class11 se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
1558 questions tagged with class11.
How many ways are there to choose (4) subjects from (6) subjects?
C. (15)
\(\binom{6}{4}=\binom{6}{2}=15\) होता है। \(\binom{n}{r}=\binom{n}{n-r}\) याद रखें। / \(\binom{6}{4}=\binom{6}{2}=15\). Remember \(\binom{n}{r}=\binom{n}{n-r}\).
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B. (78)
समिति के लिए केवल चयन है इसलिए \(\binom{13}{2}=78\) होगा। पद अलग न हों तो क्रम नहीं गिना जाता। / It is only selection for a committee, so \(\binom{13}{2}=78\). If posts are not different, order is not counted.
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How many ways are there to select (3) chairs from (7) chairs?
B. (35)
चयन में व्यवस्था नहीं करनी है इसलिए \(\binom{7}{3}=35\) है। कुर्सियां चुनना संयोजन है। / No arrangement is required, so \(\binom{7}{3}=35\). Choosing chairs is a combination.
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From (5) red balls and (3) black balls, how many ways are there to choose (2) red balls?
B. (10)
केवल लाल गेंदों में से चयन है। इसलिए \(\binom{5}{2}=10\) सही है। / Only red balls are being selected. Hence \(\binom{5}{2}=10\) is correct.
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From (6) red balls and (4) blue balls, in how many ways can (1) red and (1) blue ball be selected?
C. (24)
तरीके \(\binom{6}{1}\binom{4}{1}=24\) हैं। अलग रंगों के चयन में गुणा नियम लगाएं। / The ways are \(\binom{6}{1}\binom{4}{1}=24\). Apply multiplication rule for selections from different colors.
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How many ways are there to choose (3) cards from (11) different cards?
C. (165)
यह \(\binom{11}{3}=165\) का सीधा प्रयोग है। कार्ड चुनने में क्रम नहीं देखा जाता। / This is a direct use of \(\binom{11}{3}=165\). Order is not considered while choosing cards.
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How many ways are there to select (2) candidates for an award from (12) candidates?
B. (66)
दो समान चयनित उम्मीदवारों का क्रम महत्वपूर्ण नहीं है। अतः \(\binom{12}{2}=66\) होगा। / The order of the two selected candidates is not important. Thus \(\binom{12}{2}=66\).
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In how many ways can a committee of (3) members be selected from (9) members?
C. (84)
समिति चयन का प्रश्न है इसलिए \(\binom{9}{3}=84\) है। समिति में पद न हों तो क्रम नहीं गिना जाता। / This is a committee selection question, so \(\binom{9}{3}=84\). If posts are not assigned, order is not counted.
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What is the total number of handshakes among (10) persons taken (2) at a time?
B. (45)
हर हाथ मिलाना (2) व्यक्तियों की एक जोड़ी है। इसलिए कुल संख्या \(\binom{10}{2}=45\) है। / Each handshake is a pair of (2) persons. Therefore the total number is \(\binom{10}{2}=45\).
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How many circles can be formed from (6) points if no (3) points are collinear?
C. (20)
एक वृत्त को (3) बिंदु निर्धारित करते हैं। इसलिए \(\binom{6}{3}=20\) सही है। / A circle is determined by (3) points. Hence \(\binom{6}{3}=20\) is correct.
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There are (7) points, no (3) of which are collinear. How many triangles can be formed?
B. (35)
त्रिभुज के लिए (3) बिंदु चाहिए इसलिए \(\binom{7}{3}=35\) है। सीध में न होने की शर्त त्रिभुज को संभव बनाती है। / A triangle needs (3) points, so \(\binom{7}{3}=35\). The non-collinear condition makes triangles possible.
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How many lines can be formed by selecting (2) points from (8) points?
B. (28)
दो बिंदु एक रेखा बनाते हैं इसलिए \(\binom{8}{2}=28\) है। बिंदुओं का क्रम नहीं गिना जाता। / Two points form one line, so \(\binom{8}{2}=28\). The order of points is not counted.
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From (4) apples and (5) mangoes, in how many ways can (1) apple and (2) mangoes be selected?
C. (40)
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग फल वर्गों के चयन को गुणा करें। / The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). Multiply selections from separate fruit groups.
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From (5) boys and (4) girls, how many ways are there to select (2) boys?
A. (10)
केवल लड़कों में से चयन है इसलिए \(\binom{5}{2}=10\) होगा। अनावश्यक सूचना से भ्रमित न हों। / Only boys are being selected, so \(\binom{5}{2}=10\). Do not get confused by extra information.
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From (4) boys and (3) girls, in how many ways can (1) boy and (1) girl be selected?
B. (12)
चयन \(\binom{4}{1}\binom{3}{1}=12\) है। अलग-अलग वर्गों से चयन में गुणा करते हैं। / The selection is \(\binom{4}{1}\binom{3}{1}=12\). Multiply choices when selecting from different groups.
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In how many ways can a pair of friends be selected from (6) friends?
B. (15)
जोड़ी के लिए \(\binom{6}{2}=15\) होगा। जोड़ी में पहला और दूसरा अलग नहीं गिने जाते। / For a pair, \(\binom{6}{2}=15\). In a pair, first and second are not counted separately.
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How many ways are there to choose all (5) colors from (5) colors?
B. (1)
सभी वस्तुएं चुनने का केवल एक तरीका होता है। इसलिए \(\binom{5}{5}=1\) है। / There is only one way to choose all objects. Hence \(\binom{5}{5}=1\).
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Why is selecting (0) pens from (10) pens considered one way?
A. क्योंकि \(\binom{10}{0}=1\)Because \(\binom{10}{0}=1\)
कुछ भी न चुनना भी एक निश्चित चयन है। इसलिए \(\binom{n}{0}=1\) होता है। / Choosing nothing is also one definite selection. Therefore \(\binom{n}{0}=1\).
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How many ways are there to select (1) question from (9) questions?
C. (9)
एक वस्तु चुनने पर \(\binom{9}{1}=9\) होता है। \(\binom{n}{1}=n\) याद रखें। / Selecting one object gives \(\binom{9}{1}=9\). Remember \(\binom{n}{1}=n\).
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In how many ways can a group of (4) players be formed from (8) players?
C. (70)
समूह में क्रम नहीं होता इसलिए \(\binom{8}{4}=70\) होगा। परीक्षा में समूह शब्द दिखे तो संयोजन लगाएं। / A group has no order, so \(\binom{8}{4}=70\). In exams, the word group usually indicates combination.
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How many ways are there to choose (2) flowers from (7) different flowers?
B. (21)
यह \(\binom{7}{2}=\frac{7\cdot6}{2}=21\) है। ऐसे प्रश्नों में वस्तुओं की जोड़ी बनती है। / It is \(\binom{7}{2}=\frac{7\cdot6}{2}=21\). Such questions form pairs of objects.
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In how many ways can a team of (3) students be selected from (6) students?
C. (20)
टीम चुनने में क्रम महत्वपूर्ण नहीं होता। इसलिए \(\binom{6}{3}=20\) सही है। / Order is not important while selecting a team. Hence \(\binom{6}{3}=20\) is correct.
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If (2) books are to be selected from (5) books, how many selections are possible?
B. (10)
यह \(\binom{5}{2}=10\) का सरल प्रयोग है। परीक्षा में चयन का क्रम नहीं गिना जाता। / This is a direct use of \(\binom{5}{2}=10\). In exams, order is not counted in selection.
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D. (12600)
पहले व्यंजनों को \(\frac{6!}{3!2!}\) तरीकों से सजाएँ और (7) खाली स्थानों में स्वरों को \(\binom{7}{5}\frac{5!}{3!2!}\) तरीकों से रखें। स्वर साथ न हों तो gap method लगाएँ। / Arrange the consonants in \(\frac{6!}{3!2!}\) ways and place the vowels in (7) gaps in \(\binom{7}{5}\frac{5!}{3!2!}\) ways. Use the gap method when vowels must not be adjacent.
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B. (604800)
कुल (10!) क्रमों में तीन विशेष वस्तुओं के (3!) आपसी क्रम समान रूप से संभव हैं, इसलिए \(\frac{10!}{3!}\) है। relative order fixed हो तो कुल को उस group के factorial से बाँटें। / Among total (10!) arrangements, the (3!) relative orders of the three special objects are equally likely, so the count is \(\frac{10!}{3!}\). If relative order is fixed, divide by that group's factorial.
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In arrangements of the letters of (SELECTION), how many have all vowels together?
A. (14400)
स्वर (E,E,I,O) एक ब्लॉक हैं, इसलिए बाहरी (6) इकाइयाँ (6!) तरीकों से और अंदर स्वर \(\frac{4!}{2!}\) तरीकों से सजते हैं। vowel block में repeated vowels का भाग देना न भूलें। / The vowels (E,E,I,O) form one block, so (6) outside units arrange in (6!) ways and the vowels arrange internally in \(\frac{4!}{2!}\) ways. Do not forget repeated vowels inside the vowel block.
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B. (8640)
दो विशेष कार्डों के स्थानों में अंतर (3) होगा, ऐसे (5) स्थान-जोड़े हैं और उनका क्रम (2) तरीकों से है, फिर (6!) क्रम हैं। exact gap में position pairs गिनें। / The positions of the two particular cards differ by (3), giving (5) position pairs and (2) orders, then the remaining (6!) arrangements. Count position pairs for exact gaps.
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B. (181440)
कुल (9!) क्रमों में कप्तान और गोलकीपर के दो आपसी क्रम बराबर हैं, इसलिए आधे क्रम मान्य हैं। relative order fixed हो तो कुल को (2) से बाँटें। / In total (9!) orders, the two relative orders of captain and goalkeeper are equally likely, so half are valid. If relative order is fixed, divide by (2).
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What is the number of distinct arrangements of the letters of (INDEPENDENT)?
A. (831600)
कुल (11) अक्षर हैं और (N,E,D) क्रमशः (3,3,2) बार आते हैं, इसलिए \(\frac{11!}{3!3!2!}\) है। repeated letters की सभी आवृत्तियाँ लिखें। / There are (11) letters with (N,E,D) repeated (3,3,2) times, so the count is \(\frac{11!}{3!3!2!}\). Write all repeated-letter frequencies.
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D. (54432)
अंतिम अंक (0) होने पर (9P5) और अंतिम अंक (2,4,6,8) होने पर \(4\cdot8\cdot8P4\) मिलते हैं। शून्य और गैर-शून्य अंतिम अंक के cases अलग करें। / If the last digit is (0), count (9P5); if it is (2,4,6,8), count \(4\cdot8\cdot8P4\). Separate zero and nonzero last-digit cases.
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