factorial_notation se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
348 questions tagged with factorial_notation.
What is the simplified form of (\frac{(n+3)!}{(n+1)!}-\frac{(n+2)!}{n!})?
A. (2(n+2))
पहला पद ((n+3)(n+2)) और दूसरा ((n+2)(n+1)) है। अंतर (2(n+2)) है। / The first term is ((n+3)(n+2)) and the second is ((n+2)(n+1)). The difference is (2(n+2)).
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What is the value of \(\frac{16!}{14!\times 8}\)?
B. (30)
\(\frac{16!}{14!\times8}=\frac{16\times15}{8}=30\) है। पहले (16) को (8) से काटें। / \(\frac{16!}{14!\times8}=\frac{16\times15}{8}=30\). First cancel (16) with (8).
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Which value is equal to \(\frac{25!}{23!}\)?
B. (600)
\(\frac{25!}{23!}=25\times24=600\) होता है। केवल बचे हुए दो पदों को गुणा करें। / \(\frac{25!}{23!}=25\times24=600\). Multiply only the two remaining terms.
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What is the highest power of (7) in (80!)?
B. (12)
घात \( \left\lfloor\frac{80}{7}\right\rfloor+\left\lfloor\frac{80}{49}\right\rfloor=11+1=12\) है। अभाज्य की घातों तक ही जोड़ें। / The exponent is \(\left\lfloor\frac{80}{7}\right\rfloor+\left\lfloor\frac{80}{49}\right\rfloor=11+1=12\). Add only up to powers of the prime.
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What is the highest power of (5) in (100!)?
C. (24)
घात \( \left\lfloor\frac{100}{5}\right\rfloor+\left\lfloor\frac{100}{25}\right\rfloor=20+4=24\) है। उच्च घात के लिए भागफल जोड़ें। / The exponent is \(\left\lfloor\frac{100}{5}\right\rfloor+\left\lfloor\frac{100}{25}\right\rfloor=20+4=24\). For highest power add the quotients.
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How many trailing zeros are there in (50!)?
C. (12)
शून्य की संख्या \(\left\lfloor\frac{50}{5}\right\rfloor+\left\lfloor\frac{50}{25}\right\rfloor=10+2=12\) है। (5) की घात गिनना सबसे तेज तरीका है। / The number of zeros is \(\left\lfloor\frac{50}{5}\right\rfloor+\left\lfloor\frac{50}{25}\right\rfloor=10+2=12\). Counting powers of (5) is the fastest method.
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What is the least (n) for which (n!) is divisible by (125)?
C. 15
यहाँ \(125=5^3\) है, इसलिए \(n!\) में कम-से-कम तीन गुणनखंड \(5\) होने चाहिए। \(14!\) में केवल \(5\) और \(10\) से दो गुणनखंड \(5\) मिलते हैं, इसलिए वह \(125\) से विभाज्य नहीं है। \(15!\) में \(5,10,15\) से तीन गुणनखंड \(5\) मिलते हैं; अतः \(15!\), \(5^3=125\) से विभाज्य है। परीक्षा टिप: \(p^k\) से विभाज्यता के लिए \(n!\) में अभाज्य \(p\) की घात गिनें। / Since \(125=5^3\), \(n!\) must contain at least three factors of \(5\). In \(14!\), only \(5\) and \(10\) contribute factors of \(5\), so it has only two such factors and is not divisible by \(125\). In \(15!\), the terms \(5,10,15\) provide three factors of \(5\); hence \(15!\) is divisible by \(5^3=125\). Exam tip: for divisibility by \(p^k\), count the exponent of the prime \(p\) in \(n!\).
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What is the least (n) for which (n!) is divisible by (72)?
B. (6)
\(72=2^3\times3^2\) है और (6!) में ये सभी गुणनखंड हैं। ऐसे प्रश्नों में अभाज्य गुणनखंड पहले लिखें। / \(72=2^3\times3^2\) and (6!) contains all these factors. In such questions write prime factors first.
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If (\frac{(n+4)!}{(n+2)!}=156) then what is the value of (n)?
B. (9)
((n+4)(n+3)=156) मिलता है। \(13\times12=156\) इसलिए (n=9) है। / We get ((n+4)(n+3)=156). Since \(13\times12=156\), (n=9).
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If (\frac{n!}{(n-2)!}=132) then what is the value of (n)?
C. (12)
समीकरण (n(n-1)=132) बनता है। \(12\times11=132\) इसलिए (n=12) है। / The equation becomes (n(n-1)=132). Since \(12\times11=132\), (n=12).
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What is the value of \(\frac{14!+13!}{12!}\)?
B. (195)
(14!+13!=13!(14+1)) है। (12!) से भाग देने पर \(13\times15=195\) मिलता है। / (14!+13!=13!(14+1)). Dividing by (12!) gives \(13\times15=195\).
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What is the value of \(\frac{10!+9!}{8!}\)?
A. (99)
(10!+9!=9!(10+1)) है। (8!) से भाग देने पर \(9\times11=99\) मिलता है। / (10!+9!=9!(10+1)). Dividing by (8!) gives \(9\times11=99\).
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What is the value of \(\frac{20!}{18!\times 19}\)?
C. (20)
\(\frac{20!}{18!\times19}=\frac{20\times19}{19}=20\) है। कटौती के बाद बचा हुआ गुणनखंड देखें। / \(\frac{20!}{18!\times19}=\frac{20\times19}{19}=20\). Look at the remaining factor after cancellation.
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If (\frac{(n+3)!}{(n+1)!}=110) then what will be the value of (n)?
B. (8)
((n+3)(n+2)=110) है। \(11\times10=110\) से (n=8) मिलता है। / ((n+3)(n+2)=110). From \(11\times10=110\), we get (n=8).
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If ((n+1)! = 72(n-1)!) then what is the value of (n)?
B. (8)
काटने पर (n(n+1)=72) मिलता है। \(8\times9=72\) इसलिए (n=8) है। / After cancellation (n(n+1)=72). Since \(8\times9=72\), (n=8).
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If (\frac{n!}{(n-3)!}=210) then what is the value of (n)?
C. (7)
यह (n(n-1)(n-2)=210) देता है और \(7\times6\times5=210\) है। इसलिए (n=7) है। / This gives (n(n-1)(n-2)=210) and \(7\times6\times5=210\). Therefore (n=7).
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What is the value of \(\frac{15!}{13!\times 3}\)?
C. (70)
\(\frac{15!}{13!\times3}=\frac{15\times14}{3}=70\) है। पहले समान भाज्य हटाने से गलती कम होती है। / \(\frac{15!}{13!\times3}=\frac{15\times14}{3}=70\). Removing common factorials first reduces mistakes.
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What is the value of \(\frac{12!}{10!\times 2!}\)?
C. (66)
\(\frac{12!}{10!\times 2!}=\frac{12\times11}{2}=66\) होता है। ऐसे प्रश्नों में पूरा गुणनफल न फैलाएं। / \(\frac{12!}{10!\times 2!}=\frac{12\times11}{2}=66\). Do not expand the whole product in such questions.
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What is the value of \(\frac{9!}{7!}+\frac{8!}{6!}\)?
A. (128)
\(\frac{9!}{7!}=72\) और \(\frac{8!}{6!}=56\) इसलिए योग (128) है। परीक्षा में बड़े गुणनखंडों को तुरंत काटें। / Here \(\frac{9!}{7!}=72\) and \(\frac{8!}{6!}=56\) so the sum is (128). In exams cancel larger factorials quickly.
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If (\frac{(n+2)!}{n!}=56) then what is the positive value of (n)?
B. (6)
सरलीकरण से ((n+2)(n+1)=56) मिलता है। अतः (n=6) सही है। / Simplification gives ((n+2)(n+1)=56). Hence (n=6) is correct.
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If (n>3) and (\frac{(n+1)!}{(n-2)!}=120) then what is the value of (n)?
A. (4)
हर को काटने पर (n(n-1)(n+1)=120) मिलता है और (n=4) संतुष्ट करता है। परीक्षा में पहले क्रमागत गुणनखंड लिखें। / After cancellation we get (n(n-1)(n+1)=120) and (n=4) satisfies it. In exams write consecutive factors first.
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What is the least natural (k) for which (k!) is divisible by (72)?
C. (6)
\(72=2^3\cdot3^2\) है और (6!) में ये गुणनखंड मिल जाते हैं। ऐसे प्रश्नों में संख्या का अभाज्य गुणनखंडन करें। / \(72=2^3\cdot3^2\), and (6!) contains these factors. In such questions, use prime factorization of the number.
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What is the value of \(\frac{8!}{4!\cdot2!\cdot2!}\)?
A. (420)
\(\frac{8!}{4!\cdot2!\cdot2!}=\frac{8\cdot7\cdot6\cdot5}{4}=420\) होता है। पहले (4!) काटकर छोटे पदों में गणना करें। / \(\frac{8!}{4!\cdot2!\cdot2!}=\frac{8\cdot7\cdot6\cdot5}{4}=420\). Cancel (4!) first and calculate with smaller factors.
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If (\frac{(m+2)!}{m!}=90), what is the value of (m)?
C. (8)
(\frac{(m+2)!}{m!}=(m+2)(m+1)), इसलिए \(10\cdot9=90\) से (m=8)। लगातार संख्याओं का गुणनफल पहचानें। / (\frac{(m+2)!}{m!}=(m+2)(m+1)), so \(10\cdot9=90\) gives (m=8). Identify the product of consecutive numbers.
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Which is the correct expansion of (\frac{(n+1)!}{(n-2)!})?
B. ((n+1)n(n-1))
((n+1)!=(n+1)n(n-1)(n-2)!) होता है। इसलिए समान ((n-2)!) कटने पर ((n+1)n(n-1)) बचता है। / ((n+1)!=(n+1)n(n-1)(n-2)!). After canceling ((n-2)!), ((n+1)n(n-1)) remains.
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What is the value of \(\frac{16!}{14!}+\frac{5!}{4!}\)?
C. (245)
\(\frac{16!}{14!}=16\cdot15=240\) और \(\frac{5!}{4!}=5\), इसलिए योग (245) है। अनुपातों को अलग-अलग सरल करें। / \(\frac{16!}{14!}=16\cdot15=240\) and \(\frac{5!}{4!}=5\), so the sum is (245). Simplify the ratios separately.
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If (a!=5040), what is the value of ((a-4)!)?
B. 6
क्योंकि \(5040=7!\), अतः \(a=7\)। इसलिए \((a-4)!=(7-4)!=3!=6\)। विकल्प 24, अर्थात \(4!\), तब मिलता यदि \((a-4)!\) में \(a-4\) को गलती से 4 मान लिया जाए। परीक्षा टिप: पहले दिए गए फैक्टोरियल से चर का मान ज्ञात करें, फिर शेष व्यंजक में उसे प्रतिस्थापित करें। / Since \(5040=7!\), we get \(a=7\). Therefore, \((a-4)!=(7-4)!=3!=6\). The option 24, which is \(4!\), may result from incorrectly taking \(a-4\) as 4. Exam tip: first determine the variable from the given factorial, then substitute it into the required expression.
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What is the simplified form of (\frac{(n+3)!-(n+2)!}{(n+2)!})?
B. (n+2)
((n+3)!-(n+2)!=(n+3)(n+2)!-(n+2)!=(n+2)(n+2)!)। इसलिए मान (n+2) है। / ((n+3)!-(n+2)!=(n+3)(n+2)!-(n+2)!=(n+2)(n+2)!). Therefore the value is (n+2).
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What is the value of \(\frac{3!+2!+1!}{0!}\)?
B. (9)
(3!+2!+1!=6+2+1=9) और (0!=1), इसलिए मान (9) है। (0!) को हमेशा (1) लें। / (3!+2!+1!=6+2+1=9) and (0!=1), so the value is (9). Always take (0!) as (1).
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How many consecutive factors remain in (\frac{(n+4)!}{(n+1)!})?
B. (3)
(\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)), इसलिए तीन गुणनखंड बचते हैं। हर के फैक्टोरियल तक अंश को घटाएं। / (\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)), so three factors remain. Reduce the numerator up to the denominator factorial.
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