The sum is even when (3) even numbers or (2) odd and (1) even number are chosen, giving (28). Count parity cases separately.
Step 2
Why this answer is correct
The correct answer is B. (28). The sum is even when (3) even numbers or (2) odd and (1) even number are chosen, giving (28). Count parity cases separately.
Step 3
Exam Tip
योग सम तब होगा जब (3) सम या (2) विषम और (1) सम चुने जाएँ, कुल (28) हैं। सम-विषम मामलों को अलग-अलग गिनें।
The formula gives \( \binom{15-4+1}{4}=\binom{12}{4}=495 \). For no-consecutive selections, take combinations from reduced positions.
Step 2
Why this answer is correct
The correct answer is A. (495). The formula gives \( \binom{15-4+1}{4}=\binom{12}{4}=495 \). For no-consecutive selections, take combinations from reduced positions.
Step 3
Exam Tip
सूत्र \( \binom{15-4+1}{4}=\binom{12}{4}=495 \) देता है। क्रमागत-वर्जित चयन में घटे हुए स्थानों से संयोजन लें।
A parallelogram needs (2) lines from each family, so \( \binom{4}{2}\binom{5}{2}=60 \). The key idea is choosing two parallel lines from each family.
Step 2
Why this answer is correct
The correct answer is D. (60). A parallelogram needs (2) lines from each family, so \( \binom{4}{2}\binom{5}{2}=60 \). The key idea is choosing two parallel lines from each family.
Step 3
Exam Tip
एक समांतर चतुर्भुज के लिए हर परिवार से (2) रेखाएँ चाहिए, इसलिए \( \binom{4}{2}\binom{5}{2}=60 \)। दो-दो समानांतर रेखाएँ चुनना मुख्य विचार है।
Choosing two horizontal lines from (6) and two vertical lines from (5) gives \( \binom{6}{2}\binom{5}{2}=150 \). A rectangle needs two horizontal and two vertical lines.
Step 2
Why this answer is correct
The correct answer is B. (150). Choosing two horizontal lines from (6) and two vertical lines from (5) gives \( \binom{6}{2}\binom{5}{2}=150 \). A rectangle needs two horizontal and two vertical lines.
Step 3
Exam Tip
(6) क्षैतिज और (5) ऊर्ध्व रेखाओं में से दो-दो चुनने पर \( \binom{6}{2}\binom{5}{2}=150 \) मिलता है। आयत के लिए दो क्षैतिज और दो ऊर्ध्व रेखाएँ चाहिए।
The condition gives (2) boys and (4) girls, so \( \binom{8}{2}\binom{6}{4}=420 \). First find the distribution of numbers.
Step 2
Why this answer is correct
The correct answer is B. (420). The condition gives (2) boys and (4) girls, so \( \binom{8}{2}\binom{6}{4}=420 \). First find the distribution of numbers.
Step 3
Exam Tip
शर्त से लड़के (2) और लड़कियाँ (4) होंगी, इसलिए \( \binom{8}{2}\binom{6}{4}=420 \)। पहले संख्या-वितरण निकालें।
Subtract \( \binom{9}{5} \), where neither (A) nor (B) is selected, from total \( \binom{11}{5} \), giving (336). For at least one, complement is simplest.
Step 2
Why this answer is correct
The correct answer is D. (336). Subtract \( \binom{9}{5} \), where neither (A) nor (B) is selected, from total \( \binom{11}{5} \), giving (336). For at least one, complement is simplest.
Step 3
Exam Tip
कुल \( \binom{11}{5} \) से (A,B) दोनों न होने वाले \( \binom{9}{5} \) घटाएँ, उत्तर (336) है। कम से कम एक के लिए पूरक सबसे सरल है।
From total \( \binom{12}{5} \), subtract \( \binom{10}{3} \) selections containing both particular persons, giving (672). For not-together conditions, subtract forbidden cases.
Step 2
Why this answer is correct
The correct answer is A. (672). From total \( \binom{12}{5} \), subtract \( \binom{10}{3} \) selections containing both particular persons, giving (672). For not-together conditions, subtract forbidden cases.
Step 3
Exam Tip
कुल \( \binom{12}{5} \) में से दोनों विशेष व्यक्तियों वाले \( \binom{10}{3} \) चयन घटते हैं, उत्तर (672) है। साथ न होने पर निषिद्ध स्थिति घटाएँ।
Odd- and even-sized subsets are equal in number, so the count is \(2^8=256\). In such questions, half the subsets have odd size.
Step 2
Why this answer is correct
The correct answer is B. (256). Odd- and even-sized subsets are equal in number, so the count is \(2^8=256\). In such questions, half the subsets have odd size.
Step 3
Exam Tip
विषम और सम आकार के उपसमुच्चय बराबर होते हैं, इसलिए संख्या \(2^{8}=256\) है। ऐसे प्रश्नों में आधे उपसमुच्चय विषम आकार के होते हैं।
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।
Choose the (3) up-step positions among (8) total steps, so \( \binom{8}{3}=56 \). In paths, selecting step positions is a combination idea.
Step 2
Why this answer is correct
The correct answer is B. (56). Choose the (3) up-step positions among (8) total steps, so \( \binom{8}{3}=56 \). In paths, selecting step positions is a combination idea.
Step 3
Exam Tip
कुल (8) स्थानों में (3) ऊपर कदम चुनने हैं, इसलिए \( \binom{8}{3}=56 \) है। रास्तों में कदमों की जगहें चुनना संयोजन है।
The three cases of color distribution (2,1,1) add up to (270). When capacities are limited, keep the cases by color separate.
Step 2
Why this answer is correct
The correct answer is D. (270). The three cases of color distribution (2,1,1) add up to (270). When capacities are limited, keep the cases by color separate.
Step 3
Exam Tip
रंग-वितरण (2,1,1) के तीन मामलों का योग (270) देता है। क्षमता सीमाएँ होने पर हर रंग के मामलों को अलग रखें।
There are \(2^6\) subsets, and removing the empty selection gives (63). For at least one, remember to subtract the empty case.
Step 2
Why this answer is correct
The correct answer is A. (63). There are \(2^6\) subsets, and removing the empty selection gives (63). For at least one, remember to subtract the empty case.
Step 3
Exam Tip
सभी उपसमुच्चय \(2^6\) हैं, खाली चयन हटाने पर (63) मिलते हैं। कम से कम एक में खाली स्थिति घटाना याद रखें।
Subtract the impossible \( \binom{5}{3} \) from total \( \binom{12}{3} \), giving (210). A triangle needs three non-collinear points.
Step 2
Why this answer is correct
The correct answer is B. (210). Subtract the impossible \( \binom{5}{3} \) from total \( \binom{12}{3} \), giving (210). A triangle needs three non-collinear points.
Step 3
Exam Tip
कुल \( \binom{12}{3} \) में से असंभव \( \binom{5}{3} \) घटाएँ, उत्तर (210) है। त्रिभुज के लिए तीन असरेखीय बिंदु चाहिए।
The selection is \( \binom{4}{2}\binom{48}{3}=103776 \). For exactly-type questions, choose special objects first and then the remaining objects.
Step 2
Why this answer is correct
The correct answer is D. (103776). The selection is \( \binom{4}{2}\binom{48}{3}=103776 \). For exactly-type questions, choose special objects first and then the remaining objects.
Step 3
Exam Tip
चयन \( \binom{4}{2}\binom{48}{3}=103776 \) है। ठीक वाले प्रश्नों में पहले विशेष वस्तुएँ और फिर बाकी वस्तुएँ चुनें।
The no-consecutive formula gives \( \binom{10-3+1}{3}=\binom{8}{3}=56 \). In such questions, the gap method is very fast.
Step 2
Why this answer is correct
The correct answer is A. (56). The no-consecutive formula gives \( \binom{10-3+1}{3}=\binom{8}{3}=56 \). In such questions, the gap method is very fast.
Step 3
Exam Tip
क्रमागत न होने का सूत्र \( \binom{10-3+1}{3}=\binom{8}{3}=56 \) देता है। ऐसे प्रश्नों में खाली स्थान विधि बहुत तेज होती है।
The total segments are \( \binom{10}{2} \), and subtracting (10) sides gives (35). For polygons, first count all vertex pairs.
Step 2
Why this answer is correct
The correct answer is A. (35). The total segments are \( \binom{10}{2} \), and subtracting (10) sides gives (35). For polygons, first count all vertex pairs.
Step 3
Exam Tip
कुल रेखाखंड \( \binom{10}{2} \) हैं और (10) भुजाएँ घटाने पर (35) बचते हैं। बहुभुज में पहले सभी शीर्ष-युग्म गिनें।
The coefficient is \( \binom{12}{5} \), so the answer is (792). In exams, for ( (1+x)^n ), the coefficient of \(x^r\) is \( \binom{n}{r} \).
Step 2
Why this answer is correct
The correct answer is D. (792). The coefficient is \( \binom{12}{5} \), so the answer is (792). In exams, for ( (1+x)^n ), the coefficient of \(x^r\) is \( \binom{n}{r} \).
Step 3
Exam Tip
गुणांक \( \binom{12}{5} \) है, इसलिए उत्तर (792) है। परीक्षा में ( (1+x)^n ) में \(x^r\) का गुणांक सीधे \( \binom{n}{r} \) लें।
The number of special objects can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335\).
Step 2
Why this answer is correct
The correct answer is C. (49335). The number of special objects can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335\).
Step 3
Exam Tip
विशेष वस्तुएं (0), (1) या (2) चुनी जा सकती हैं। कुल \(\binom{7}{0}\binom{13}{8}+\binom{7}{1}\binom{13}{7}+\binom{7}{2}\binom{13}{6}=49335\) है।
(14) खिलाड़ियों में से (7) खिलाड़ियों का चयन करना है। एक कप्तान पहले से तय है और उसे चुना नहीं जाना है लेकिन उपकप्तान अवश्य चुना जाना है। कितने तरीके हैं?
The captain is excluded and the vice-captain is fixed. The remaining (6) players are chosen from (12), so \(\binom{12}{6}=924\).
Step 2
Why this answer is correct
The correct answer is D. (924). The captain is excluded and the vice-captain is fixed. The remaining (6) players are chosen from (12), so \(\binom{12}{6}=924\).
Step 3
Exam Tip
कप्तान हट गया और उपकप्तान तय है। बाकी (6) खिलाड़ी (12) में से चुने जाएंगे इसलिए \(\binom{12}{6}=924\) है।
The number of odd selections is \(2^{15-1}=16384\). Remember that even and odd selections are equal in such questions.
Step 2
Why this answer is correct
The correct answer is C. (16384). The number of odd selections is \(2^{15-1}=16384\). Remember that even and odd selections are equal in such questions.
Step 3
Exam Tip
विषम चयन की संख्या \(2^{15-1}=16384\) होती है। परीक्षा में सम और विषम चयन की संख्या बराबर याद रखें।
Choose (1) of the two fixed cards and (5) cards from the remaining (13). The ways are \(\binom{2}{1}\binom{13}{5}=2574\).
Step 2
Why this answer is correct
The correct answer is A. (2574). Choose (1) of the two fixed cards and (5) cards from the remaining (13). The ways are \(\binom{2}{1}\binom{13}{5}=2574\).
Step 3
Exam Tip
दो निश्चित कार्डों में से (1) चुनें और बाकी (5) कार्ड (13) में से चुनें। तरीके \(\binom{2}{1}\binom{13}{5}=2574\) हैं।
The number of seniors can be (0), (1), (2), or (3). The total is \(\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072\).
Step 2
Why this answer is correct
The correct answer is D. (105072). The number of seniors can be (0), (1), (2), or (3). The total is \(\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0), (1), (2) या (3) हो सकती है। कुल \(\binom{10}{0}\binom{12}{7}+\binom{10}{1}\binom{12}{6}+\binom{10}{2}\binom{12}{5}+\binom{10}{3}\binom{12}{4}=105072\) है।
