permutations se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
967 questions tagged with permutations.
B. (30)
व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। repeated letters के factorial से भाग देना जरूरी है। / The arrangements are \(\frac{5!}{2!2!}=30\). It is necessary to divide by factorials of repeated letters.
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How many ways are there to arrange (3) charts from (11) different charts in order?
A. (990)
\({}^{11}P_{3}=11\times10\times9=990\) है। क्रम वाले चयन में \({}^{n}P_{r}\) लगाएं। / \({}^{11}P_{3}=11\times10\times9=990\). Use \({}^{n}P_{r}\) in ordered selection.
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How many (2)-digit even numbers can be formed from digits (1,2,3,4,6) without repetition?
D. (12)
इकाई स्थान पर (2,4,6) में से (3) विकल्प और दहाई पर (4) विकल्प हैं, कुल (12)। even number में इकाई स्थान पहले देखें। / The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.
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In how many ways can (7) different postcards be arranged from left to right?
C. (5040)
बाएँ से दाएँ (7) अलग पोस्टकार्डों की व्यवस्था (7!=5040) है। linear arrangement में factorial लें। / The left-to-right arrangement of (7) distinct postcards is (7!=5040). Use factorial in linear arrangement.
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What will be the value of \({}^{12}P_{0}+{}^{12}P_{1}\)?
B. (13)
\({}^{12}P_{0}=1\) और \({}^{12}P_{1}=12\), इसलिए योग (13) है। standard values याद रखें। / \({}^{12}P_{0}=1\) and \({}^{12}P_{1}=12\), so the sum is (13). Remember standard values.
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A. (30)
व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। समान अक्षरों को अलग-अलग न गिनें। / The arrangements are \(\frac{5!}{2!2!}=30\). Do not count identical letters as different.
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In how many ways can first, second and third performance order be chosen from (9) singers?
D. (504)
प्रदर्शन क्रम महत्त्वपूर्ण है इसलिए \({}^{9}P_{3}=9\times8\times7=504\)। order बदलने से कार्यक्रम बदल जाता है। / Performance order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Changing order changes the programme.
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How many (4)-digit numbers can be formed from digits (4,5,6,7,8,9) without repetition?
C. (360)
चार स्थानों के लिए \(6\times5\times4\times3=360\) तरीके हैं। बिना पुनरावृत्ति में विकल्प घटते हैं। / For four places, there are \(6\times5\times4\times3=360\) ways. Without repetition, choices decrease.
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In how many ways can (5) different bells be placed at (5) fixed positions?
B. (120)
(5) अलग घंटियाँ (5) स्थानों पर (5!=120) तरीकों से लगेंगी। निर्धारित स्थानों की arrangement factorial से होती है। / (5) different bells can be placed at (5) positions in (5!=120) ways. Arrangements at fixed positions use factorial.
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How many distinct words can be formed using all letters of (MOUSE)?
A. (120)
(MOUSE) में (5) अलग अक्षर हैं इसलिए (5!=120)। अलग अक्षरों में factorial ही कुल व्यवस्था देता है। / The word (MOUSE) has (5) distinct letters, so (5!=120). For distinct letters, factorial gives total arrangements.
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D. (72)
पहले स्लॉट के लिए (9) और दूसरे के लिए (8) विकल्प हैं इसलिए (72)। स्लॉट अलग हों तो permutation गिनें। / There are (9) choices for the first slot and (8) for the second, so (72). Count permutation when slots are distinct.
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What is the value of \({}^{5}P_{0}+{}^{5}P_{2}\)?
C. (21)
\({}^{5}P_{0}=1\) और \({}^{5}P_{2}=20\), इसलिए योग (21) है। पहले अलग-अलग values निकालें। / \({}^{5}P_{0}=1\) and \({}^{5}P_{2}=20\), so the sum is (21). Find the values separately first.
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How many ways are there to arrange (6) different beads in a straight embroidery line?
B. (720)
सीधी रेखा में (6) अलग मोतियों की व्यवस्था (6!=720) है। इसे circular arrangement न मानें। / In a straight line, the arrangement of (6) distinct beads is (6!=720). Do not treat it as a circular arrangement.
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How many (2)-digit numbers can be formed from digits (3,6,9) if repetition is allowed?
A. (9)
दोनों स्थानों पर (3) विकल्प हैं इसलिए \(3^2=9\)। repetition allowed हो तो विकल्प वही रहते हैं। / There are (3) choices at both places, so \(3^2=9\). With repetition allowed, choices remain the same.
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In how many ways can first, second and third ranks be given among (15) students?
D. (2730)
तीन rank क्रम वाले हैं इसलिए \({}^{15}P_{3}=15\times14\times13=2730\)। rank वाले प्रश्न permutation के होते हैं। / The three ranks are ordered, so \({}^{15}P_{3}=15\times14\times13=2730\). Rank questions use permutation.
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How many distinct arrangements are possible for the word (COCO)?
C. (6)
(C) दो बार और (O) दो बार हैं इसलिए \(\frac{4!}{2!2!}=6\)। दो repeated letters हों तो दोनों से भाग दें। / (C) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated letters.
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How many (2)-letter codes can be formed from (P,Q,R,S,T,U) without repetition?
B. (30)
\({}^{6}P_{2}=6\times5=30\) कोड बनेंगे। कोड में अक्षर का स्थान बदलने से कोड बदलता है। / \({}^{6}P_{2}=6\times5=30\) codes are possible. In a code, changing the position of letters changes the code.
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How many ways are there to place (2) plates from (8) different plates on the upper and lower racks?
A. (56)
ऊपरी और निचला रैक अलग हैं इसलिए \({}^{8}P_{2}=56\)। अलग स्थानों में order गिना जाता है। / The upper and lower racks are distinct, so \({}^{8}P_{2}=56\). Order is counted for distinct positions.
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What is the value of \({}^{6}P_{5}\)?
D. (720)
\({}^{6}P_{5}=6\times5\times4\times3\times2=720\) है। पाँच स्थान हों तो पाँच घटते गुणनखंड लें। / \({}^{6}P_{5}=6\times5\times4\times3\times2=720\). For five places, take five decreasing factors.
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How many arrangements of the letters of (CODE) are possible if the last letter is (E)?
C. (6)
अंतिम स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed position को पहले अलग करें। / The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate the fixed position first.
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In how many ways can first and second speakers be chosen from (14) participants?
B. (182)
पहला और दूसरा वक्ता अलग क्रम वाले स्थान हैं इसलिए \({}^{14}P_{2}=182\)। speaking order में permutation लगता है। / First and second speakers are ordered positions, so \({}^{14}P_{2}=182\). Speaking order uses permutation.
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In how many ways can (4) different direction cards be placed from top to bottom?
A. (24)
ऊपर से नीचे (4) अलग कार्डों की व्यवस्था (4!=24) है। linear order में factorial प्रयोग करें। / The top-to-bottom arrangement of (4) distinct cards is (4!=24). Use factorial in linear order.
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How many (3)-digit numbers can be formed from digits (0,1,5,7) without repetition?
D. (18)
सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प मिलते हैं, कुल (18)। पहले स्थान पर (0) की अनुमति नहीं है। / There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Zero is not allowed in the first place.
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In how many ways can a starting station and ending station be chosen from (6) different stations?
C. (30)
शुरू और समाप्ति क्रम वाले हैं इसलिए \({}^{6}P_{2}=30\)। route में starting और ending बदलने से मामला बदलता है। / Starting and ending are ordered, so \({}^{6}P_{2}=30\). In a route, changing start and end changes the case.
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How many distinct letter arrangements are possible for the word (BANANA)?
B. (60)
(A) तीन बार और (N) दो बार है इसलिए \(\frac{6!}{3!2!}=60\)। repeated groups के factorial से भाग दें। / (A) occurs three times and (N) twice, so \(\frac{6!}{3!2!}=60\). Divide by factorials of repeated groups.
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In how many ways can (5) different stamps be pasted on (5) different envelopes?
A. (120)
(5) अलग टिकट (5) अलग लिफाफों पर (5!=120) तरीकों से लगेंगे। one-to-one matching में factorial उपयोगी है। / (5) different stamps can be pasted on (5) different envelopes in (5!=120) ways. Factorial is useful in one-to-one matching.
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What is the value of \({}^{15}P_{0}\)?
D. (1)
\({}^{n}P_{0}=1\) इसलिए \({}^{15}P_{0}=1\)। शून्य वस्तु चुनने का एक तरीका माना जाता है। / \({}^{n}P_{0}=1\), so \({}^{15}P_{0}=1\). Choosing zero objects is counted in one way.
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How many (3)-digit odd numbers can be formed from digits (2,4,6,8,9) without repetition?
C. (12)
इकाई स्थान पर केवल (9) आएगा और बाकी दो स्थान \(4\times3=12\) तरीकों से भरेंगे। odd number में इकाई स्थान पहले देखें। / Only (9) can be in the units place, and the other two places can be filled in \(4\times3=12\) ways. For odd numbers, check the units place first.
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In how many ways can (5) children from (8) children be seated on a bench in order?
B. (6720)
उत्तर \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\) है। बेंच पर बाएँ-दाएँ क्रम महत्त्वपूर्ण होता है। / The answer is \({}^{8}P_{5}=8\times7\times6\times5\times4=6720\). Left-right order on a bench matters.
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How many distinct arrangements are there for the word (DAD)?
A. (3)
(D) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements हैं। repeated letters को अलग-अलग न गिनें। / (D) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Do not count repeated letters as distinct.
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