Level 55 • 50/50 questions • 35 seconds per question.
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If (\frac{(n+1)!}{n!}=8), what is the value of (n)?
B. (7)
(\frac{(n+1)!}{n!}=n+1), इसलिए (n+1=8) और (n=7)। पहले फैक्टोरियल अनुपात सरल करें। / (\frac{(n+1)!}{n!}=n+1), so (n+1=8) and (n=7). Simplify the factorial ratio first.
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What is the value of \(\frac{9!}{7!}-5!\)?
A. (-48)
\(\frac{9!}{7!}=9\times8=72\) और (5!=120), इसलिए मान (-48) है। घटाव में चिह्न ध्यान से रखें। / \(\frac{9!}{7!}=9\times8=72\) and (5!=120), so the value is (-48). Keep the sign carefully in subtraction.
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If (n=3), what will be the value of (\frac{(n+2)!}{n!})?
B. (20)
(n=3) रखने पर \(\frac{5!}{3!}=5\times4=20\)। पहले चर का मान रखें। / Putting (n=3), \(\frac{5!}{3!}=5\times4=20\). Substitute the variable first.
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If ((n+1)!=720), what is the value of (n)?
B. (5)
(720=6!), इसलिए (n+1=6) और (n=5)। छोटे फैक्टोरियल मान याद रखें। / Since (720=6!), (n+1=6) and (n=5). Remember small factorial values.
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What is the value of \(\frac{6!}{3!}+4!\)?
C. (144)
\(\frac{6!}{3!}=120\) और (4!=24), इसलिए कुल (144) है। पहले भिन्न को सरल करें। / \(\frac{6!}{3!}=120\) and (4!=24), so the total is (144). Simplify the fraction first.
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What is the value of \(\frac{5!}{3!}+\frac{5!}{4!}\)?
B. (25)
\(\frac{5!}{3!}=20\) और \(\frac{5!}{4!}=5\), इसलिए योग (25) है। हर अनुपात अलग-अलग सरल करें। / \(\frac{5!}{3!}=20\) and \(\frac{5!}{4!}=5\), so the sum is (25). Simplify each ratio separately.
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What is the simplified value of (\frac{(n+2)!}{(n+1)!}-\frac{n!}{(n-1)!})?
B. (2)
पहला अनुपात (n+2) और दूसरा (n) है। अंतर ((n+2)-n=2) होगा। / The first ratio is (n+2) and the second is (n). The difference is ((n+2)-n=2).
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What is the value of \(\frac{8!}{5!,3!}+\frac{6!}{4!,2!}\)?
C. (71)
\(\frac{8!}{5!,3!}=56\) और \(\frac{6!}{4!,2!}=15\), इसलिए कुल (71) है। दोनों पदों को अलग हल करें। / \(\frac{8!}{5!,3!}=56\) and \(\frac{6!}{4!,2!}=15\), so the total is (71). Solve both terms separately.
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If (\frac{n!}{(n-2)!}=42), what is the value of (n)?
B. (7)
(\frac{n!}{(n-2)!}=n(n-1)), इसलिए (n(n-1)=42)। \(7\times6=42\), अतः (n=7)। / (\frac{n!}{(n-2)!}=n(n-1)), so (n(n-1)=42). Since \(7\times6=42\), (n=7).
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What is the value of \(\frac{10!}{8!}+\frac{4!}{2!}\)?
C. (102)
\(\frac{10!}{8!}=90\) और \(\frac{4!}{2!}=12\), इसलिए योग (102) है। समान फैक्टोरियल भाग काटें। / \(\frac{10!}{8!}=90\) and \(\frac{4!}{2!}=12\), so the sum is (102). Cancel common factorial parts.
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What is the value of \(\frac{7!-6!}{6!}\)?
B. (6)
\(7!=7\times6!\), इसलिए (\frac{7!-6!}{6!}=\frac{6!(7-1)}{6!}=6)। समान फैक्टोरियल बाहर लें। / Since \(7!=7\times6!\), (\frac{7!-6!}{6!}=\frac{6!(7-1)}{6!}=6). Take the common factorial out.
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What is the simplified form of (\frac{(n+3)!}{n!})?
A. ((n+3)(n+2)(n+1))
((n+3)!=(n+3)(n+2)(n+1)n!), इसलिए (n!) कट जाता है। क्रम से तीन गुणक बचते हैं। / ((n+3)!=(n+3)(n+2)(n+1)n!), so (n!) cancels. Three consecutive factors remain.
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If (\frac{(n+2)!}{n!}=30), what is the value of (n)?
B. (4)
(\frac{(n+2)!}{n!}=(n+2)(n+1))। \(6\times5=30\), इसलिए (n=4)। / (\frac{(n+2)!}{n!}=(n+2)(n+1)). Since \(6\times5=30\), (n=4).
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What is the value of \(\frac{9!}{6!,3!}-\frac{7!}{5!,2!}\)?
A. (63)
\(\frac{9!}{6!,3!}=84\) और \(\frac{7!}{5!,2!}=21\), इसलिए अंतर (63) है। दोनों संयोजन-जैसे पद अलग निकालें। / \(\frac{9!}{6!,3!}=84\) and \(\frac{7!}{5!,2!}=21\), so the difference is (63). Evaluate both combination-like terms separately.
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What is the value of \(\frac{5!+3!}{4!}\)?
C. \(\frac{21}{4}\)
अंश (120+6=126) है और (4!=24)। इसलिए मान \(\frac{126}{24}=\frac{21}{4}\) है। / The numerator is (120+6=126) and (4!=24). Hence the value is \(\frac{126}{24}=\frac{21}{4}\).
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What is the value of \(\frac{8!}{4!}-\frac{7!}{4!}\)?
B. (1470)
(\frac{8!-7!}{4!}=\frac{7!(8-1)}{4!}=7\times7\times6\times5=1470)। समान हर होने पर अंश मिलाकर देखें। / (\frac{8!-7!}{4!}=\frac{7!(8-1)}{4!}=7\times7\times6\times5=1470). With a common denominator, combine the numerator.
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What is the value of \(\frac{11!}{9!,2!}\)?
B. (55)
\(\frac{11!}{9!,2!}=\frac{11\times10}{2}=55\)। बड़े फैक्टोरियल को (9!) तक लिखें। / \(\frac{11!}{9!,2!}=\frac{11\times10}{2}=55\). Write the larger factorial up to (9!).
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Which expression is equal to \(\frac{12!}{9!}\)?
B. \(12\times11\times10\)
\(\frac{12!}{9!}=12\times11\times10\)। हर में (9!) होने से (9!) तक का भाग कट जाता है। / \(\frac{12!}{9!}=12\times11\times10\). Since the denominator is (9!), the part up to (9!) cancels.
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What is the simplified form of (\frac{(n+1)!-n!}{n!})?
B. (n)
((n+1)!=(n+1)n!), इसलिए अंश (n!{(n+1)-1}=nn!) है। भाग देने पर (n) मिलता है। / Since ((n+1)!=(n+1)n!), the numerator is (n!{(n+1)-1}=nn!). Dividing gives (n).
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If (\frac{n!}{(n-1)!}+2=9), what is the value of (n)?
C. (7)
(\frac{n!}{(n-1)!}=n), इसलिए (n+2=9)। अतः (n=7) है। / (\frac{n!}{(n-1)!}=n), so (n+2=9). Therefore, (n=7).
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What is the value of \(\frac{6!}{2!,4!}+\frac{5!}{2!,3!}\)?
B. (25)
\(\frac{6!}{2!,4!}=15\) और \(\frac{5!}{2!,3!}=10\), इसलिए कुल (25) है। छोटे पदों को अलग-अलग गिनें। / \(\frac{6!}{2!,4!}=15\) and \(\frac{5!}{2!,3!}=10\), so the total is (25). Count the smaller terms separately.
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What is the value of \(\frac{7!}{3!,4!}\times2!\)?
B. (70)
\(\frac{7!}{3!,4!}=35\) और (2!=2), इसलिए गुणनफल (70) है। पहले भिन्न को सरल करें। / \(\frac{7!}{3!,4!}=35\) and (2!=2), so the product is (70). Simplify the fraction first.
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What is the value of \(\frac{10!-9!}{8!}\)?
B. (81)
अंश (9!(10-1)=9\cdot9!) है। \(\frac{9\cdot9!}{8!}=9\times9=81\) मिलता है। / The numerator is (9!(10-1)=9\cdot9!). Thus \(\frac{9\cdot9!}{8!}=9\times9=81\).
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What is the simplified form of (\frac{(n+2)!-(n+1)!}{n!})?
A. ((n+1)2)
अंश ((n+1)![(n+2)-1]=(n+1)2n!) है। (n!) कटने पर ((n+1)2) मिलता है। / The numerator is ((n+1)![(n+2)-1]=(n+1)2n!). After cancelling (n!), the result is ((n+1)2).
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If (\frac{(n+3)!}{(n+1)!}=56), what is the value of (n)?
B. (5)
(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2))। \(8\times7=56\), इसलिए (n=5)। / (\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)). Since \(8\times7=56\), (n=5).
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What is the value of \(\frac{5!,3!}{4!}\)?
B. (30)
\(\frac{5!}{4!}=5\) और (3!=6), इसलिए मान (30) है। गुणन और भाग को क्रम से सरल करें। / \(\frac{5!}{4!}=5\) and (3!=6), so the value is (30). Simplify multiplication and division step by step.
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What is the value of \(\frac{8!}{7!}+\frac{7!}{6!}+\frac{6!}{5!}\)?
C. (21)
तीनों पद क्रमशः (8), (7) और (6) हैं। योग (8+7+6=21) है। / The three terms are (8), (7), and (6) respectively. The sum is (8+7+6=21).
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What is the value of \(\frac{9!}{5!,4!}\div\frac{6!}{4!,2!}\)?
B. \( \frac{42}{5}\)
पहला पद (126) और दूसरा (15) है। \(\frac{126}{15}=\frac{42}{5}\) मिलता है। / The first term is (126) and the second is (15). Thus \(\frac{126}{15}=\frac{42}{5}\).
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What is the simplified form of (\frac{(n+4)!}{(n+2)!})?
A. ((n+4)(n+3))
((n+4)!=(n+4)(n+3)(n+2)!)। ((n+2)!) कटने पर ((n+4)(n+3)) बचता है। / ((n+4)!=(n+4)(n+3)(n+2)!). After cancelling ((n+2)!), ((n+4)(n+3)) remains.
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If (\frac{n!}{(n-3)!}=120), what is the value of (n)?
C. (6)
(\frac{n!}{(n-3)!}=n(n-1)(n-2))। \(6\times5\times4=120\), इसलिए (n=6)। / (\frac{n!}{(n-3)!}=n(n-1)(n-2)). Since \(6\times5\times4=120\), (n=6).
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What is the value of \(\frac{7!}{5!}-\frac{6!}{4!}\)?
A. (12)
\(\frac{7!}{5!}=42\) और \(\frac{6!}{4!}=30\), इसलिए अंतर (12) है। छोटे अनुपात सीधे निकालें। / \(\frac{7!}{5!}=42\) and \(\frac{6!}{4!}=30\), so the difference is (12). Evaluate smaller ratios directly.
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What is the value of \(\frac{4!+5!}{3!}\)?
B. (24)
अंश (24+120=144) है और (3!=6)। इसलिए मान (24) है। / The numerator is (24+120=144) and (3!=6). Therefore, the value is (24).
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What is the value of \(\frac{12!}{10!,2!}-\frac{5!}{3!,2!}\)?
C. (56)
\(\frac{12!}{10!,2!}=66\) और \(\frac{5!}{3!,2!}=10\), इसलिए अंतर (56) है। बड़े और छोटे पद अलग रखें। / \(\frac{12!}{10!,2!}=66\) and \(\frac{5!}{3!,2!}=10\), so the difference is (56). Keep the larger and smaller terms separate.
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If (n(n-1)=72), what will be the value of (\frac{n!}{(n-2)!})?
B. (72)
(\frac{n!}{(n-2)!}=n(n-1))। दिए गए अनुसार इसका मान (72) है। / (\frac{n!}{(n-2)!}=n(n-1)). From the given condition, its value is (72).
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What is the value of \(\frac{6!}{5!}\times\frac{5!}{3!}\)?
C. (120)
पहला अनुपात (6) और दूसरा \(5\times4=20\) है। गुणनफल \(6\times20=120\) है। / The first ratio is (6) and the second is \(5\times4=20\). The product is \(6\times20=120\).
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What is the value of \(\frac{10!}{7!,3!}\)?
C. (120)
\(\frac{10!}{7!,3!}=\frac{10\times9\times8}{6}=120\)। (3!=6) रखना न भूलें। / \(\frac{10!}{7!,3!}=\frac{10\times9\times8}{6}=120\). Do not forget that (3!=6).
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What is the value of \(\frac{7!+5!}{6!}\)?
A. \(7+\frac{1}{6}\)
\(\frac{7!}{6!}=7\) और \(\frac{5!}{6!}=\frac{1}{6}\), इसलिए मान \(7+\frac{1}{6}\) है। पदों को हर से अलग-अलग भाग दें। / \(\frac{7!}{6!}=7\) and \(\frac{5!}{6!}=\frac{1}{6}\), so the value is \(7+\frac{1}{6}\). Divide each term by the denominator separately.
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If \(a=\frac{8!}{6!}\) and (b=3!), what is the value of (a-b)?
B. (50)
\(a=8\times7=56\) और (b=6), इसलिए (a-b=50)। पहले दोनों मान अलग निकालें। / \(a=8\times7=56\) and (b=6), so (a-b=50). Find both values separately first.
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What is the simplified form of (\frac{(n+2)!}{(n-1)!})?
A. ((n+2)(n+1)n)
((n+2)!=(n+2)(n+1)n(n-1)!)। इसलिए ((n-1)!) कटने पर तीन गुणक बचते हैं। / ((n+2)!=(n+2)(n+1)n(n-1)!). Hence after cancelling ((n-1)!), three factors remain.
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What is the value of \(\frac{9!}{8!}+\frac{8!}{7!}-\frac{7!}{6!}\)?
C. (10)
तीन अनुपात क्रमशः (9), (8) और (7) हैं। इसलिए (9+8-7=10) है। / The three ratios are (9), (8), and (7) respectively. Hence (9+8-7=10).
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If (\frac{(n+1)!}{(n-1)!}=20), what is the value of (n)?
B. (4)
यह अनुपात (n(n+1)) है। \(4\times5=20\), इसलिए (n=4)। / This ratio is (n(n+1)). Since \(4\times5=20\), (n=4).
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What is the value of \(\frac{8!}{6!}-\frac{5!}{4!}+2!\)?
C. (53)
\(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\) और (2!=2)। इसलिए (56-5+2=53) है। / \(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\), and (2!=2). Thus (56-5+2=53).
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What is the value of \(\frac{7!}{4!,3!}+\frac{7!}{6!,1!}\)?
C. (42)
पहला पद (35) और दूसरा पद (7) है। कुल (42) मिलता है। / The first term is (35) and the second term is (7). The total is (42).
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How many consecutive factors remain in (\frac{(n+3)!}{n!})?
B. (3)
(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1))। इसलिए तीन क्रमागत गुणक बचते हैं। / (\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)). Therefore, three consecutive factors remain.
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If \(x=\frac{9!}{7!}\) and \(y=\frac{6!}{4!}\), what is the value of \(\frac{x}{y}\)?
A. \(\frac{12}{5}\)
(x=72) और (y=30), इसलिए \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\)। अनुपात को अंत में सरल करें। / (x=72) and (y=30), so \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\). Simplify the ratio at the end.
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What is the simplified form of (\frac{(n+2)!}{n!}+\frac{(n+1)!}{n!})?
A. ((n+1)(n+3))
पहला पद ((n+2)(n+1)) और दूसरा (n+1) है। समान ((n+1)) लेने पर ((n+1)(n+3)) मिलता है। / The first term is ((n+2)(n+1)) and the second is (n+1). Taking common ((n+1)) gives ((n+1)(n+3)).
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What is the value of \(\frac{13!}{11!}-\frac{10!}{8!}\)?
B. (66)
\(\frac{13!}{11!}=13\times12=156\) और \(\frac{10!}{8!}=10\times9=90\), इसलिए अंतर (66) है। पहले दोनों फैक्टोरियल अनुपात अलग-अलग सरल करें। / \(\frac{13!}{11!}=13\times12=156\) and \(\frac{10!}{8!}=10\times9=90\), so the difference is (66). Simplify both factorial ratios separately first.
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If (\frac{(n+2)!}{(n-1)!}=210), what is the value of (n)?
B. (5)
(\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n)। \(7\times6\times5=210\), इसलिए (n=5) है। / (\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n). Since \(7\times6\times5=210\), (n=5).
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What is the value of \(\frac{9!-8!}{7!}\)?
B. (64)
अंश (8!(9-1)=8\cdot8!) है। \(\frac{8\cdot8!}{7!}=8\times8=64\) मिलता है। / The numerator is (8!(9-1)=8\cdot8!). Thus \(\frac{8\cdot8!}{7!}=8\times8=64\).
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What is the simplified form of (\frac{(n+4)!}{(n+1)!})?
B. ((n+4)(n+3)(n+2))
((n+4)!=(n+4)(n+3)(n+2)(n+1)!), इसलिए ((n+1)!) कट जाता है। तीन क्रमागत गुणक बचते हैं। / ((n+4)!=(n+4)(n+3)(n+2)(n+1)!), so ((n+1)!) cancels. Three consecutive factors remain.
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