(\binom{7}{2}\) ways and derange the remaining (5), giving (!5=44). Exam tip: use derangement for exactly correct cases.
Step 2
Why this answer is correct
The correct answer is B. (1854). Choose the (2) correctly assigned prizes in \(\binom{7}{2}\) ways and derange the remaining (5), giving (!5=44). Exam tip: use derangement for exactly correct cases.
Step 3
Exam Tip
सही जाने वाले (2) पुरस्कार \(\binom{7}{2}\) तरीकों से चुनें और बाकी (5) का derangement (!5=44) है। परीक्षा में exactly correct cases के लिए derangement का प्रयोग करें।
There are (11) letters; (I) occurs four times, (S) occurs four times, and (P) occurs twice, so the count is \(\frac{11!}{4!,4!,2!}\). Exam tip: recheck repeated-letter counts.
Step 2
Why this answer is correct
The correct answer is A. (34650). There are (11) letters; (I) occurs four times, (S) occurs four times, and (P) occurs twice, so the count is \(\frac{11!}{4!,4!,2!}\). Exam tip: recheck repeated-letter counts.
Step 3
Exam Tip
कुल (11) अक्षर हैं; (I) चार, (S) चार और (P) दो बार आते हैं, इसलिए \(\frac{11!}{4!,4!,2!}\) है। परीक्षा में समान अक्षरों की गिनती दोबारा जांचें।
Fix (A); the opposite seat for (B) is fixed, and the remaining (8) people can be arranged in (8!) ways. Exam tip: fix one person to remove circular symmetry.
Step 2
Why this answer is correct
The correct answer is A. (80640). Fix (A); the opposite seat for (B) is fixed, and the remaining (8) people can be arranged in (8!) ways. Exam tip: fix one person to remove circular symmetry.
Step 3
Exam Tip
(A) को स्थिर करें, (B) की विपरीत सीट निश्चित है, और बाकी (8) लोग (8!) तरीकों से बैठते हैं। परीक्षा में circular symmetry हटाने के लिए एक व्यक्ति को स्थिर करें।
We assign (4) of the (6) rings to (4) distinct fingers with order, so \(^{6}P_{4}=360\). Exam tip: distinct fingers require permutations.
Step 2
Why this answer is correct
The correct answer is A. (360). We assign (4) of the (6) rings to (4) distinct fingers with order, so \(^{6}P_{4}=360\). Exam tip: distinct fingers require permutations.
Step 3
Exam Tip
(6) अंगूठियों में से (4) को (4) अलग-अलग उंगलियों पर क्रम सहित रखना है, इसलिए \(^{6}P_{4}=360\)। परीक्षा में distinct fingers होने पर permutation लगती है।
Choose the other (4) digits with (0) and (8) in \(\binom{7}{4}\) ways; for each selection subtract (5!) leading-zero cases from (6!). Exam tip: leading zero correction is mandatory.
Step 2
Why this answer is correct
The correct answer is B. (13440). Choose the other (4) digits with (0) and (8) in \(\binom{7}{4}\) ways; for each selection subtract (5!) leading-zero cases from (6!). Exam tip: leading zero correction is mandatory.
Step 3
Exam Tip
(0) और (8) के साथ बाकी (4) अंक \(\binom{7}{4}\) तरीकों से चुनें; प्रत्येक चयन में कुल (6!) से first zero वाले (5!) घटाएं। परीक्षा में leading zero का correction अनिवार्य है।
There are (2) alternate patterns, and in each pattern Indians can be arranged in (5!) ways and foreigners in (5!) ways. Exam tip: equal groups in alternate arrangement give (2) patterns.
Step 2
Why this answer is correct
The correct answer is A. (28800). There are (2) alternate patterns, and in each pattern Indians can be arranged in (5!) ways and foreigners in (5!) ways. Exam tip: equal groups in alternate arrangement give (2) patterns.
Step 3
Exam Tip
दो alternate patterns संभव हैं और प्रत्येक में भारतीय (5!) तथा विदेशी (5!) तरीकों से बैठते हैं। परीक्षा में बराबर समूहों के alternate arrangement में (2) patterns होते हैं।
In total (7!) arrangements, the two relative orders of (A) and (B) are equally likely, so half are valid. Exam tip: use symmetry for relative order conditions like left of.
Step 2
Why this answer is correct
The correct answer is A. (2520). In total (7!) arrangements, the two relative orders of (A) and (B) are equally likely, so half are valid. Exam tip: use symmetry for relative order conditions like left of.
Step 3
Exam Tip
कुल (7!) व्यवस्थाओं में (A) और (B) के दो सापेक्ष क्रम समान रूप से संभव हैं, इसलिए आधी व्यवस्थाएं मान्य हैं। परीक्षा में left of जैसी relative order शर्त के लिए symmetry प्रयोग करें।
Total arrangements are \(\frac{7!}{2!}\), and among the (4) relative orders of vowels only (1) order is allowed, so divide by (4!). Exam tip: divide by the factorial when relative order is fixed.
Step 2
Why this answer is correct
The correct answer is D. (210). Total arrangements are \(\frac{7!}{2!}\), and among the (4) relative orders of vowels only (1) order is allowed, so divide by (4!). Exam tip: divide by the factorial when relative order is fixed.
Step 3
Exam Tip
कुल व्यवस्थाएं \(\frac{7!}{2!}\) हैं और (4) स्वरों के सापेक्ष क्रमों में केवल (1) क्रम स्वीकार्य है, इसलिए (4!) से भाग दें। परीक्षा में relative order fixed हो तो संबंधित factorial से divide करें।
First choose ordered seat pairs for (A,B) with distance at least (3); there are (42) such pairs, then seat the remaining (4) people in \(^{6}P_{4}\) ways. Exam tip: position-pair counting is useful for numbered seats.
Step 2
Why this answer is correct
The correct answer is B. (16800). First choose ordered seat pairs for (A,B) with distance at least (3); there are (42) such pairs, then seat the remaining (4) people in \(^{6}P_{4}\) ways. Exam tip: position-pair counting is useful for numbered seats.
Step 3
Exam Tip
पहले (A,B) के सीट-जोड़ों में दूरी कम से कम (3) रखें; ऐसे ordered pairs (42) हैं, फिर बाकी (4) लोग \(^{6}P_{4}\) तरीकों से बैठते हैं। परीक्षा में numbered seats में position-pair गिनना उपयोगी है।
Arrange the (5) mathematics books in (5!) ways, then place the (4) physics books in \(^{6}P_{4}\) gaps. Exam tip: gaps are the safest method for no two together.
Step 2
Why this answer is correct
The correct answer is A. (86400). Arrange the (5) mathematics books in (5!) ways, then place the (4) physics books in \(^{6}P_{4}\) gaps. Exam tip: gaps are the safest method for no two together.
Step 3
Exam Tip
पहले (5) गणित पुस्तकें (5!) तरीकों से रखें, फिर (6) gaps में (4) भौतिकी पुस्तकें \(^{6}P_{4}\) तरीकों से रखें। परीक्षा में no two together के लिए gaps सबसे सुरक्षित तरीका है।
Use inclusion-exclusion on total (8!), subtracting cases with (T) first and (E) last: (8!-7!-7!+6!). Exam tip: add back the overlap in two forbidden conditions.
Step 2
Why this answer is correct
The correct answer is B. (30960). Use inclusion-exclusion on total (8!), subtracting cases with (T) first and (E) last: (8!-7!-7!+6!). Exam tip: add back the overlap in two forbidden conditions.
Step 3
Exam Tip
कुल (8!) से (T) first और (E) last वाले मामलों को inclusion-exclusion से घटाएं, यानी (8!-7!-7!+6!)। परीक्षा में दो निषेध शर्तों में overlap जोड़ना न भूलें।
