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.
What will be the value of \({}^{8}P_{0}\)?
C. (1)
\({}^{n}P_{0}=1\) इसलिए \({}^{8}P_{0}=1\)। शून्य वस्तुओं की arrangement एक मानी जाती है। / \({}^{n}P_{0}=1\), so \({}^{8}P_{0}=1\). The arrangement of zero objects is counted as one.
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How many (2)-digit numbers can be formed from digits (0,1,2) without repetition?
B. (4)
दहाई स्थान पर (0) नहीं आ सकता इसलिए (2) विकल्प और इकाई पर (2) विकल्प हैं, कुल (4)। संख्या बनाते समय पहले स्थान पर (0) की जाँच करें। / Zero cannot be in the tens place, so there are (2) choices for tens and (2) for units, total (4). In number formation, check zero in the first place.
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How many ways are possible to place (6) different balls in a row?
A. (720)
(6) अलग गेंदों की व्यवस्था (6!=720) है। सभी वस्तुएँ अलग हों तो कुल व्यवस्था factorial होती है। / The arrangement of (6) distinct balls is (6!=720). If all objects are distinct, total arrangements are factorial.
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In how many ways can (3) people be made to stand in a row from (4) people?
D. (24)
यह \({}^{4}P_{3}=4\times3\times2=24\) है। पंक्ति में खड़ा करना ordered arrangement है। / This is \({}^{4}P_{3}=4\times3\times2=24\). Standing in a row is an ordered arrangement.
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The word (BOOK) has (4) letters and (O) occurs twice. How many distinct arrangements are possible?
C. (12)
दो समान (O) होने से व्यवस्थाएँ \(\frac{4!}{2!}=12\) होंगी। समान वस्तुओं के लिए factorial से भाग दें। / Because two (O)'s are identical, arrangements are \(\frac{4!}{2!}=12\). For identical objects, divide by their factorial.
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How many ways are there to arrange (3) cards selected from (5) different cards in order?
B. (60)
यह \({}^{5}P_{3}=5\times4\times3=60\) है। कार्डों का क्रम बदलने से व्यवस्था बदलती है। / This is \({}^{5}P_{3}=5\times4\times3=60\). Changing the order of cards changes the arrangement.
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If (n=5) and (r=2), what is the value of \({}^{n}P_{r}\)?
A. (20)
\({}^{5}P_{2}=5\times4=20\)। (n) से शुरू करके (r) तक घटते गुणनखंड लें। / \({}^{5}P_{2}=5\times4=20\). Start from (n) and take (r) decreasing factors.
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In how many ways can (2) pictures from (6) pictures be placed in first and second positions?
D. (30)
पहले स्थान के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए \(6\times5=30\)। स्थान अलग-अलग हों तो क्रम गिनें। / There are (6) choices for the first position and (5) for the second, so \(6\times5=30\). Count order when positions are distinct.
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What is the value of \({}^{10}P_{1}\)?
C. (10)
\({}^{n}P_{1}=n\) होता है इसलिए \({}^{10}P_{1}=10\)। एक पद की व्यवस्था में सीधे (n) लिखें। / Since \({}^{n}P_{1}=n\), \({}^{10}P_{1}=10\). For one position, write (n) directly.
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How many arrangements are formed from the letters of the word (SUN)?
B. (6)
(SUN) में (3) अलग अक्षर हैं इसलिए (3!=6)। छोटे शब्दों में अक्षर गिनकर factorial लगाएं। / The word (SUN) has (3) distinct letters, so (3!=6). Count the letters and apply factorial for small words.
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How many ways are there to arrange (4) pens selected from (7) different pens in order?
A. (840)
\({}^{7}P_{4}=7\times6\times5\times4=840\)। (r) स्थानों के लिए (r) घटते गुणनखंड लें। / \({}^{7}P_{4}=7\times6\times5\times4=840\). For (r) places, multiply (r) decreasing factors.
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How many (1)-digit numbers can be formed from digits (1,3,5,7,9) without repetition?
D. (5)
एक स्थान के लिए (5) विकल्प हैं इसलिए उत्तर (5) है। (1)-स्थान वाली व्यवस्था में उपलब्ध विकल्प ही उत्तर होते हैं। / There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.
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How many ordered arrangements can be made by taking (3) letters from (A,B,C,D)?
C. (24)
उत्तर \({}^{4}P_{3}=4\times3\times2=24\) है। चुनना और क्रम में रखना permutation है। / The answer is \({}^{4}P_{3}=4\times3\times2=24\). Choosing and arranging in order is permutation.
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In how many ways can (4) friends be seated on (2) special seats?
B. (12)
दो विशेष सीटों पर क्रम महत्त्वपूर्ण है इसलिए \({}^{4}P_{2}=12\)। सीटें अलग हों तो permutation लें। / Order matters on two special seats, so \({}^{4}P_{2}=12\). If seats are distinct, use permutation.
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If \({}^{n}P_{0}\) is asked, what is its value?
A. (1)
\({}^{n}P_{0}=1\) होता है क्योंकि कोई वस्तु न चुनने का एक तरीका होता है। यह standard result याद रखें। / \({}^{n}P_{0}=1\) because there is one way to choose nothing. Remember this standard result.
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In how many ways can all letters of the word (MATH) be ordered?
D. (24)
(MATH) में (4) अलग अक्षर हैं इसलिए (4!=24)। सभी अलग अक्षरों की पूरी व्यवस्था factorial से मिलती है। / The word (MATH) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.
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C. (60)
पुरस्कार अलग हैं और विद्यार्थी दोहराए नहीं जा सकते इसलिए \({}^{5}P_{3}=60\)। अलग पुरस्कारों के वितरण में क्रम महत्त्वपूर्ण होता है। / The prizes are distinct and students cannot repeat, so \({}^{5}P_{3}=60\). Distribution of distinct prizes makes order important.
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What is the value of \({}^{6}P_{3}\)?
B. (120)
\({}^{6}P_{3}=6\times5\times4=120\) है। (r=3) हो तो तीन घटते गुणनखंड लें। / \({}^{6}P_{3}=6\times5\times4=120\). When (r=3), take three decreasing factors.
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In a race with (7) runners, in how many ways can first and second places occur?
D. (42)
प्रथम और द्वितीय स्थान क्रम वाले हैं इसलिए \({}^{7}P_{2}=42\) होगा। रैंकिंग में permutation लगता है। / First and second places are ordered, so \({}^{7}P_{2}=42\). Ranking questions use permutation.
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How many ways are there to arrange (2) objects selected from (9) objects in order?
C. (72)
उत्तर \({}^{9}P_{2}=9\times8=72\) है। पहले स्थान के बाद दूसरा विकल्प एक कम हो जाता है। / The answer is \({}^{9}P_{2}=9\times8=72\). After the first place, the second choice decreases by one.
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In how many ways can the letters of the word (DOG) be arranged?
B. (6)
(DOG) में (3) अलग अक्षर हैं इसलिए (3!=6) व्यवस्थाएँ हैं। अक्षर अलग हों तो सीधे factorial प्रयोग करें। / The word (DOG) has (3) distinct letters, so there are (3!=6) arrangements. Use factorial when letters are distinct.
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What will be the value of \({}^{5}P_{1}\)?
A. (5)
\({}^{n}P_{1}=n\) होता है इसलिए \({}^{5}P_{1}=5\)। एक वस्तु चुनकर व्यवस्थित करने में (n) तरीके होते हैं। / Since \({}^{n}P_{1}=n\), \({}^{5}P_{1}=5\). Choosing and arranging one object gives (n) ways.
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In how many ways can (5) different people be seated on (5) chairs?
D. (120)
(5) लोगों को (5) स्थानों पर बैठाने के तरीके (5!=120) हैं। seating in a row में factorial याद रखें। / The number of ways to seat (5) people in (5) places is (5!=120). Remember factorial for seating in a row.
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How many (3)-digit numbers can be formed from digits (1,2,3) without repetition?
C. (6)
तीनों अलग अंकों की व्यवस्था (3!=6) है। बिना पुनरावृत्ति में हर स्थान के विकल्प घटते हैं। / The arrangement of three distinct digits is (3!=6). Without repetition, choices decrease place by place.
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In how many ways can a captain and vice-captain be chosen from (8) players?
B. (56)
कप्तान और उपकप्तान अलग पद हैं इसलिए \({}^{8}P_{2}=56\) होगा। जब जिम्मेदारी अलग हो तो क्रम महत्त्वपूर्ण होता है। / Captain and vice-captain are distinct posts, so \({}^{8}P_{2}=56\). When roles differ, order matters.
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What is the value of \({}^{7}P_{2}\)?
A. (42)
\({}^{7}P_{2}=7\times6=42\) होता है। छोटे (r) के लिए घटते हुए (r) गुणनखंड लें। / \({}^{7}P_{2}=7\times6=42\). For small (r), multiply (r) decreasing factors.
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In how many ways can (4) different coloured flags be placed in a straight line?
D. (24)
(4) अलग झंडों की व्यवस्था (4!=24) होती है। सीधी रेखा में लगाना linear permutation है। / The arrangement of (4) distinct flags is (4!=24). Placing in a straight line is a linear permutation.
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How many different words can be formed using all letters of the word (CAT)?
C. (6)
(CAT) में (3) अलग अक्षर हैं इसलिए (3!=6) शब्द बनेंगे। छोटे शब्दों में सभी अक्षरों की व्यवस्था सीधे factorial से करें। / The word (CAT) has (3) distinct letters, so (3!=6) words are possible. For small words, arrange all letters using factorial.
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B. (120)
यहाँ पद अलग हैं इसलिए क्रम महत्त्वपूर्ण है और उत्तर \({}^{6}P_{3}=120\) है। परीक्षा में पद अलग हों तो permutation लें। / Here the posts are different, so order matters and the answer is \({}^{6}P_{3}=120\). In exams, use permutation when posts are distinct.
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If (5) different books are arranged in a row, how many ways are possible?
A. (120)
(5) अलग वस्तुओं की व्यवस्था (5!) होती है। परीक्षा में अलग-अलग वस्तुओं के लिए factorial लगाएं। / The arrangement of (5) distinct objects is (5!). In exams, use factorial for distinct objects.
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