\(\frac{12!}{10!,2!}+\frac{8!}{6!,2!}\) का मान क्या है?
What is the value of \(\frac{12!}{10!,2!}+\frac{8!}{6!,2!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (84)
B (88)
C (90)
D (94)
Explanation opens after your attempt
Step 1
Concept
The first term is (66) and the second term is (28). Adding them gives (94).
Step 2
Why this answer is correct
The correct answer is D. (94). The first term is (66) and the second term is (28). Adding them gives (94).
Step 3
Exam Tip
पहला पद (66) और दूसरा पद (28) है। दोनों को जोड़ने पर (94) मिलता है।
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यदि (\frac{(n+2)!}{n!}=42), तो (n) का मान क्या है?
If (\frac{(n+2)!}{n!}=42), what is the value of (n)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
(\frac{(n+2)!}{n!}=(n+2)(n+1)). Since \(7\times6=42\), (n=5).
Step 2
Why this answer is correct
The correct answer is C. (5). (\frac{(n+2)!}{n!}=(n+2)(n+1)). Since \(7\times6=42\), (n=5).
Step 3
Exam Tip
(\frac{(n+2)!}{n!}=(n+2)(n+1))। \(7\times6=42\), इसलिए (n=5) है।
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(\frac{(n+3)!}{(n+1)!}+\frac{(n+2)!}{n!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+3)!}{(n+1)!}+\frac{(n+2)!}{n!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A ((n+2)2 )
B (2(n+2)2 )
C (2n+4)
D ((n+3)(n+1))
Explanation opens after your attempt
Correct Answer
B. (2(n+2)2 )
Step 1
Concept
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 ).
Step 2
Why this answer is correct
The correct answer is B. (2(n+2)2 ). 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 ).
Step 3
Exam Tip
पहला पद ((n+3)(n+2)) और दूसरा ((n+2)(n+1)) है। समान ((n+2)) लेने पर (2(n+2)2 ) मिलता है।
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\(\frac{9!}{6!}-\frac{7!}{5!}\) का मान क्या है?
What is the value of \(\frac{9!}{6!}-\frac{7!}{5!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (462)
B (480)
C (504)
D (546)
Explanation opens after your attempt
Step 1
Concept
\(\frac{9!}{6!}=504\) and \(\frac{7!}{5!}=42\), so the difference is (462). Simplify both ratios separately.
Step 2
Why this answer is correct
The correct answer is A. (462). \(\frac{9!}{6!}=504\) and \(\frac{7!}{5!}=42\), so the difference is (462). Simplify both ratios separately.
Step 3
Exam Tip
\(\frac{9!}{6!}=504\) और \(\frac{7!}{5!}=42\), इसलिए अंतर (462) है। दोनों अनुपात अलग-अलग सरल करें।
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\(\frac{6!+5!}{4!}\) का मान क्या है?
What is the value of \(\frac{6!+5!}{4!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (25)
B (30)
C (32)
D (35)
Explanation opens after your attempt
Step 1
Concept
The numerator is (720+120=840) and (4!=24). Therefore the value is \(\frac{840}{24}=35\).
Step 2
Why this answer is correct
The correct answer is D. (35). The numerator is (720+120=840) and (4!=24). Therefore the value is \(\frac{840}{24}=35\).
Step 3
Exam Tip
अंश (720+120=840) है और (4!=24)। इसलिए मान \(\frac{840}{24}=35\) है।
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यदि (\frac{n!}{(n-2)!}=56), तो (n) का मान क्या है?
If (\frac{n!}{(n-2)!}=56), what is the value of (n)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
(\frac{n!}{(n-2)!}=n(n-1)). Since \(8\times7=56\), (n=8).
Step 2
Why this answer is correct
The correct answer is B. (8). (\frac{n!}{(n-2)!}=n(n-1)). Since \(8\times7=56\), (n=8).
Step 3
Exam Tip
(\frac{n!}{(n-2)!}=n(n-1))। \(8\times7=56\), इसलिए (n=8) है।
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\(\frac{8!-7!}{6!}\) का मान क्या है?
What is the value of \(\frac{8!-7!}{6!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (42)
B (48)
C (56)
D (49)
Explanation opens after your attempt
Step 1
Concept
The numerator is (7!(8-1)=7\cdot7!). Thus \(\frac{7\cdot7!}{6!}=7\times7=49\).
Step 2
Why this answer is correct
The correct answer is D. (49). The numerator is (7!(8-1)=7\cdot7!). Thus \(\frac{7\cdot7!}{6!}=7\times7=49\).
Step 3
Exam Tip
अंश (7!(8-1)=7\cdot7!) है। \(\frac{7\cdot7!}{6!}=7\times7=49\) मिलेगा।
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यदि (n=4), तो (\frac{(n+2)!}{n!}) का मान क्या है?
If (n=4), what is the value of (\frac{(n+2)!}{n!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (20)
B (24)
C (30)
D (36)
Explanation opens after your attempt
Step 1
Concept
Putting (n=4), \(\frac{6!}{4!}=6\times5=30\). Substitute the variable first and simplify the factorial ratio.
Step 2
Why this answer is correct
The correct answer is C. (30). Putting (n=4), \(\frac{6!}{4!}=6\times5=30\). Substitute the variable first and simplify the factorial ratio.
Step 3
Exam Tip
(n=4) रखने पर \(\frac{6!}{4!}=6\times5=30\)। पहले चर का मान रखकर फैक्टोरियल अनुपात सरल करें।
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(\frac{(n+4)!}{(n+1)!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+4)!}{(n+1)!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A ((n+4)(n+3))
B ((n+4)(n+3)(n+2))
C ((n+3)(n+2)(n+1))
D ((n+4)(n+2))
Explanation opens after your attempt
Correct Answer
B. ((n+4)(n+3)(n+2))
Step 1
Concept
((n+4)!=(n+4)(n+3)(n+2)(n+1)!), so ((n+1)!) cancels. Three consecutive factors remain.
Step 2
Why this answer is correct
The correct answer is B. ((n+4)(n+3)(n+2)). ((n+4)!=(n+4)(n+3)(n+2)(n+1)!), so ((n+1)!) cancels. Three consecutive factors remain.
Step 3
Exam Tip
((n+4)!=(n+4)(n+3)(n+2)(n+1)!), इसलिए ((n+1)!) कट जाता है। तीन क्रमागत गुणक बचते हैं।
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\(\frac{9!-8!}{7!}\) का मान क्या है?
What is the value of \(\frac{9!-8!}{7!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (56)
B (64)
C (72)
D (80)
Explanation opens after your attempt
Step 1
Concept
The numerator is (8!(9-1)=8\cdot8!). Thus \(\frac{8\cdot8!}{7!}=8\times8=64\).
Step 2
Why this answer is correct
The correct answer is B. (64). The numerator is (8!(9-1)=8\cdot8!). Thus \(\frac{8\cdot8!}{7!}=8\times8=64\).
Step 3
Exam Tip
अंश (8!(9-1)=8\cdot8!) है। \(\frac{8\cdot8!}{7!}=8\times8=64\) मिलता है।
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यदि (\frac{(n+2)!}{(n-1)!}=210), तो (n) का मान क्या है?
If (\frac{(n+2)!}{(n-1)!}=210), what is the value of (n)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
(\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n). Since \(7\times6\times5=210\), (n=5).
Step 2
Why this answer is correct
The correct answer is B. (5). (\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n). Since \(7\times6\times5=210\), (n=5).
Step 3
Exam Tip
(\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n)। \(7\times6\times5=210\), इसलिए (n=5) है।
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\(\frac{13!}{11!}-\frac{10!}{8!}\) का मान क्या है?
What is the value of \(\frac{13!}{11!}-\frac{10!}{8!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (42)
B (66)
C (72)
D (90)
Explanation opens after your attempt
Step 1
Concept
\(\frac{13!}{11!}=13\times12=156\) and \(\frac{10!}{8!}=10\times9=90\), so the difference is (66). Simplify both factorial ratios separately first.
Step 2
Why this answer is correct
The correct answer is B. (66). \(\frac{13!}{11!}=13\times12=156\) and \(\frac{10!}{8!}=10\times9=90\), so the difference is (66). Simplify both factorial ratios separately first.
Step 3
Exam Tip
\(\frac{13!}{11!}=13\times12=156\) और \(\frac{10!}{8!}=10\times9=90\), इसलिए अंतर (66) है। पहले दोनों फैक्टोरियल अनुपात अलग-अलग सरल करें।
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(\frac{(n+2)!}{n!}+\frac{(n+1)!}{n!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+2)!}{n!}+\frac{(n+1)!}{n!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A ((n+1)(n+3))
B ((n+1)(n+2))
C (2n+3)
D \(n^2+n+1\)
Explanation opens after your attempt
Correct Answer
A. ((n+1)(n+3))
Step 1
Concept
The first term is ((n+2)(n+1)) and the second is (n+1). Taking common ((n+1)) gives ((n+1)(n+3)).
Step 2
Why this answer is correct
The correct answer is A. ((n+1)(n+3)). The first term is ((n+2)(n+1)) and the second is (n+1). Taking common ((n+1)) gives ((n+1)(n+3)).
Step 3
Exam Tip
पहला पद ((n+2)(n+1)) और दूसरा (n+1) है। समान ((n+1)) लेने पर ((n+1)(n+3)) मिलता है।
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यदि \(x=\frac{9!}{7!}\) और \(y=\frac{6!}{4!}\), तो \(\frac{x}{y}\) का मान क्या है?
If \(x=\frac{9!}{7!}\) and \(y=\frac{6!}{4!}\), what is the value of \(\frac{x}{y}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A \(\frac{12}{5}\)
B \(\frac{5}{12}\)
C \(\frac{6}{5}\)
D \(\frac{5}{6}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{12}{5}\)
Step 1
Concept
(x=72) and (y=30), so \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\). Simplify the ratio at the end.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{12}{5}\). (x=72) and (y=30), so \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\). Simplify the ratio at the end.
Step 3
Exam Tip
(x=72) और (y=30), इसलिए \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\)। अनुपात को अंत में सरल करें।
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(\frac{(n+3)!}{n!}) में कुल कितने क्रमागत गुणक बचते हैं?
How many consecutive factors remain in (\frac{(n+3)!}{n!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)). Therefore, three consecutive factors remain.
Step 2
Why this answer is correct
The correct answer is B. (3). (\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)). Therefore, three consecutive factors remain.
Step 3
Exam Tip
(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1))। इसलिए तीन क्रमागत गुणक बचते हैं।
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\(\frac{7!}{4!,3!}+\frac{7!}{6!,1!}\) का मान क्या है?
What is the value of \(\frac{7!}{4!,3!}+\frac{7!}{6!,1!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (28)
B (35)
C (42)
D (49)
Explanation opens after your attempt
Step 1
Concept
The first term is (35) and the second term is (7). The total is (42).
Step 2
Why this answer is correct
The correct answer is C. (42). The first term is (35) and the second term is (7). The total is (42).
Step 3
Exam Tip
पहला पद (35) और दूसरा पद (7) है। कुल (42) मिलता है।
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\(\frac{8!}{6!}-\frac{5!}{4!}+2!\) का मान क्या है?
What is the value of \(\frac{8!}{6!}-\frac{5!}{4!}+2!\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (51)
B (52)
C (53)
D (54)
Explanation opens after your attempt
Step 1
Concept
\(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\), and (2!=2). Thus (56-5+2=53).
Step 2
Why this answer is correct
The correct answer is C. (53). \(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\), and (2!=2). Thus (56-5+2=53).
Step 3
Exam Tip
\(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\) और (2!=2)। इसलिए (56-5+2=53) है।
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यदि (\frac{(n+1)!}{(n-1)!}=20), तो (n) का मान क्या है?
If (\frac{(n+1)!}{(n-1)!}=20), what is the value of (n)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
This ratio is (n(n+1)). Since \(4\times5=20\), (n=4).
Step 2
Why this answer is correct
The correct answer is B. (4). This ratio is (n(n+1)). Since \(4\times5=20\), (n=4).
Step 3
Exam Tip
यह अनुपात (n(n+1)) है। \(4\times5=20\), इसलिए (n=4)।
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\(\frac{9!}{8!}+\frac{8!}{7!}-\frac{7!}{6!}\) का मान क्या है?
What is the value of \(\frac{9!}{8!}+\frac{8!}{7!}-\frac{7!}{6!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
The three ratios are (9), (8), and (7) respectively. Hence (9+8-7=10).
Step 2
Why this answer is correct
The correct answer is C. (10). The three ratios are (9), (8), and (7) respectively. Hence (9+8-7=10).
Step 3
Exam Tip
तीन अनुपात क्रमशः (9), (8) और (7) हैं। इसलिए (9+8-7=10) है।
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(\frac{(n+2)!}{(n-1)!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+2)!}{(n-1)!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A ((n+2)(n+1)n)
B ((n+2)(n+1))
C ((n+2)n)
D (n+2)
Explanation opens after your attempt
Correct Answer
A. ((n+2)(n+1)n)
Step 1
Concept
((n+2)!=(n+2)(n+1)n(n-1)!). Hence after cancelling ((n-1)!), three factors remain.
Step 2
Why this answer is correct
The correct answer is A. ((n+2)(n+1)n). ((n+2)!=(n+2)(n+1)n(n-1)!). Hence after cancelling ((n-1)!), three factors remain.
Step 3
Exam Tip
((n+2)!=(n+2)(n+1)n(n-1)!)। इसलिए ((n-1)!) कटने पर तीन गुणक बचते हैं।
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यदि \(a=\frac{8!}{6!}\) और (b=3!), तो (a-b) का मान क्या है?
If \(a=\frac{8!}{6!}\) and (b=3!), what is the value of (a-b)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (48)
B (50)
C (54)
D (56)
Explanation opens after your attempt
Step 1
Concept
\(a=8\times7=56\) and (b=6), so (a-b=50). Find both values separately first.
Step 2
Why this answer is correct
The correct answer is B. (50). \(a=8\times7=56\) and (b=6), so (a-b=50). Find both values separately first.
Step 3
Exam Tip
\(a=8\times7=56\) और (b=6), इसलिए (a-b=50)। पहले दोनों मान अलग निकालें।
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\(\frac{7!+5!}{6!}\) का मान क्या है?
What is the value of \(\frac{7!+5!}{6!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A \(7+\frac{1}{6}\)
B (8)
C \(\frac{43}{6}\)
D \(6+\frac{1}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(7+\frac{1}{6}\)
Step 1
Concept
\(\frac{7!}{6!}=7\) and \(\frac{5!}{6!}=\frac{1}{6}\), so the value is \(7+\frac{1}{6}\). Divide each term by the denominator separately.
Step 2
Why this answer is correct
The correct answer is A. \(7+\frac{1}{6}\). \(\frac{7!}{6!}=7\) and \(\frac{5!}{6!}=\frac{1}{6}\), so the value is \(7+\frac{1}{6}\). Divide each term by the denominator separately.
Step 3
Exam Tip
\(\frac{7!}{6!}=7\) और \(\frac{5!}{6!}=\frac{1}{6}\), इसलिए मान \(7+\frac{1}{6}\) है। पदों को हर से अलग-अलग भाग दें।
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\(\frac{10!}{7!,3!}\) का मान क्या है?
What is the value of \(\frac{10!}{7!,3!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (100)
B (110)
C (120)
D (130)
Explanation opens after your attempt
Step 1
Concept
\(\frac{10!}{7!,3!}=\frac{10\times9\times8}{6}=120\). Do not forget that (3!=6).
Step 2
Why this answer is correct
The correct answer is C. (120). \(\frac{10!}{7!,3!}=\frac{10\times9\times8}{6}=120\). Do not forget that (3!=6).
Step 3
Exam Tip
\(\frac{10!}{7!,3!}=\frac{10\times9\times8}{6}=120\)। (3!=6) रखना न भूलें।
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\(\frac{6!}{5!}\times\frac{5!}{3!}\) का मान क्या है?
What is the value of \(\frac{6!}{5!}\times\frac{5!}{3!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (90)
B (100)
C (120)
D (150)
Explanation opens after your attempt
Step 1
Concept
The first ratio is (6) and the second is \(5\times4=20\). The product is \(6\times20=120\).
Step 2
Why this answer is correct
The correct answer is C. (120). The first ratio is (6) and the second is \(5\times4=20\). The product is \(6\times20=120\).
Step 3
Exam Tip
पहला अनुपात (6) और दूसरा \(5\times4=20\) है। गुणनफल \(6\times20=120\) है।
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यदि (n(n-1)=72), तो (\frac{n!}{(n-2)!}) का मान क्या होगा?
If (n(n-1)=72), what will be the value of (\frac{n!}{(n-2)!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (36)
B (72)
C (144)
D (216)
Explanation opens after your attempt
Step 1
Concept
(\frac{n!}{(n-2)!}=n(n-1)). From the given condition, its value is (72).
Step 2
Why this answer is correct
The correct answer is B. (72). (\frac{n!}{(n-2)!}=n(n-1)). From the given condition, its value is (72).
Step 3
Exam Tip
(\frac{n!}{(n-2)!}=n(n-1))। दिए गए अनुसार इसका मान (72) है।
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\(\frac{12!}{10!,2!}-\frac{5!}{3!,2!}\) का मान क्या है?
What is the value of \(\frac{12!}{10!,2!}-\frac{5!}{3!,2!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (46)
B (50)
C (56)
D (60)
Explanation opens after your attempt
Step 1
Concept
\(\frac{12!}{10!,2!}=66\) and \(\frac{5!}{3!,2!}=10\), so the difference is (56). Keep the larger and smaller terms separate.
Step 2
Why this answer is correct
The correct answer is C. (56). \(\frac{12!}{10!,2!}=66\) and \(\frac{5!}{3!,2!}=10\), so the difference is (56). Keep the larger and smaller terms separate.
Step 3
Exam Tip
\(\frac{12!}{10!,2!}=66\) और \(\frac{5!}{3!,2!}=10\), इसलिए अंतर (56) है। बड़े और छोटे पद अलग रखें।
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\(\frac{4!+5!}{3!}\) का मान क्या है?
What is the value of \(\frac{4!+5!}{3!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (20)
B (24)
C (28)
D (30)
Explanation opens after your attempt
Step 1
Concept
The numerator is (24+120=144) and (3!=6). Therefore, the value is (24).
Step 2
Why this answer is correct
The correct answer is B. (24). The numerator is (24+120=144) and (3!=6). Therefore, the value is (24).
Step 3
Exam Tip
अंश (24+120=144) है और (3!=6)। इसलिए मान (24) है।
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\(\frac{7!}{5!}-\frac{6!}{4!}\) का मान क्या है?
What is the value of \(\frac{7!}{5!}-\frac{6!}{4!}\)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (12)
B (18)
C (24)
D (30)
Explanation opens after your attempt
Step 1
Concept
\(\frac{7!}{5!}=42\) and \(\frac{6!}{4!}=30\), so the difference is (12). Evaluate smaller ratios directly.
Step 2
Why this answer is correct
The correct answer is A. (12). \(\frac{7!}{5!}=42\) and \(\frac{6!}{4!}=30\), so the difference is (12). Evaluate smaller ratios directly.
Step 3
Exam Tip
\(\frac{7!}{5!}=42\) और \(\frac{6!}{4!}=30\), इसलिए अंतर (12) है। छोटे अनुपात सीधे निकालें।
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यदि (\frac{n!}{(n-3)!}=120), तो (n) का मान क्या है?
If (\frac{n!}{(n-3)!}=120), what is the value of (n)?
#factorial_notation
#permutations_combinations
#class_11
#medium
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
(\frac{n!}{(n-3)!}=n(n-1)(n-2)). Since \(6\times5\times4=120\), (n=6).
Step 2
Why this answer is correct
The correct answer is C. (6). (\frac{n!}{(n-3)!}=n(n-1)(n-2)). Since \(6\times5\times4=120\), (n=6).
Step 3
Exam Tip
(\frac{n!}{(n-3)!}=n(n-1)(n-2))। \(6\times5\times4=120\), इसलिए (n=6)।
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(\frac{(n+4)!}{(n+2)!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+4)!}{(n+2)!})?
#factorial_notation
#permutations_combinations
#class_11
#medium
A ((n+4)(n+3))
B ((n+4)(n+2))
C ((n+3)(n+2))
D (n+4)
Explanation opens after your attempt
Correct Answer
A. ((n+4)(n+3))
Step 1
Concept
((n+4)!=(n+4)(n+3)(n+2)!). After cancelling ((n+2)!), ((n+4)(n+3)) remains.
Step 2
Why this answer is correct
The correct answer is A. ((n+4)(n+3)). ((n+4)!=(n+4)(n+3)(n+2)!). After cancelling ((n+2)!), ((n+4)(n+3)) remains.
Step 3
Exam Tip
((n+4)!=(n+4)(n+3)(n+2)!)। ((n+2)!) कटने पर ((n+4)(n+3)) बचता है।
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