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.
A. (42)
पहले डिब्बे के लिए (7) और दूसरे के लिए (6) विकल्प हैं इसलिए (42)। अलग डिब्बों में क्रमचय गिना जाता है। / There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.
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What is the value of \({}^{4}P_{0}+{}^{4}P_{1}\)?
D. (5)
\({}^{4}P_{0}=1\) और \({}^{4}P_{1}=4\), इसलिए योग (5) है। standard values याद रखें। / \({}^{4}P_{0}=1\) and \({}^{4}P_{1}=4\), so the sum is (5). Remember standard values.
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How many ways are there to arrange (5) different beads on a straight string?
C. (120)
सीधी डोरी पर (5) अलग मोतियों की व्यवस्था (5!=120) है। इसे circular arrangement न मानें। / On a straight string, the arrangement of (5) distinct beads is (5!=120). Do not treat it as a circular arrangement.
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How many (3)-digit numbers can be formed from digits (4,5,6) if repetition is allowed?
B. (27)
तीनों स्थानों पर (3) विकल्प हैं इसलिए \(3^3=27\)। repetition allowed हो तो हर स्थान पर समान विकल्प रहते हैं। / There are (3) choices for each of the three places, so \(3^3=27\). With repetition allowed, choices remain the same at every place.
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How many ways are there to choose the first speaker and second speaker from (12) persons?
A. (132)
पहला और दूसरा वक्ता क्रम वाले पद हैं इसलिए \({}^{12}P_{2}=132\)। बोलने के क्रम में permutation लगता है। / First and second speakers are ordered positions, so \({}^{12}P_{2}=132\). Speaking order uses permutation.
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How many distinct arrangements are possible for the word (EYE)?
D. (3)
(E) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements होंगे। repeated letters को अलग-अलग न गिनें। / (E) occurs twice, so \(\frac{3!}{2!}=3\) arrangements are possible. Do not count repeated letters as distinct.
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How many (3)-letter codes can be formed from letters (K,L,M,N) without repetition?
C. (24)
\({}^{4}P_{3}=4\times3\times2=24\) कोड बनेंगे। कोड में अक्षरों का स्थान बदलने से कोड बदल जाता है। / \({}^{4}P_{3}=4\times3\times2=24\) codes are possible. In a code, changing positions changes the code.
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How many ways are there to place (2) books from (6) different books in right and left positions?
B. (30)
दायाँ और बायाँ स्थान अलग हैं इसलिए \({}^{6}P_{2}=30\)। स्थानों की पहचान अलग हो तो क्रम गिना जाता है। / Right and left positions are distinct, so \({}^{6}P_{2}=30\). When positions are identified, order is counted.
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What is the value of \({}^{5}P_{4}\)?
A. (120)
\({}^{5}P_{4}=5\times4\times3\times2=120\) है। (r=4) हो तो (4) घटते गुणनखंड लें। / \({}^{5}P_{4}=5\times4\times3\times2=120\). When (r=4), take (4) decreasing factors.
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How many arrangements of the letters of (TEAM) are possible if the last letter is (M)?
D. (6)
अंतिम स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed condition को पहले लागू करें। / The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.
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In how many ways can a president and secretary be chosen from (11) candidates?
C. (110)
दो पद अलग हैं इसलिए \({}^{11}P_{2}=11\times10=110\)। अलग पदों के लिए क्रमचय लगाएं। / The two posts are distinct, so \({}^{11}P_{2}=11\times10=110\). Use permutation for distinct posts.
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In how many ways can (3) distinct red, blue and green flags be placed on a pole from top to bottom?
B. (6)
(3) अलग झंडों की क्रमबद्ध व्यवस्था (3!=6) है। ऊपर-नीचे का क्रम बदलने से arrangement बदलती है। / The ordered arrangement of (3) distinct flags is (3!=6). Changing top-bottom order changes the arrangement.
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How many (3)-digit numbers can be formed from digits (0,1,2,3) without repetition?
A. (18)
सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प हैं, कुल (18)। पहले स्थान पर (0) न रखें। / There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Do not put (0) in the first place.
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How many ways are there to choose an ordered trip of (2) cities from (5) different cities?
D. (20)
पहले शहर के लिए (5) और दूसरे के लिए (4) विकल्प हैं इसलिए (20)। यात्रा क्रम बदलने से योजना बदलती है। / There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.
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The word (SCHOOL) has (O) twice. How many distinct letter arrangements are possible?
C. (360)
कुल (6) अक्षर हैं और (O) दो समान हैं इसलिए \(\frac{6!}{2!}=360\)। repeated letter के लिए denominator लगाएं। / There are (6) letters and two identical (O)'s, so \(\frac{6!}{2!}=360\). Use a denominator for a repeated letter.
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In how many ways can (4) different labels be pasted on (4) different boxes?
B. (24)
(4) अलग लेबल (4) अलग डिब्बों पर (4!=24) तरीकों से लगेंगे। matching arrangements में factorial काम आता है। / (4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.
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What is the value of \({}^{7}P_{0}\)?
A. (1)
\({}^{n}P_{0}=1\) होता है इसलिए \({}^{7}P_{0}=1\)। शून्य वस्तु चुनने का एक ही तरीका माना जाता है। / \({}^{n}P_{0}=1\), so \({}^{7}P_{0}=1\). Choosing zero objects is counted in one way.
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How many (3)-digit numbers can be formed from digits (1,4,7,8,9) without repetition?
D. (60)
तीन स्थानों के लिए \(5\times4\times3=60\) तरीके हैं। बिना पुनरावृत्ति में हर अगला विकल्प कम होता है। / For three places, there are \(5\times4\times3=60\) ways. Without repetition, each next choice decreases.
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In how many ways can (4) children from (6) children be made to stand in the front row?
C. (360)
उत्तर \({}^{6}P_{4}=6\times5\times4\times3=360\) है। पंक्ति में क्रम महत्त्वपूर्ण होता है। / The answer is \({}^{6}P_{4}=6\times5\times4\times3=360\). Order matters in a row.
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How many distinct arrangements are there for the word (MOM)?
A. (3)
(M) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements हैं। समान अक्षर होने पर कुल arrangements घट जाती हैं। / (M) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Repeated letters reduce total arrangements.
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If an order of (3) letters is to be made from (A,B,C,D,E), how many total ways are possible?
B. (60)
\({}^{5}P_{3}=5\times4\times3=60\) तरीके होंगे। अक्षरों का क्रम बदलने से arrangement बदलती है। / There will be \({}^{5}P_{3}=5\times4\times3=60\) ways. Changing the order of letters changes the arrangement.
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How many ways are there to arrange (4) notebooks from (8) different notebooks on a table in order?
A. (1680)
यह \({}^{8}P_{4}=8\times7\times6\times5=1680\) है। (4) स्थानों के लिए (4) घटते गुणनखंड लें। / This is \({}^{8}P_{4}=8\times7\times6\times5=1680\). Take (4) decreasing factors for (4) positions.
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How many distinct words can be formed using all letters of (DELHI)?
B. (120)
(DELHI) में (5) अलग अक्षर हैं इसलिए (5!=120)। अलग अक्षरों की संख्या ही factorial का आधार होती है। / The word (DELHI) has (5) distinct letters, so (5!=120). The number of distinct letters is the base of the factorial.
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What is the value of \({}^{9}P_{1}\)?
A. (9)
\({}^{n}P_{1}=n\) होता है इसलिए \({}^{9}P_{1}=9\)। एक स्थान के लिए सीधे उपलब्ध वस्तुएँ गिनें। / Since \({}^{n}P_{1}=n\), \({}^{9}P_{1}=9\). For one position, count the available objects directly.
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D. (24)
पहला स्थान निश्चित है और शेष (4) बैग (4!=24) तरीकों से लगेंगे। fixed item के बाद बचे items की व्यवस्था करें। / The first position is fixed and the remaining (4) bags can be arranged in (4!=24) ways. Arrange the remaining items after fixing the item.
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How many (2)-digit numbers can be formed from digits (2,5,8) if repetition is allowed?
C. (9)
दोनों स्थानों पर (3) विकल्प हैं इसलिए \(3\times3=9\)। पुनरावृत्ति हो तो विकल्प कम नहीं होते। / There are (3) choices for each place, so \(3\times3=9\). With repetition, choices do not decrease.
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In how many ways can a captain, vice-captain and wicketkeeper be chosen from (10) players?
B. (720)
तीन पद अलग हैं इसलिए \({}^{10}P_{3}=10\times9\times8=720\)। अलग पदों में order महत्त्वपूर्ण होता है। / The three posts are distinct, so \({}^{10}P_{3}=10\times9\times8=720\). Order matters for different posts.
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How many arrangements of the letters of (NOTE) are possible if the first letter is (N)?
A. (6)
पहला स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed position को पहले अलग करें। / The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate fixed positions first.
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How many ways are there to place (3) colours from (7) colours in top, middle and bottom positions?
D. (210)
तीनों स्थान अलग हैं इसलिए \({}^{7}P_{3}=7\times6\times5=210\)। स्थानों के नाम अलग हों तो क्रमचय लें। / The three positions are distinct, so \({}^{7}P_{3}=7\times6\times5=210\). Use permutation when positions have different names.
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In how many ways can (4) different flowers be placed one each in (4) vases?
C. (24)
(4) अलग फूल (4) अलग स्थानों पर (4!=24) तरीकों से रखे जा सकते हैं। एक-एक वस्तु रखने में व्यवस्था गिनी जाती है। / (4) distinct flowers can be placed in (4) distinct positions in (4!=24) ways. Placing one each counts arrangements.
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