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.
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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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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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 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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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 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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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+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 \(\frac{13!}{11!,2!}-\frac{6!}{4!,2!}\)?
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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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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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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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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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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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 \(\frac{9!}{7!}\div\frac{6!}{5!}\)?
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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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 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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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 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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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 \(\frac{12!}{10!,2!}+\frac{8!}{6!,2!}\)?
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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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 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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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 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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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 \(\frac{8!-7!}{6!}\)?
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 (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 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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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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