permutation_connection se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
12 questions tagged with permutation_connection.
If \(^{n}C_r=\frac{^{n}P_r}{5040}\), what is the value of (r)?
B. (7)
\(^{n}C_r=\frac{^{n}P_r}{r!}\) और (7!=5040) है। परीक्षा में divisor को (r!) से मिलाएँ। / \(^{n}C_r=\frac{^{n}P_r}{r!}\), and (7!=5040). In exams match the divisor with (r!).
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Which relation correctly connects \(^{14}C_{11}\) with \(^{14}P_3\)?
B. \(^{14}C_{11}=\frac{^{14}P_3}{3!}\)
\(^{14}C_{11}=^{14}C_3\) और \(^{14}C_3=\frac{^{14}P_3}{3!}\) है। परीक्षा में पहले complement से छोटा index लें। / \(^{14}C_{11}=^{14}C_3\), and \(^{14}C_3=\frac{^{14}P_3}{3!}\). In exams first use the complement to get a smaller index.
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If \(^{n}C_r=\frac{^{n}P_r}{120}\), what is the value of (r)?
B. (5)
\(^{n}C_r=\frac{^{n}P_r}{r!}\) और (5!=120) है। परीक्षा में divisor को (r!) से मिलाएँ। / \(^{n}C_r=\frac{^{n}P_r}{r!}\) and (5!=120). In exams match the divisor with (r!).
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How can \(^{11}C_8\) be connected with \(^{11}P_3\)?
A. \(^{11}C_8=\frac{^{11}P_3}{3!}\)
\(^{11}C_8=^{11}C_3\) और \(^{11}C_3=\frac{^{11}P_3}{3!}\) है। परीक्षा में पहले complement से छोटा index लें। / \(^{11}C_8=^{11}C_3\) and \(^{11}C_3=\frac{^{11}P_3}{3!}\). In exams first use complement to get a smaller index.
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By which formula connection can \(^{10}P_3\div3!\) be directly called \(^{10}C_3\)?
A. \(^{n}C_r=\frac{^{n}P_r}{r!}\)
Combination permutation में गिने गए (r!) internal orders को हटाकर मिलता है। परीक्षा में ordered से unordered जाने पर divide करें। / Combination is obtained by removing (r!) internal orders counted in permutation. In exams divide when going from ordered to unordered.
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To obtain \(^{8}C_3\) from \(^{8}P_3\), by what should we divide?
A. (3!)
तीन चुनी वस्तुओं के (3!) orders permutation में extra गिने जाते हैं। परीक्षा में combination पाने के लिए (r!) से भाग दें। / The (3!) orders of three chosen objects are counted extra in permutation. In exams divide by (r!) to get combination.
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If (r) objects are first selected and then ordered which expression gives the total arrangement?
C. \(^{n}C_r\times r!\)
पहले चयन \(^{n}C_r\) और फिर हर चयन की (r!) व्यवस्थाएँ होती हैं। परीक्षा में choose then arrange को multiply करें। / First there are \(^{n}C_r\) selections and then (r!) arrangements for each selection. In exams multiply for choose then arrange.
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To obtain \(^{n}C_4\) from \(^{n}P_4\) by what should we divide?
A. (4!)
चार चुनी वस्तुओं के (4!) internal orders हटाने होते हैं। परीक्षा में \(^{n}C_r=\frac{^{n}P_r}{r!}\) लगाएँ। / The (4!) internal orders of four selected objects must be removed. In exams apply \(^{n}C_r=\frac{^{n}P_r}{r!}\).
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Which relation obtains \(^{n}C_3\) from \(^{n}P_3\)?
C. \(^{n}C_3=\frac{^{n}P_3}{3!}\)
Permutation में (3!) आंतरिक क्रम extra गिने जाते हैं। परीक्षा में combination पाने के लिए इन क्रमों से भाग दें। / Permutation counts (3!) internal orders extra. In exams divide by these orders to get combination.
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Which expression shows \(^{n}P_r\) as selection first and arrangement later?
C. \(^{n}C_r\times r!\)
पहले (r) वस्तुएँ चुनने के \(^{n}C_r\) तरीके हैं और फिर (r!) व्यवस्थाएँ हैं। परीक्षा में ordered selection को choose and arrange लिखें। / There are \(^{n}C_r\) ways to choose (r) objects and then (r!) arrangements. In exams write ordered selection as choose and arrange.
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How can the value of \(^{8}C_2\) be obtained from \(^{8}P_2\)?
B. \(^{8}C_2=\frac{^{8}P_2}{2!}\)
Combination पाने के लिए ordered arrangements को (2!) से भाग दिया जाता है। परीक्षा में internal orders हटाने के लिए भाग दें। / To get combination ordered arrangements are divided by (2!). In exams divide to remove internal orders.
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Why do we divide by (r!) in \(^{n}C_r=\frac{^{n}P_r}{r!}\)?
D. क्योंकि क्रम को हटाना होता हैBecause order has to be removed
Combination में क्रम नहीं गिना जाता इसलिए चुनी गई (r) वस्तुओं के (r!) क्रम हटते हैं। परीक्षा में combination को बिना क्रम का चयन मानें। / In combination order is not counted so (r!) orders of selected objects are removed. In exams treat combination as selection without order.
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