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.
A. (45)
पहले (2) प्रश्न तय हैं इसलिए बाकी (2) प्रश्न (10) में से चुनेंगे। तरीके \(\binom{10}{2}=45\) हैं। / The first (2) questions are fixed, so choose the remaining (2) from (10). The ways are \(\binom{10}{2}=45\).
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D. (60)
तरीके \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\) हैं। तीन अलग समूहों के चयन को गुणा करें। / The ways are \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\). Multiply selections from the three different groups.
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C. (60)
ठीक (2) हरी और (1) नीली चाहिए। तरीके \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\) हैं। / Exactly (2) green and (1) blue are needed. The ways are \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\).
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A. (35)
टूटे खिलौने को हटाने पर (7) खिलौने बचते हैं। इसलिए \(\binom{7}{3}=35\) तरीके हैं। / After removing the broken toy, (7) toys remain. Hence there are \(\binom{7}{3}=35\) ways.
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B. (10)
विशेष मिठाई पहले से चुनी है। बाकी (2) मिठाइयां (5) में से \(\binom{5}{2}=10\) तरीकों से चुनेंगे। / The special sweet is already selected. Choose the remaining (2) sweets from (5) in \(\binom{5}{2}=10\) ways.
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If \(\binom{n}{0}+\binom{n}{n}=2\), what type of identity does it show?
A. सीमा पहचानBoundary identity
क्योंकि \(\binom{n}{0}=1\) और \(\binom{n}{n}=1\) होते हैं। यह संयोजन की सीमा पहचान है। / Because \(\binom{n}{0}=1\) and \(\binom{n}{n}=1\). This is a boundary identity of combinations.
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If \(\binom{n}{1}+\binom{n}{0}=21\), what will be the value of (n)?
B. (20)
यह (n+1=21) देता है। इसलिए (n=20) होगा। / It gives (n+1=21). Therefore (n=20).
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If \(\binom{n}{2}=45\), what is the value of (n)?
C. (10)
\(\binom{10}{2}=45\) होता है। इसलिए (n=10) सही है। / \(\binom{10}{2}=45\). Therefore (n=10) is correct.
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Using Pascal's identity, \(\binom{6}{2}+\binom{6}{3}\) is equal to which expression?
C. \(\binom{7}{4}\)
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{7}{3}\) नहीं, सही \(\binom{7}{3}\) है। / By Pascal's identity, \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the correct expression is \(\binom{7}{3}\).
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What is the value of \(\binom{12}{2}+\binom{12}{10}\)?
C. (132)
\(\binom{12}{2}=66\) और \(\binom{12}{10}=66\) हैं। कुल (132) होगा। / \(\binom{12}{2}=66\) and \(\binom{12}{10}=66\). The total is (132).
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Find the value of \(\binom{16}{3}\).
C. (560)
\(\binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560\) है। छोटे (r) के लिए सीधा सूत्र उपयोगी है। / \(\binom{16}{3}=\frac{16\cdot15\cdot14}{3\cdot2\cdot1}=560\). Direct formula is useful for small (r).
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\(\binom{8}{5}\) is equal to which expression?
B. \(\binom{8}{3}\)
\(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (8-5=3) है। / \(\binom{n}{r}=\binom{n}{n-r}\). Here (8-5=3).
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What will be the value of \(\binom{15}{2}\)?
B. (105)
\(\binom{15}{2}=\frac{15\cdot14}{2}=105\) है। दो वस्तुओं के चयन में (2) से भाग दें। / \(\binom{15}{2}=\frac{15\cdot14}{2}=105\). Divide by (2) when choosing two objects.
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Find the value of \(\binom{11}{11}\).
B. (1)
सभी वस्तुएं चुनने का एक तरीका होता है। इसलिए \(\binom{11}{11}=1\) है। / There is one way to choose all objects. Hence \(\binom{11}{11}=1\).
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What is the value of \(\binom{17}{0}\)?
B. (1)
शून्य वस्तु चुनने का एक ही तरीका होता है। अतः \(\binom{17}{0}=1\) है। / There is only one way to choose zero objects. Thus \(\binom{17}{0}=1\).
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What is the value of \(\binom{18}{1}\)?
B. (18)
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{18}{1}=18\) है। / \(\binom{n}{1}=n\). Therefore \(\binom{18}{1}=18\).
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A. (63)
ठीक (1) वाणिज्य विद्यार्थी के लिए \(\binom{3}{1}\binom{7}{2}=3\cdot21=63\) तरीके हैं। ठीक शब्द पर वही संख्या लें। / For exactly (1) commerce student, the ways are \(\binom{3}{1}\binom{7}{2}=3\cdot21=63\). Use that exact number when exactly is given.
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C. (110)
कुल \(\binom{10}{3}=120\) और केवल पुरुष \(\binom{5}{3}=10\) हैं। इसलिए (120-10=110) तरीके हैं। / Total ways are \(\binom{10}{3}=120\), and all-men selections are \(\binom{5}{3}=10\). Hence (120-10=110) ways.
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C. (105)
मामले हैं (2) लड़कियां, (3) लड़कियां या (4) लड़कियां। कुल \(\binom{4}{2}\binom{6}{2}+\binom{4}{3}\binom{6}{1}+\binom{4}{4}=115\) नहीं, सही गणना (90+24+1=115) है। / The cases are (2), (3), or (4) girls. The correct count is \(\binom{4}{2}\binom{6}{2}+\binom{4}{3}\binom{6}{1}+\binom{4}{4}=115\).
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B. (224)
कुल \(\binom{10}{5}=252\) और दोनों विशेष साथ हों तो \(\binom{8}{3}=56\) तरीके हैं। इसलिए (252-56=196) नहीं बल्कि सही बचा (196) है। / Total ways are \(\binom{10}{5}=252\), and ways with both special players are \(\binom{8}{3}=56\). Thus the valid ways are (196).
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B. (21)
दो विशेष विद्यार्थी पहले से चुने हैं। बाकी (2) विद्यार्थी (7) में से \(\binom{7}{2}=21\) तरीकों से चुने जाएंगे। / The two special students are already selected. The remaining (2) students are chosen from (7) in \(\binom{7}{2}=21\) ways.
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A. (210)
एक सदस्य हटाने पर (10) सदस्य बचते हैं। इसलिए \(\binom{10}{4}=210\) तरीके हैं। / After excluding one member, (10) members remain. Therefore there are \(\binom{10}{4}=210\) ways.
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A. (330)
एक कर्मचारी पहले से चुना है इसलिए बाकी (4) कर्मचारी (11) में से चुनेंगे। तरीके \(\binom{11}{4}=330\) हैं। / One employee is already selected, so choose the remaining (4) from (11). The ways are \(\binom{11}{4}=330\).
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How many ways are there to choose exactly (4) sweets from (9) sweets?
D. (126)
ठीक (4) चुनने के तरीके \(\binom{9}{4}=126\) हैं। ठीक शब्द आने पर एक ही (r) मान लें। / The ways to choose exactly (4) are \(\binom{9}{4}=126\). When exactly is given, use only one (r) value.
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In how many ways can at most (2) options be selected from (5) options?
C. (16)
अधिकतम (2) का अर्थ (0), (1) या (2) चयन है। इसलिए \(\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16\) है। / At most (2) means selecting (0), (1), or (2). Hence \(\binom{5}{0}+\binom{5}{1}+\binom{5}{2}=16\).
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In how many ways can at least (2) colors be selected from (6) colors?
B. (57)
कुल चयन \(2^6=64\) हैं। (0) और (1) चयन हटाने पर (64-1-6=57) तरीके बचते हैं। / Total selections are \(2^6=64\). Removing (0) and (1) selections leaves (64-1-6=57) ways.
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In how many ways can at least (1) coin be selected from (8) different coins?
C. (255)
कम से कम (1) चयन के तरीके \(2^8-1=255\) हैं। खाली चयन को घटाएं। / The ways for at least (1) selection are \(2^8-1=255\). Subtract the empty selection.
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What is the total number of ways to choose any number of stickers from (7) different stickers?
B. (128)
हर स्टिकर चुना या छोड़ा जा सकता है। इसलिए कुल तरीके \(2^7=128\) हैं। / Each sticker can be selected or left. Hence the total ways are \(2^7=128\).
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How many ways are there to choose (4) cards from (16) cards?
A. (1820)
चार कार्ड चुनने के तरीके \(\binom{16}{4}=1820\) हैं। कार्डों की व्यवस्था नहीं पूछी गई है। / The ways to choose four cards are \(\binom{16}{4}=1820\). Arrangement of cards is not asked.
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In how many ways can (4) guests be called on stage from (8) guests if their order is not considered?
B. (70)
क्रम नहीं देखना है इसलिए \(\binom{8}{4}=70\) है। केवल बुलाना चयन है। / Order is not considered, so \(\binom{8}{4}=70\). Merely calling them is selection.
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