Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 121-130 of 608 questions.
What is the value of \({}^{14}P_{1}\)?
A. (14)
\({}^{n}P_{1}=n\) होता है इसलिए \({}^{14}P_{1}=14\)। एक स्थान के लिए सीधे उपलब्ध वस्तुएँ गिनें। / Since \({}^{n}P_{1}=n\), \({}^{14}P_{1}=14\). For one position, count available objects directly.
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How many distinct arrangements can be formed using all letters of (SMART)?
B. (120)
(SMART) में (5) अलग अक्षर हैं इसलिए (5!=120)। distinct letters की पूरी arrangement factorial होती है। / The word (SMART) has (5) distinct letters, so (5!=120). The full arrangement of distinct letters is factorial.
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How many ways are there to arrange (4) badges from (10) different badges in order?
A. 5040
यह क्रमचय का प्रश्न है, क्योंकि बैजों का क्रम महत्वपूर्ण है। अतः \({}^{10}P_{4}=10\times9\times8\times7=5040\)। विकल्प 210, \({}^{10}C_{4}\) है, जो केवल चयन के लिए होता है, क्रम लगाने के लिए नहीं। परीक्षा टिप: जब क्रम बदलने पर परिणाम बदलता हो, तो क्रमचय का प्रयोग करें। / This is a permutation question because the order of the badges matters. Therefore, \({}^{10}P_{4}=10\times9\times8\times7=5040\). The value 210 is \({}^{10}C_{4}\), which counts only selections, not arrangements. Exam tip: use permutations whenever changing the order changes the outcome.
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If an order of (2) letters is to be made from (A,E,I,O,U), how many ways are possible?
D. (20)
यह \({}^{5}P_{2}=5\times4=20\) है। ordered letters में (AE) और (EA) अलग माने जाते हैं। / This is \({}^{5}P_{2}=5\times4=20\). In ordered letters, (AE) and (EA) are considered different.
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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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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 (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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What is the value of \({}^{15}P_{0}\)?
B. 1
क्रमचय का सूत्र \,\({}^{n}P_r=\frac{n!}{(n-r)!}\) है। अतः \,\({}^{15}P_0=\frac{15!}{15!}=1\)। शून्य वस्तुओं को चुनकर व्यवस्थित करने का केवल एक तरीका होता है—कुछ भी न चुनना। विकल्प 0 गलत है, क्योंकि यह चयन के तरीकों की संख्या नहीं बल्कि चुनी गई वस्तुओं की संख्या दर्शाता है। परीक्षा टिप: जब \,\(r=0\) हो, तो हमेशा \,\({}^{n}P_0=1\) होता है। / Using the permutation formula, \({}^{n}P_r=\frac{n!}{(n-r)!}\). Therefore, \({}^{15}P_0=\frac{15!}{15!}=1\). There is exactly one way to select and arrange zero objects: choose nothing. Option 0 is incorrect because it represents the number of objects selected, not the number of possible arrangements. Exam tip: whenever \(r=0\), \({}^{n}P_0=1\).
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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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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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