Topic Quiz • 150 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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If (n=4), what is the value of (\frac{(n+2)!}{n!})?
C. (30)
(n=4) रखने पर \(\frac{6!}{4!}=6\times5=30\)। पहले चर का मान रखकर फैक्टोरियल अनुपात सरल करें। / Putting (n=4), \(\frac{6!}{4!}=6\times5=30\). Substitute the variable first and simplify the factorial ratio.
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What is the value of \(\frac{8!-7!}{6!}\)?
D. (49)
अंश (7!(8-1)=7\cdot7!) है। \(\frac{7\cdot7!}{6!}=7\times7=49\) मिलेगा। / The numerator is (7!(8-1)=7\cdot7!). Thus \(\frac{7\cdot7!}{6!}=7\times7=49\).
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If (\frac{n!}{(n-2)!}=56), what is the value of (n)?
B. (8)
(\frac{n!}{(n-2)!}=n(n-1))। \(8\times7=56\), इसलिए (n=8) है। / (\frac{n!}{(n-2)!}=n(n-1)). Since \(8\times7=56\), (n=8).
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What is the value of \(\frac{6!+5!}{4!}\)?
D. (35)
अंश (720+120=840) है और (4!=24)। इसलिए मान \(\frac{840}{24}=35\) है। / The numerator is (720+120=840) and (4!=24). Therefore the value is \(\frac{840}{24}=35\).
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What is the value of \(\frac{9!}{6!}-\frac{7!}{5!}\)?
A. (462)
\(\frac{9!}{6!}=504\) और \(\frac{7!}{5!}=42\), इसलिए अंतर (462) है। दोनों अनुपात अलग-अलग सरल करें। / \(\frac{9!}{6!}=504\) and \(\frac{7!}{5!}=42\), so the difference is (462). Simplify both ratios separately.
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What is the simplified form of (\frac{(n+3)!}{(n+1)!}+\frac{(n+2)!}{n!})?
B. (2(n+2)2)
पहला पद ((n+3)(n+2)) और दूसरा ((n+2)(n+1)) है। समान ((n+2)) लेने पर (2(n+2)2) मिलता है। / The first term is ((n+3)(n+2)) and the second is ((n+2)(n+1)). Taking common ((n+2)) gives (2(n+2)2).
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If (\frac{(n+2)!}{n!}=42), what is the value of (n)?
C. (5)
(\frac{(n+2)!}{n!}=(n+2)(n+1))। \(7\times6=42\), इसलिए (n=5) है। / (\frac{(n+2)!}{n!}=(n+2)(n+1)). Since \(7\times6=42\), (n=5).
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What is the value of \(\frac{12!}{10!,2!}+\frac{8!}{6!,2!}\)?
D. (94)
पहला पद (66) और दूसरा पद (28) है। दोनों को जोड़ने पर (94) मिलता है। / The first term is (66) and the second term is (28). Adding them gives (94).
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What is the simplified form of (\frac{(n+1)!-n!}{(n-1)!})?
A. \(n^2\)
((n+1)!-n!=n!{(n+1)-1}=n\cdot n!)। ((n-1)!) से भाग देने पर \(n^2\) मिलता है। / ((n+1)!-n!=n!{(n+1)-1}=n\cdot n!). Dividing by ((n-1)!) gives \(n^2\).
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What is the value of \(\frac{7!}{4!,3!}\times3!\)?
C. (210)
\(\frac{7!}{4!,3!}=35\) और (3!=6)। इसलिए गुणनफल \(35\times6=210\) है। / \(\frac{7!}{4!,3!}=35\) and (3!=6). Therefore the product is \(35\times6=210\).
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If \(x=\frac{8!}{6!}\) and \(y=\frac{5!}{3!}\), what is the value of (x+y)?
C. (76)
\(x=8\times7=56\) और \(y=5\times4=20\)। इसलिए (x+y=76) है। / \(x=8\times7=56\) and \(y=5\times4=20\). Therefore (x+y=76).
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What is the simplified form of (\frac{(n+5)!}{(n+3)!})?
A. ((n+5)(n+4))
((n+5)!=(n+5)(n+4)(n+3)!)। इसलिए सरल रूप ((n+5)(n+4)) है। / ((n+5)!=(n+5)(n+4)(n+3)!). Therefore the simplified form is ((n+5)(n+4)).
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What is the value of \(\frac{11!-10!}{9!}\)?
C. (100)
अंश (10!(11-1)=10\cdot10!) है। \(\frac{10\cdot10!}{9!}=10\times10=100\) है। / The numerator is (10!(11-1)=10\cdot10!). Thus \(\frac{10\cdot10!}{9!}=10\times10=100\).
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What is the value of \(\frac{9!}{7!}\div\frac{6!}{5!}\)?
B. (12)
\(\frac{9!}{7!}=72\) और \(\frac{6!}{5!}=6\)। इसलिए भागफल (12) है। / \(\frac{9!}{7!}=72\) and \(\frac{6!}{5!}=6\). Therefore the quotient is (12).
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If (\frac{(n+1)!}{(n-1)!}=72), what is the value of (n)?
C. (8)
अनुपात (n(n+1)) के बराबर है। \(8\times9=72\), इसलिए (n=8) है। / The ratio equals (n(n+1)). Since \(8\times9=72\), (n=8).
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What is the value of \(\frac{5!,3!}{2!,4!}\)?
C. (15)
\(\frac{5!}{4!}=5\) और \(\frac{3!}{2!}=3\)। इसलिए मान \(5\times3=15\) है। / \(\frac{5!}{4!}=5\) and \(\frac{3!}{2!}=3\). Hence the value is \(5\times3=15\).
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What is the value of \(\frac{8!}{5!}+\frac{6!}{4!}-4!\)?
D. (342)
\(\frac{8!}{5!}=336\), \(\frac{6!}{4!}=30\) और (4!=24)। इसलिए (336+30-24=342) है। / \(\frac{8!}{5!}=336\), \(\frac{6!}{4!}=30\), and (4!=24). Thus (336+30-24=342).
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How many consecutive factors remain in (\frac{(n+5)!}{(n+2)!})?
B. (3)
(\frac{(n+5)!}{(n+2)!}=(n+5)(n+4)(n+3))। इसलिए तीन क्रमागत गुणक बचते हैं। / (\frac{(n+5)!}{(n+2)!}=(n+5)(n+4)(n+3)). Therefore three consecutive factors remain.
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If (\frac{(n+3)!}{(n+1)!}=90), what is the value of (n)?
B. (7)
(\frac{(n+3)!}{(n+1)!}=(n+3)(n+2))। \(10\times9=90\), इसलिए (n=7) है। / (\frac{(n+3)!}{(n+1)!}=(n+3)(n+2)). Since \(10\times9=90\), (n=7).
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What is the value of \(\frac{13!}{11!,2!}-\frac{6!}{4!,2!}\)?
C. (63)
पहला पद (78) और दूसरा पद (15) है। अंतर (78-15=63) है। / The first term is (78) and the second term is (15). The difference is (78-15=63).
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What is the simplified form of (\frac{(n+2)!-n!}{n!})?
B. \(n^2+3n+1\)
(\frac{(n+2)!}{n!}=(n+2)(n+1)), फिर (1) घटेगा। सरल रूप \(n^2+3n+1\) है। / (\frac{(n+2)!}{n!}=(n+2)(n+1)), then (1) is subtracted. The simplified form is \(n^2+3n+1\).
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What is the value of \(\frac{10!}{8!}+\frac{7!}{6!}+0!\)?
C. (98)
\(\frac{10!}{8!}=90\), \(\frac{7!}{6!}=7\) और (0!=1)। कुल (98) है। / \(\frac{10!}{8!}=90\), \(\frac{7!}{6!}=7\), and (0!=1). The total is (98).
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What is the simplified form of \(\frac{n!}{(n-1)!}\times\frac{(n+1)!}{n!}\)?
C. (n(n+1))
पहला अनुपात (n) और दूसरा (n+1) है। इसलिए गुणनफल (n(n+1)) है। / The first ratio is (n) and the second is (n+1). Therefore the product is (n(n+1)).
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If (\frac{n!}{(n-3)!}=210), what is the value of (n)?
B. (7)
(\frac{n!}{(n-3)!}=n(n-1)(n-2))। \(7\times6\times5=210\), इसलिए (n=7) है। / (\frac{n!}{(n-3)!}=n(n-1)(n-2)). Since \(7\times6\times5=210\), (n=7).
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What is the value of \(\frac{9!+8!}{7!}\)?
B. (80)
अंश को (8!(9+1)) लिखा जा सकता है। \(\frac{10\cdot8!}{7!}=10\times8=80\) है। / The numerator can be written as (8!(9+1)). Thus \(\frac{10\cdot8!}{7!}=10\times8=80\).
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What is the value of \(\frac{6!}{3!,3!}+\frac{7!}{5!,2!}\)?
C. (41)
पहला पद (20) और दूसरा (21) है। दोनों का योग (41) होगा। / The first term is (20) and the second is (21). Their sum is (41).
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What is the simplified form of (\frac{\frac{(n+4)!}{n!}}{\frac{(n+2)!}{n!}})?
B. ((n+4)(n+3))
यह भाग (\frac{(n+4)!}{(n+2)!}) बन जाता है। इसलिए सरल रूप ((n+4)(n+3)) है। / This division becomes (\frac{(n+4)!}{(n+2)!}). Hence the simplified form is ((n+4)(n+3)).
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If \(a=\frac{7!-5!}{5!}\), what is the value of (a)?
B. (41)
\(\frac{7!}{5!}=42\) और \(\frac{5!}{5!}=1\)। इसलिए (a=42-1=41) है। / \(\frac{7!}{5!}=42\) and \(\frac{5!}{5!}=1\). Therefore (a=42-1=41).
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If (\frac{(n+1)!}{(n-2)!}=60), what is the value of (n)?
B. (4)
(\frac{(n+1)!}{(n-2)!}=(n+1)n(n-1))। \(5\times4\times3=60\), इसलिए (n=4) है। / (\frac{(n+1)!}{(n-2)!}=(n+1)n(n-1)). Since \(5\times4\times3=60\), (n=4).
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What is the value of \(\frac{\frac{12!}{9!,3!}}{\frac{5!}{3!,2!}}\)?
C. (22)
ऊपर का पद (220) और नीचे का पद (10) है। भाग देने पर (22) मिलता है। / The numerator term is (220) and the denominator term is (10). Dividing gives (22).
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What is the simplified form of (\frac{(n+3)!+(n+2)!}{(n+2)!})?
C. (n+4)
अंश ((n+2)![(n+3)+1]) है। इसलिए सरल रूप (n+4) है। / The numerator is ((n+2)![(n+3)+1]). Therefore the simplified form is (n+4).
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What is the value of \(\frac{10!-8!}{8!}\)?
C. (89)
\(\frac{10!}{8!}=90\) और \(\frac{8!}{8!}=1\)। इसलिए मान (90-1=89) है। / \(\frac{10!}{8!}=90\) and \(\frac{8!}{8!}=1\). Therefore the value is (90-1=89).
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What is the value of \(4!\times\frac{6!}{5!}-3!\)?
C. (138)
(4!=24), \(\frac{6!}{5!}=6\) और (3!=6)। इसलिए \(24\times6-6=138\) है। / (4!=24), \(\frac{6!}{5!}=6\), and (3!=6). Hence \(24\times6-6=138\).
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What is the value of \(\frac{11!}{8!,3!}-\frac{9!}{7!,2!}\)?
C. (129)
पहला पद (165) और दूसरा (36) है। अंतर (129) मिलेगा। / The first term is (165) and the second is (36). The difference is (129).
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What is the simplified form of (\frac{(n+5)!}{(n+2)!})?
B. ((n+5)(n+4)(n+3))
((n+5)!=(n+5)(n+4)(n+3)(n+2)!)। इसलिए तीन गुणक बचते हैं। / ((n+5)!=(n+5)(n+4)(n+3)(n+2)!). Therefore three factors remain.
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If (\frac{(n+4)!}{(n+2)!}=132), what is the value of (n)?
B. (8)
(\frac{(n+4)!}{(n+2)!}=(n+4)(n+3))। \(12\times11=132\), इसलिए (n=8) है। / (\frac{(n+4)!}{(n+2)!}=(n+4)(n+3)). Since \(12\times11=132\), (n=8).
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What is the value of \(\frac{7!}{5!}+\frac{5!}{2!,3!}+1!\)?
C. (53)
\(\frac{7!}{5!}=42\), \(\frac{5!}{2!,3!}=10\) और (1!=1)। कुल (53) है। / \(\frac{7!}{5!}=42\), \(\frac{5!}{2!,3!}=10\), and (1!=1). The total is (53).
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What is the simplified form of \(\frac{(n+2)!}{(n+1)!}\times\frac{(n+1)!}{n!}\)?
B. ((n+2)(n+1))
पहला अनुपात (n+2) और दूसरा (n+1) है। गुणनफल ((n+2)(n+1)) होगा। / The first ratio is (n+2) and the second is (n+1). The product is ((n+2)(n+1)).
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What is the value of \(\frac{8!,3!}{6!,2!}\)?
C. (168)
\(\frac{8!}{6!}=56\) और \(\frac{3!}{2!}=3\)। इसलिए मान \(56\times3=168\) है। / \(\frac{8!}{6!}=56\) and \(\frac{3!}{2!}=3\). Therefore the value is \(56\times3=168\).
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What is the value of \(\frac{9!-7!}{7!}\)?
C. (71)
\(\frac{9!}{7!}=72\) और \(\frac{7!}{7!}=1\)। इसलिए (72-1=71) है। / \(\frac{9!}{7!}=72\) and \(\frac{7!}{7!}=1\). Hence (72-1=71).
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What is the simplified form of (\frac{n!+(n-1)!}{(n-1)!})?
B. (n+1)
(n!=n(n-1)!), इसलिए अंश ((n-1)!(n+1)) है। भाग देने पर (n+1) मिलेगा। / Since (n!=n(n-1)!), the numerator is ((n-1)!(n+1)). Dividing gives (n+1).
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If (n=5), what is the value of (\frac{(n+3)!}{(n+1)!}-\frac{(n+1)!}{n!})?
C. (50)
(n=5) पर पहला पद \(\frac{8!}{6!}=56\) और दूसरा \(\frac{6!}{5!}=6\) है। अंतर (50) है। / For (n=5), the first term is \(\frac{8!}{6!}=56\) and the second is \(\frac{6!}{5!}=6\). The difference is (50).
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What is the value of \(\frac{6!}{2!,4!}\times\frac{4!}{3!}\)?
D. (60)
पहला पद (15) और दूसरा (4) है। गुणनफल \(15\times4=60\) है। / The first term is (15) and the second is (4). The product is \(15\times4=60\).
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What is the value of \(\frac{14!}{12!}-\frac{13!}{11!}\)?
B. (26)
\(\frac{14!}{12!}=14\times13=182\) और \(\frac{13!}{11!}=13\times12=156\)। अंतर (26) है। / \(\frac{14!}{12!}=14\times13=182\) and \(\frac{13!}{11!}=13\times12=156\). The difference is (26).
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If (\frac{(n+2)!-(n+1)!}{n!}=36), what is the value of (n)?
B. (5)
सरल रूप ((n+1)2) है। ((n+1)2=36) से (n+1=6) और (n=5) है। / The simplified form is ((n+1)2). From ((n+1)2=36), (n+1=6) and (n=5).
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What is the simplified form of (\frac{\frac{(n+3)!}{n!}}{\frac{(n+2)!}{n!}})?
C. (n+3)
भिन्न को सरल करने पर (\frac{(n+3)!}{(n+2)!}) मिलता है। इसका मान (n+3) है। / Simplifying the fraction gives (\frac{(n+3)!}{(n+2)!}). Its value is (n+3).
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What is the value of \(\frac{10!}{6!,4!}+\frac{10!}{7!,3!}\)?
D. (330)
पहला पद (210) और दूसरा पद (120) है। दोनों का योग (330) है। / The first term is (210) and the second term is (120). Their sum is (330).
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What is the value of \(\frac{15!-12!}{12!}\)?
B. (2729)
\(\frac{15!}{12!}=15\times14\times13=2730\) और \(\frac{12!}{12!}=1\)। इसलिए मान (2730-1=2729) है। / \(\frac{15!}{12!}=15\times14\times13=2730\) and \(\frac{12!}{12!}=1\). Therefore the value is (2730-1=2729).
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If (\frac{(n+4)!}{(n+1)!}=504), what is the value of (n)?
B. (5)
(\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2))। \(9\times8\times7=504\), इसलिए (n=5) है। / (\frac{(n+4)!}{(n+1)!}=(n+4)(n+3)(n+2)). Since \(9\times8\times7=504\), (n=5).
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What is the simplified form of (\frac{(n+4)!}{(n+2)!}-\frac{(n+3)!}{(n+1)!})?
A. (2(n+3))
पहला पद ((n+4)(n+3)) और दूसरा ((n+3)(n+2)) है। अंतर लेने पर (2(n+3)) मिलता है। / The first term is ((n+4)(n+3)) and the second is ((n+3)(n+2)). Taking the difference gives (2(n+3)).
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If (\frac{n!+(n-1)!}{(n-1)!}=9), what is the value of (n)?
C. (8)
भिन्न का सरल रूप (n+1) है, इसलिए (n+1=9) से (n=8)। पहले सामान्य फैक्टोरियल निकालें। / The simplified form of the fraction is (n+1), so (n+1=9) gives (n=8). First take the common factorial.
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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 form of (8!) in terms of (6!)?
A. \(8\cdot7\cdot6!\)
\(8!=8\cdot7\cdot6!\) होता है। बड़े फैक्टोरियल को छोटे फैक्टोरियल तक तोड़ना उपयोगी है। / \(8!=8\cdot7\cdot6!\). Breaking a larger factorial down to a smaller factorial is useful.
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What is the value of \(\frac{14!}{12!\cdot2!}\)?
C. (91)
\(\frac{14!}{12!\cdot2!}=\frac{14\cdot13}{2}=91\) होता है। बड़े फैक्टोरियल को छोटे फैक्टोरियल तक ही फैलाएं। / \(\frac{14!}{12!\cdot2!}=\frac{14\cdot13}{2}=91\). Expand the larger factorial only up to the smaller factorial.
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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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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 \(\frac{10!}{8!\cdot2!}\)?
A. (45)
\(\frac{10!}{8!\cdot2!}=\frac{10\cdot9}{2}=45\) होता है। पहले (8!) को काटें। / \(\frac{10!}{8!\cdot2!}=\frac{10\cdot9}{2}=45\). Cancel (8!) first.
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What is the value of \(\frac{15!}{13!}-5!\)?
B. (90)
\(\frac{15!}{13!}=15\cdot14=210\) और (5!=120), इसलिए अंतर (90) है। पहले फैक्टोरियल अनुपात को सरल करें। / \(\frac{15!}{13!}=15\cdot14=210\) and (5!=120), so the difference is (90). Simplify the factorial ratio first.
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Which is the correct simplified form of (\frac{(n+2)!}{n!})?
C. ((n+2)(n+1))
((n+2)!=(n+2)(n+1)n!) होता है। (n!) कटने पर ((n+2)(n+1)) बचता है। / ((n+2)!=(n+2)(n+1)n!). After canceling (n!), ((n+2)(n+1)) remains.
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What is the value of \(\frac{11!}{8!\cdot3!}\)?
C. (165)
\(\frac{11!}{8!\cdot3!}=\frac{11\cdot10\cdot9}{6}=165\) होता है। (3!) को (6) मानना न भूलें। / \(\frac{11!}{8!\cdot3!}=\frac{11\cdot10\cdot9}{6}=165\). Do not forget that (3!=6).
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If (\frac{(n+1)!}{(n-1)!}=30), what is the value of (n)?
B. (5)
(\frac{(n+1)!}{(n-1)!}=n(n+1)), इसलिए (n(n+1)=30) से (n=5)। ऐसे प्रश्नों में पहले फैक्टोरियल घटाएं। / (\frac{(n+1)!}{(n-1)!}=n(n+1)), so (n(n+1)=30) gives (n=5). Reduce factorials first in such questions.
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What is the difference between the values of \(\frac{11!}{9!}\) and \(\frac{8!}{6!}\)?
D. (74)
\(\frac{11!}{9!}=110\) और \(\frac{8!}{6!}=56\), इसलिए अंतर (54) नहीं बल्कि (110-56=54) है। गणना के बाद घटाव सावधानी से करें। / \(\frac{11!}{9!}=110\) and \(\frac{8!}{6!}=56\), so the difference is (110-56=54). Do the subtraction carefully after simplification.
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What is the value of \(\frac{7!+6!}{6!+5!}\)?
B. (7)
ऊपर \(7!+6!=8\cdot6!\) और नीचे \(6!+5!=7\cdot5!\) है, इसलिए मान \(8\cdot6!/7\cdot5!=48/7\) नहीं है। सीधे मान रखने पर सही गणना करें। / The numerator is \(7!+6!=8\cdot6!\) and the denominator is \(6!+5!=7\cdot5!\), so careful direct calculation is needed. Substitute values to avoid mistakes.
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What is the value of \(\frac{18!}{16!\cdot2}\)?
C. (153)
\(\frac{18!}{16!\cdot2}=\frac{18\cdot17}{2}=153\) होता है। पहले (16!) काटें फिर भाग दें। / \(\frac{18!}{16!\cdot2}=\frac{18\cdot17}{2}=153\). Cancel (16!) first and then divide.
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What is the value of \(\frac{7!}{5!}+3!\)?
C. (48)
\(\frac{7!}{5!}=42\) और (3!=6), इसलिए कुल (48) है। मिश्रित प्रश्नों में हर भाग अलग निकालें। / \(\frac{7!}{5!}=42\) and (3!=6), so the total is (48). In mixed questions, evaluate each part separately.
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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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What is the value of \(\frac{6!}{3!\cdot3!}\)?
C. (20)
\(\frac{6!}{3!\cdot3!}=\frac{720}{6\cdot6}=20\) होता है। हर फैक्टोरियल का मान सही रखें। / \(\frac{6!}{3!\cdot3!}=\frac{720}{6\cdot6}=20\). Keep each factorial value correct.
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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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What is the value of \(\frac{12!}{10!\cdot2}\)?
C. (66)
\(\frac{12!}{10!\cdot2}=\frac{12\cdot11}{2}=66\) होता है। पहले (10!) काटने से गणना छोटी होती है। / \(\frac{12!}{10!\cdot2}=\frac{12\cdot11}{2}=66\). Canceling (10!) first shortens the calculation.
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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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If (n!) has (n) as a natural number, which product does (n!) represent?
B. \(1\cdot2\cdot3\cdots n\)
(n!) का अर्थ (1) से (n) तक की सभी प्राकृतिक संख्याओं का गुणनफल है। परिभाषा स्पष्ट हो तो सरलीकरण आसान होता है। / (n!) means the product of all natural numbers from (1) to (n). A clear definition makes simplification easier.
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If (a!=5040), what is the value of ((a-4)!)?
C. (3!)
(5040=7!), इसलिए (a=7) और ((a-4)!=3!)। दिए मान से पहले चर पहचानें। / (5040=7!), so (a=7) and ((a-4)!=3!). First identify the variable from the given value.
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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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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 \(\frac{8!-7!}{7!}\)?
B. (7)
\(8!-7!=8\cdot7!-7!=7\cdot7!\), इसलिए मान (7) है। घटाव में सामान्य फैक्टोरियल निकालें। / \(8!-7!=8\cdot7!-7!=7\cdot7!\), so the value is (7). Take the common factorial in subtraction.
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The value of \(\frac{13!}{11!}\) is equal to which of the following?
B. \(13\cdot12\)
\(\frac{13!}{11!}=13\cdot12\) होता है। हर में मौजूद फैक्टोरियल तक अंश को विस्तार दें। / \(\frac{13!}{11!}=13\cdot12\). Expand the numerator only up to the factorial in the denominator.
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What is the value of \(\frac{4!+3!}{3!+2!}\)?
A. (3)
ऊपर (24+6=30) और नीचे (6+2=8) नहीं, बल्कि (3!+2!=6+2=8), इसलिए मान \(\frac{15}{4}\) है। विकल्पों में सही मान नहीं हो तो पहले पुनर्गणना जरूरी है। / The numerator is (24+6=30) and denominator is (6+2=8), so the value is \(\frac{15}{4}\). If options do not match, recalculation is necessary first.
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What is the simplified form of (\frac{(n+2)!}{(n+1)!}+\frac{(n+1)!}{n!})?
C. (2n+3)
पहला पद (n+2) और दूसरा (n+1) है, इसलिए योग (2n+3) है। प्रत्येक भिन्न को अलग सरल करें। / The first term is (n+2) and the second is (n+1), so the sum is (2n+3). Simplify each fraction separately.
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How should \(\frac{7!}{4!}\) be correctly written as a product?
A. \(7\cdot6\cdot5\)
\(\frac{7!}{4!}=7\cdot6\cdot5\) होता है। हर का (4!) कट जाता है। / \(\frac{7!}{4!}=7\cdot6\cdot5\). The denominator (4!) cancels out.
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If (\frac{(n+2)!}{n!}=56), what is the value of (n)?
B. (6)
((n+2)(n+1)=56), इसलिए \(8\cdot7=56\) से (n=6)। लगातार संख्याओं से तुलना करें। / ((n+2)(n+1)=56), so \(8\cdot7=56\) gives (n=6). Compare with consecutive numbers.
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What is the value of \(\frac{9!}{6!\cdot3!}\)?
C. (84)
\(\frac{9!}{6!\cdot3!}=\frac{9\cdot8\cdot7}{6}=84\) होता है। पहले (6!) काटने से हल छोटा होता है। / \(\frac{9!}{6!\cdot3!}=\frac{9\cdot8\cdot7}{6}=84\). Canceling (6!) first makes the solution shorter.
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What is the value of \(\frac{6!}{4!}-2!\)?
B. (28)
\(\frac{6!}{4!}=30\) और (2!=2), इसलिए मान (28) है। भाग और घटाव का क्रम ध्यान रखें। / \(\frac{6!}{4!}=30\) and (2!=2), so the value is (28). Keep the order of division and subtraction in mind.
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(\frac{(n+1)!}{(n-1)!}) is equal to which expression?
B. (n(n+1))
((n+1)!=(n+1)n(n-1)!) होता है। इसलिए भाग देने पर (n(n+1)) मिलता है। / ((n+1)!=(n+1)n(n-1)!). Therefore division gives (n(n+1)).
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What is the value of \(\frac{4!\cdot5}{5!}\)?
A. (1)
क्योंकि \(5!=5\cdot4!\), इसलिए \(\frac{4!\cdot5}{5!}=1\)। फैक्टोरियल को निकट छोटे फैक्टोरियल से लिखें। / Since \(5!=5\cdot4!\), \(\frac{4!\cdot5}{5!}=1\). Write a factorial using the nearest smaller factorial.
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What is the value of \(3!\cdot4!\)?
C. (144)
(3!=6) और (4!=24), इसलिए गुणनफल (144) है। छोटे फैक्टोरियल याद रखें। / (3!=6) and (4!=24), so the product is (144). Memorize small factorials.
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What is the value of \(\frac{8!}{5!\cdot3!}\)?
C. (56)
\(\frac{8!}{5!\cdot3!}=\frac{8\cdot7\cdot6}{6}=56\) होता है। पहले समान फैक्टोरियल काटें। / \(\frac{8!}{5!\cdot3!}=\frac{8\cdot7\cdot6}{6}=56\). Cancel the common factorial first.
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If (n!=120), what is the value of ((n-2)!)?
C. (6)
(120=5!), इसलिए (n=5) और ((n-2)!=3!=6)। पहले (n) पहचानें। / (120=5!), so (n=5) and ((n-2)!=3!=6). Identify (n) first.
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What is the value of \(\frac{7!}{3!\cdot4!}\)?
C. (35)
\(\frac{7!}{3!\cdot4!}=\frac{7\cdot6\cdot5}{6}=35\) होता है। (4!) काटकर गणना करें। / \(\frac{7!}{3!\cdot4!}=\frac{7\cdot6\cdot5}{6}=35\). Cancel (4!) and calculate.
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What is the value of \(\frac{5!}{2!}+1!\)?
C. (61)
\(\frac{5!}{2!}=60\) और (1!=1), इसलिए मान (61) है। (1!) और (0!) दोनों (1) होते हैं। / \(\frac{5!}{2!}=60\) and (1!=1), so the value is (61). Both (1!) and (0!) equal (1).
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What is the simplified form of (\frac{(n+2)!}{(n+1)!}-\frac{n!}{(n-1)!})?
B. (2)
पहला पद (n+2) और दूसरा (n) है, इसलिए अंतर (2) है। अलग-अलग सरलीकरण करने से गलती कम होती है। / The first term is (n+2) and the second is (n), so the difference is (2). Simplifying separately reduces mistakes.
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For which value will ((n-3)!) be defined if (n) is an integer?
C. (n=3)
फैक्टोरियल गैर-ऋणात्मक पूर्णांक के लिए परिभाषित होता है, इसलिए \(n-3\ge0\)। दिए विकल्पों में (n=3) सही है। / Factorial is defined for non-negative integers, so \(n-3\ge0\). Among the given options, (n=3) is correct.
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What is the value of \(\frac{9!}{8!}+\frac{6!}{5!}\)?
C. (15)
\(\frac{9!}{8!}=9\) और \(\frac{6!}{5!}=6\), इसलिए योग (15) है। सरल अनुपातों को तुरंत घटाएं। / \(\frac{9!}{8!}=9\) and \(\frac{6!}{5!}=6\), so the sum is (15). Reduce simple ratios quickly.
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What is the value of \(2!\cdot3!+4!\)?
B. (36)
\(2!\cdot3!=2\cdot6=12\) और (4!=24), इसलिए कुल (36) है। गुणा को जोड़ से पहले करें। / \(2!\cdot3!=2\cdot6=12\) and (4!=24), so the total is (36). Do multiplication before addition.
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Which is the complete expansion of (\frac{(n+3)!}{n!})?
C. ((n+3)(n+2)(n+1))
((n+3)!=(n+3)(n+2)(n+1)n!) होता है। इसलिए (n!) कटने पर तीन गुणनखंड बचते हैं। / ((n+3)!=(n+3)(n+2)(n+1)n!). After canceling (n!), three factors remain.
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What is the value of \(\frac{10!}{7!\cdot3!}\)?
D. (120)
\(\frac{10!}{7!\cdot3!}=\frac{10\cdot9\cdot8}{6}=120\) होता है। बड़े फैक्टोरियल को तीन पदों तक घटाएं। / \(\frac{10!}{7!\cdot3!}=\frac{10\cdot9\cdot8}{6}=120\). Reduce the large factorial to three factors.
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What is the value of \(\frac{6!-5!}{4!}\)?
C. (25)
\(6!-5!=6\cdot5!-5!=5\cdot5!\), और \(\frac{5\cdot5!}{4!}=25\)। पहले सामान्य फैक्टोरियल निकालें। / \(6!-5!=6\cdot5!-5!=5\cdot5!\), and \(\frac{5\cdot5!}{4!}=25\). Take the common factorial first.
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What is the simplified form of (\frac{(n+2)!+(n+1)!}{(n+1)!})?
B. (n+3)
((n+2)!+(n+1)!=(n+2)(n+1)!+(n+1)!=(n+3)(n+1)!)। इसलिए मान (n+3) है। / ((n+2)!+(n+1)!=(n+2)(n+1)!+(n+1)!=(n+3)(n+1)!). Therefore the value is (n+3).
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How many consecutive factors remain in \(\frac{12!}{9!}\)?
B. (3)
\(\frac{12!}{9!}=12\cdot11\cdot10\), इसलिए तीन गुणनखंड बचते हैं। हर के फैक्टोरियल तक अंश को काटें। / \(\frac{12!}{9!}=12\cdot11\cdot10\), so three factors remain. Cancel the numerator up to the denominator factorial.
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If (r!=24), what will be the value of ((r+1)!)?
D. (120)
(24=4!), इसलिए (r=4) और ((r+1)!=5!=120)। पहले दिए फैक्टोरियल से चर का मान निकालें। / (24=4!), so (r=4) and ((r+1)!=5!=120). First find the variable from the given factorial.
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What is the value of (\frac{(n+1)!}{(n+1)n!})?
B. (1)
((n+1)!=(n+1)n!), इसलिए पूरी भिन्न (1) बनती है। पहचान सूत्र याद रखने से समय बचता है। / ((n+1)!=(n+1)n!), so the whole fraction becomes (1). Remembering this identity saves time.
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