(7!) का मान क्या है?
What is the value of (7!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (720)
B (5040)
C (40320)
D (840)
Explanation opens after your attempt
Step 1
Concept
\(7!=7\times6\times5\times4\times3\times2\times1=5040\). Find a larger factorial by multiplying the previous factorial.
Step 2
Why this answer is correct
The correct answer is B. (5040). \(7!=7\times6\times5\times4\times3\times2\times1=5040\). Find a larger factorial by multiplying the previous factorial.
Step 3
Exam Tip
\(7!=7\times6\times5\times4\times3\times2\times1=5040\)। बड़े क्रमगुणित को पिछले क्रमगुणित से गुणा करके निकालें।
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(8!) को (6!) की सहायता से कैसे लिखेंगे?
How can (8!) be written using (6!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A \(8\times7\times6!\)
B (8+7+6!)
C \(8\times6!\)
D \(7\times6!\)
Explanation opens after your attempt
Correct Answer
A. \(8\times7\times6!\)
Step 1
Concept
\(8!=8\times7\times6!\). Breaking a factorial down to a smaller factorial is an easy method.
Step 2
Why this answer is correct
The correct answer is A. \(8\times7\times6!\). \(8!=8\times7\times6!\). Breaking a factorial down to a smaller factorial is an easy method.
Step 3
Exam Tip
\(8!=8\times7\times6!\) होता है। क्रमगुणित को छोटे क्रमगुणित तक तोड़ना आसान तरीका है।
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\(\frac{9!}{7!}\) का मान क्या है?
What is the value of \(\frac{9!}{7!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (63)
B (72)
C (90)
D (126)
Explanation opens after your attempt
Step 1
Concept
\(\frac{9!}{7!}=9\times8=72\). Cancel the common (7!) part.
Step 2
Why this answer is correct
The correct answer is B. (72). \(\frac{9!}{7!}=9\times8=72\). Cancel the common (7!) part.
Step 3
Exam Tip
\(\frac{9!}{7!}=9\times8=72\)। समान (7!) भाग को काट दें।
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(6!+1!) का मान क्या होगा?
What will be the value of (6!+1!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (720)
B (721)
C (719)
D (722)
Explanation opens after your attempt
Step 1
Concept
(6!=720) and (1!=1), so the sum is (721). Find both factorials before adding.
Step 2
Why this answer is correct
The correct answer is B. (721). (6!=720) and (1!=1), so the sum is (721). Find both factorials before adding.
Step 3
Exam Tip
(6!=720) और (1!=1), इसलिए योग (721) है। जोड़ से पहले दोनों क्रमगुणित निकालें।
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\(\frac{8!}{6!,2!}\) का मान क्या है?
What is the value of \(\frac{8!}{6!,2!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (14)
B (28)
C (56)
D (64)
Explanation opens after your attempt
Step 1
Concept
\(\frac{8!}{6!,2!}=\frac{8\times7}{2}=28\). Expand the larger factorial up to the smaller one first.
Step 2
Why this answer is correct
The correct answer is B. (28). \(\frac{8!}{6!,2!}=\frac{8\times7}{2}=28\). Expand the larger factorial up to the smaller one first.
Step 3
Exam Tip
\(\frac{8!}{6!,2!}=\frac{8\times7}{2}=28\)। पहले बड़े क्रमगुणित को छोटे तक फैलाएं।
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यदि (n=4), तो (n!) का मान क्या है?
If (n=4), what is the value of (n!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (16)
B (20)
C (24)
D (32)
Explanation opens after your attempt
Step 1
Concept
Putting (n=4), (n!=4!=24). Substitute the variable first and then evaluate the factorial.
Step 2
Why this answer is correct
The correct answer is C. (24). Putting (n=4), (n!=4!=24). Substitute the variable first and then evaluate the factorial.
Step 3
Exam Tip
(n=4) रखने पर (n!=4!=24)। पहले चर का मान रखकर फिर क्रमगुणित निकालें।
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(2!+4!-1!) का मान क्या है?
What is the value of (2!+4!-1!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (23)
B (24)
C (25)
D (26)
Explanation opens after your attempt
Step 1
Concept
(2!=2), (4!=24), and (1!=1), so the value is (25). Convert factorials into ordinary numbers.
Step 2
Why this answer is correct
The correct answer is C. (25). (2!=2), (4!=24), and (1!=1), so the value is (25). Convert factorials into ordinary numbers.
Step 3
Exam Tip
(2!=2), (4!=24) और (1!=1), इसलिए मान (25) है। क्रमगुणित को साधारण संख्याओं में बदलें।
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\(\frac{10!}{9!}\) का मान क्या है?
What is the value of \(\frac{10!}{9!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (9)
B (10)
C (19)
D (90)
Explanation opens after your attempt
Step 1
Concept
\(\frac{10!}{9!}=10\) because \(10!=10\times9!\). In such questions, full expansion is not needed.
Step 2
Why this answer is correct
The correct answer is B. (10). \(\frac{10!}{9!}=10\) because \(10!=10\times9!\). In such questions, full expansion is not needed.
Step 3
Exam Tip
\(\frac{10!}{9!}=10\) क्योंकि \(10!=10\times9!\)। ऐसे सवाल में पूरा विस्तार लिखने की जरूरत नहीं होती।
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\(4!\times3!\) का मान क्या है?
What is the value of \(4!\times3!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (120)
B (132)
C (144)
D (156)
Explanation opens after your attempt
Step 1
Concept
(4!=24) and (3!=6), so the product is (144). First evaluate each factorial separately.
Step 2
Why this answer is correct
The correct answer is C. (144). (4!=24) and (3!=6), so the product is (144). First evaluate each factorial separately.
Step 3
Exam Tip
(4!=24) और (3!=6), इसलिए गुणनफल (144) है। पहले प्रत्येक क्रमगुणित अलग निकालें।
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\(6\times5\times4\times3!\) किसके बराबर है?
\(6\times5\times4\times3!\) is equal to which of the following?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (6!)
B (5!)
C (4!)
D (7!)
Explanation opens after your attempt
Step 1
Concept
\(6!=6\times5\times4\times3!\), so the given form is (6!). Recognizing expansions is important in factorial questions.
Step 2
Why this answer is correct
The correct answer is A. (6!). \(6!=6\times5\times4\times3!\), so the given form is (6!). Recognizing expansions is important in factorial questions.
Step 3
Exam Tip
\(6!=6\times5\times4\times3!\), इसलिए दिया गया रूप (6!) है। विस्तार पहचानना क्रमगुणित प्रश्नों में जरूरी है।
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\(\frac{5!+4!}{4!}\) का मान क्या है?
What is the value of \(\frac{5!+4!}{4!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(\frac{5!+4!}{4!}=\frac{120+24}{24}=6\). Add the numerator correctly and then divide.
Step 2
Why this answer is correct
The correct answer is B. (6). \(\frac{5!+4!}{4!}=\frac{120+24}{24}=6\). Add the numerator correctly and then divide.
Step 3
Exam Tip
\(\frac{5!+4!}{4!}=\frac{120+24}{24}=6\)। अंश को सही तरह से जोड़कर भाग दें।
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यदि (n!=720), तो (n) का मान क्या है?
If (n!=720), what is the value of (n)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
(6!=720), so (n=6). Remembering small factorial values helps answer quickly.
Step 2
Why this answer is correct
The correct answer is B. (6). (6!=720), so (n=6). Remembering small factorial values helps answer quickly.
Step 3
Exam Tip
(6!=720), इसलिए (n=6)। छोटे क्रमगुणित मान याद रखने से उत्तर जल्दी मिलता है।
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((2+3)!) का मान क्या है?
What is the value of ((2+3)!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (24)
B (60)
C (120)
D (720)
Explanation opens after your attempt
Step 1
Concept
First (2+3=5), then (5!=120). Evaluate the bracket first.
Step 2
Why this answer is correct
The correct answer is C. (120). First (2+3=5), then (5!=120). Evaluate the bracket first.
Step 3
Exam Tip
पहले (2+3=5), फिर (5!=120)। कोष्ठक की गणना पहले करें।
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(7!-6!) का मान क्या है?
What is the value of (7!-6!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (4320)
B (5040)
C (720)
D (3600)
Explanation opens after your attempt
Step 1
Concept
(7!=5040) and (6!=720), so the difference is (4320). Write both factorials separately while subtracting.
Step 2
Why this answer is correct
The correct answer is A. (4320). (7!=5040) and (6!=720), so the difference is (4320). Write both factorials separately while subtracting.
Step 3
Exam Tip
(7!=5040) और (6!=720), इसलिए अंतर (4320) है। घटाते समय दोनों क्रमगुणित अलग लिखें।
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\(\frac{6!}{5!}+2!\) का मान क्या है?
What is the value of \(\frac{6!}{5!}+2!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\frac{6!}{5!}=6\) and (2!=2), so the total is (8). Simplify the quotient first.
Step 2
Why this answer is correct
The correct answer is C. (8). \(\frac{6!}{5!}=6\) and (2!=2), so the total is (8). Simplify the quotient first.
Step 3
Exam Tip
\(\frac{6!}{5!}=6\) और (2!=2), इसलिए कुल (8) है। पहले भाग को सरल करें।
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\(3!\times3!\) का मान क्या है?
What is the value of \(3!\times3!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (18)
B (24)
C (30)
D (36)
Explanation opens after your attempt
Step 1
Concept
(3!=6), so \(3!\times3!=6\times6=36\). Evaluate the repeated factorial once first.
Step 2
Why this answer is correct
The correct answer is D. (36). (3!=6), so \(3!\times3!=6\times6=36\). Evaluate the repeated factorial once first.
Step 3
Exam Tip
(3!=6), इसलिए \(3!\times3!=6\times6=36\)। समान क्रमगुणित को पहले एक बार निकालें।
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\(\frac{7!}{4!}\) का मान क्या है?
What is the value of \(\frac{7!}{4!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (210)
B (240)
C (280)
D (320)
Explanation opens after your attempt
Step 1
Concept
\(\frac{7!}{4!}=7\times6\times5=210\). Cancel the smaller factorial to reduce calculation.
Step 2
Why this answer is correct
The correct answer is A. (210). \(\frac{7!}{4!}=7\times6\times5=210\). Cancel the smaller factorial to reduce calculation.
Step 3
Exam Tip
\(\frac{7!}{4!}=7\times6\times5=210\)। छोटे क्रमगुणित को काटकर गणना कम करें।
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(1!+2!+3!) का मान क्या है?
What is the value of (1!+2!+3!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
(1!=1), (2!=2), and (3!=6), so the sum is (9). Evaluate factorials before adding them.
Step 2
Why this answer is correct
The correct answer is B. (9). (1!=1), (2!=2), and (3!=6), so the sum is (9). Evaluate factorials before adding them.
Step 3
Exam Tip
(1!=1), (2!=2) और (3!=6), इसलिए योग (9) है। क्रमगुणित को जोड़ने से पहले उनका मान निकालें।
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(\frac{(n+3)!}{(n+2)!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+3)!}{(n+2)!})?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (n+1)
B (n+2)
C (n+3)
D (1)
Explanation opens after your attempt
Step 1
Concept
((n+3)!=(n+3)(n+2)!), so division leaves (n+3). Nearby factorials cancel directly.
Step 2
Why this answer is correct
The correct answer is C. (n+3). ((n+3)!=(n+3)(n+2)!), so division leaves (n+3). Nearby factorials cancel directly.
Step 3
Exam Tip
((n+3)!=(n+3)(n+2)!), इसलिए भाग देने पर (n+3) बचेगा। पास-पास वाले क्रमगुणित सीधे कटते हैं।
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\(8\times7\times6\times5!\) किसके बराबर है?
\(8\times7\times6\times5!\) is equal to which of the following?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (7!)
B (8!)
C (9!)
D (6!)
Explanation opens after your attempt
Step 1
Concept
\(8!=8\times7\times6\times5!\). Match the factorial expansion with the last factorial part.
Step 2
Why this answer is correct
The correct answer is B. (8!). \(8!=8\times7\times6\times5!\). Match the factorial expansion with the last factorial part.
Step 3
Exam Tip
\(8!=8\times7\times6\times5!\) होता है। क्रमगुणित विस्तार को अंतिम फैक्टोरियल से जोड़ें।
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\(\frac{4!+2!}{2!}\) का मान क्या है?
What is the value of \(\frac{4!+2!}{2!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (11)
B (12)
C (13)
D (14)
Explanation opens after your attempt
Step 1
Concept
\(\frac{4!+2!}{2!}=\frac{24+2}{2}=13\). Complete the addition in the numerator first.
Step 2
Why this answer is correct
The correct answer is C. (13). \(\frac{4!+2!}{2!}=\frac{24+2}{2}=13\). Complete the addition in the numerator first.
Step 3
Exam Tip
\(\frac{4!+2!}{2!}=\frac{24+2}{2}=13\)। जोड़ वाले अंश को पहले पूरा करें।
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\(5!\div3!\) का मान क्या है?
What is the value of \(5!\div3!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (10)
B (15)
C (20)
D (25)
Explanation opens after your attempt
Step 1
Concept
\(5!\div3!=\frac{5!}{3!}=5\times4=20\). Write division as a fraction to solve.
Step 2
Why this answer is correct
The correct answer is C. (20). \(5!\div3!=\frac{5!}{3!}=5\times4=20\). Write division as a fraction to solve.
Step 3
Exam Tip
\(5!\div3!=\frac{5!}{3!}=5\times4=20\)। भाग को भिन्न के रूप में लिखकर हल करें।
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यदि (a=3!) और (b=2!), तो (a+b) का मान क्या है?
If (a=3!) and (b=2!), what is the value of (a+b)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
(a=6) and (b=2), so (a+b=8). In variable questions, find the values first.
Step 2
Why this answer is correct
The correct answer is C. (8). (a=6) and (b=2), so (a+b=8). In variable questions, find the values first.
Step 3
Exam Tip
(a=6) और (b=2), इसलिए (a+b=8)। चर वाले प्रश्नों में पहले मान निकालें।
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\(\frac{9!}{6!}\) का सरल मान क्या है?
What is the simplified value of \(\frac{9!}{6!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (504)
B (432)
C (360)
D (3024)
Explanation opens after your attempt
Step 1
Concept
\(\frac{9!}{6!}=9\times8\times7=504\). Expand down to the smaller factorial and cancel.
Step 2
Why this answer is correct
The correct answer is A. (504). \(\frac{9!}{6!}=9\times8\times7=504\). Expand down to the smaller factorial and cancel.
Step 3
Exam Tip
\(\frac{9!}{6!}=9\times8\times7=504\)। छोटे क्रमगुणित तक विस्तार करें और काटें।
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(0!+4!) का मान क्या है?
What is the value of (0!+4!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (24)
B (25)
C (26)
D (23)
Explanation opens after your attempt
Step 1
Concept
(0!=1) and (4!=24), so the sum is (25). Remember to take (0!) as (1).
Step 2
Why this answer is correct
The correct answer is B. (25). (0!=1) and (4!=24), so the sum is (25). Remember to take (0!) as (1).
Step 3
Exam Tip
(0!=1) और (4!=24), इसलिए योग (25) है। (0!) को (1) लेना याद रखें।
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(10!) को (8!) की सहायता से कैसे लिखेंगे?
How can (10!) be written using (8!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A \(10\times8!\)
B \(10\times9\times8!\)
C \(9\times8!\)
D (10+9+8!)
Explanation opens after your attempt
Correct Answer
B. \(10\times9\times8!\)
Step 1
Concept
\(10!=10\times9\times8!\). Write the larger factorial in terms of the smaller factorial.
Step 2
Why this answer is correct
The correct answer is B. \(10\times9\times8!\). \(10!=10\times9\times8!\). Write the larger factorial in terms of the smaller factorial.
Step 3
Exam Tip
\(10!=10\times9\times8!\) होता है। बड़े क्रमगुणित को छोटे क्रमगुणित से जोड़कर लिखें।
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(\frac{(n+1)!}{(n-1)!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+1)!}{(n-1)!})?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (n+1)
B (n(n+1))
C (n-1)
D ((n+1)(n-1))
Explanation opens after your attempt
Correct Answer
B. (n(n+1))
Step 1
Concept
((n+1)!=(n+1)n(n-1)!), so the remaining part is (n(n+1)). Cancel the common factorial part.
Step 2
Why this answer is correct
The correct answer is B. (n(n+1)). ((n+1)!=(n+1)n(n-1)!), so the remaining part is (n(n+1)). Cancel the common factorial part.
Step 3
Exam Tip
((n+1)!=(n+1)n(n-1)!), इसलिए शेष (n(n+1)) है। समान क्रमगुणित भाग काटें।
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\(2!\times5!\) का मान क्या है?
What is the value of \(2!\times5!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (120)
B (180)
C (240)
D (300)
Explanation opens after your attempt
Step 1
Concept
(2!=2) and (5!=120), so the product is (240). First find the factorial values.
Step 2
Why this answer is correct
The correct answer is C. (240). (2!=2) and (5!=120), so the product is (240). First find the factorial values.
Step 3
Exam Tip
(2!=2) और (5!=120), इसलिए गुणनफल (240) है। पहले क्रमगुणित मान निकालें।
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\(\frac{6!}{4!}+0!\) का मान क्या है?
What is the value of \(\frac{6!}{4!}+0!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (29)
B (30)
C (31)
D (32)
Explanation opens after your attempt
Step 1
Concept
\(\frac{6!}{4!}=6\times5=30\) and (0!=1), so the total is (31). Use the correct value of (0!).
Step 2
Why this answer is correct
The correct answer is C. (31). \(\frac{6!}{4!}=6\times5=30\) and (0!=1), so the total is (31). Use the correct value of (0!).
Step 3
Exam Tip
\(\frac{6!}{4!}=6\times5=30\) और (0!=1), इसलिए कुल (31) है। (0!) को सही मान दें।
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(4!) और (3!) का अनुपात क्या है?
What is the ratio of (4!) to (3!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (4:1)
B (3:1)
C (2:1)
D (1:4)
Explanation opens after your attempt
Step 1
Concept
\(\frac{4!}{3!}=4\), so the ratio is (4:1). For the ratio, cancel both terms by (3!).
Step 2
Why this answer is correct
The correct answer is A. (4:1). \(\frac{4!}{3!}=4\), so the ratio is (4:1). For the ratio, cancel both terms by (3!).
Step 3
Exam Tip
\(\frac{4!}{3!}=4\), इसलिए अनुपात (4:1) है। अनुपात के लिए दोनों पदों को (3!) से काटें।
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(5!-3!) का मान क्या है?
What is the value of (5!-3!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (108)
B (114)
C (120)
D (126)
Explanation opens after your attempt
Step 1
Concept
(5!=120) and (3!=6), so the difference is (114). Find factorial values before subtracting.
Step 2
Why this answer is correct
The correct answer is B. (114). (5!=120) and (3!=6), so the difference is (114). Find factorial values before subtracting.
Step 3
Exam Tip
(5!=120) और (3!=6), इसलिए अंतर (114) है। घटाव करते समय क्रमगुणित का मान पहले निकालें।
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\(\frac{3!+0!}{1!}\) का मान क्या है?
What is the value of \(\frac{3!+0!}{1!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
(3!+0!=6+1=7) and (1!=1), so the value is (7). Both (0!) and (1!) are (1).
Step 2
Why this answer is correct
The correct answer is B. (7). (3!+0!=6+1=7) and (1!=1), so the value is (7). Both (0!) and (1!) are (1).
Step 3
Exam Tip
(3!+0!=6+1=7) और (1!=1), इसलिए मान (7) है। (0!) और (1!) दोनों (1) होते हैं।
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किसका मान (24) है?
Which one has value (24)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (3!)
B (4!)
C (5!)
D (6!)
Explanation opens after your attempt
Step 1
Concept
\(4!=4\times3\times2\times1=24\). Remembering small factorial values is useful.
Step 2
Why this answer is correct
The correct answer is B. (4!). \(4!=4\times3\times2\times1=24\). Remembering small factorial values is useful.
Step 3
Exam Tip
\(4!=4\times3\times2\times1=24\)। छोटे क्रमगुणित मान याद रखना उपयोगी है।
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(9!) को (10!) की सहायता से कैसे लिखेंगे?
How can (9!) be written using (10!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A \(10\times10!\)
B \(\frac{10!}{10}\)
C \(\frac{10!}{9}\)
D (10!-1)
Explanation opens after your attempt
Correct Answer
B. \(\frac{10!}{10}\)
Step 1
Concept
\(10!=10\times9!\), so \(9!=\frac{10!}{10}\). The reverse relation is also useful in factorials.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10!}{10}\). \(10!=10\times9!\), so \(9!=\frac{10!}{10}\). The reverse relation is also useful in factorials.
Step 3
Exam Tip
\(10!=10\times9!\), इसलिए \(9!=\frac{10!}{10}\)। उलटा संबंध भी क्रमगुणित में काम आता है।
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\(\frac{8!}{4!,4!}\) का मान क्या है?
What is the value of \(\frac{8!}{4!,4!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (56)
B (70)
C (84)
D (96)
Explanation opens after your attempt
Step 1
Concept
\(\frac{8!}{4!,4!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\). Cancel factorials to make the numbers smaller.
Step 2
Why this answer is correct
The correct answer is B. (70). \(\frac{8!}{4!,4!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\). Cancel factorials to make the numbers smaller.
Step 3
Exam Tip
\(\frac{8!}{4!,4!}=\frac{8\times7\times6\times5}{4\times3\times2\times1}=70\)। क्रमगुणित काटकर संख्या छोटी करें।
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(6!) में (6) से (1) तक कितने गुणक होते हैं?
How many factors from (6) to (1) are present in (6!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(6!=6\times5\times4\times3\times2\times1\), which has (6) factors. (n!) has (n) factors.
Step 2
Why this answer is correct
The correct answer is B. (6). \(6!=6\times5\times4\times3\times2\times1\), which has (6) factors. (n!) has (n) factors.
Step 3
Exam Tip
\(6!=6\times5\times4\times3\times2\times1\), इसमें (6) गुणक हैं। (n!) में (n) गुणक होते हैं।
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\(\frac{7!}{5!}+1!\) का मान क्या है?
What is the value of \(\frac{7!}{5!}+1!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (42)
B (43)
C (44)
D (45)
Explanation opens after your attempt
Step 1
Concept
\(\frac{7!}{5!}=7\times6=42\) and (1!=1), so the total is (43). Simplify the quotient first.
Step 2
Why this answer is correct
The correct answer is B. (43). \(\frac{7!}{5!}=7\times6=42\) and (1!=1), so the total is (43). Simplify the quotient first.
Step 3
Exam Tip
\(\frac{7!}{5!}=7\times6=42\) और (1!=1), इसलिए कुल (43) है। पहले भाग को सरल करें।
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\(3!\times4!-4!\) का मान क्या है?
What is the value of \(3!\times4!-4!\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (100)
B (108)
C (120)
D (144)
Explanation opens after your attempt
Step 1
Concept
\(3!\times4!-4!=6\times24-24=120\). You may also factor out the common term (4!).
Step 2
Why this answer is correct
The correct answer is C. (120). \(3!\times4!-4!=6\times24-24=120\). You may also factor out the common term (4!).
Step 3
Exam Tip
\(3!\times4!-4!=6\times24-24=120\)। समान पद (4!) को अलग लेकर भी हल कर सकते हैं।
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यदि (x=5!) और (y=4!), तो (x-y) का मान क्या है?
If (x=5!) and (y=4!), what is the value of (x-y)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (96)
B (100)
C (108)
D (120)
Explanation opens after your attempt
Step 1
Concept
(x=120) and (y=24), so (x-y=96). Replace variables by their factorial values first.
Step 2
Why this answer is correct
The correct answer is A. (96). (x=120) and (y=24), so (x-y=96). Replace variables by their factorial values first.
Step 3
Exam Tip
(x=120) और (y=24), इसलिए (x-y=96)। चर को पहले उनके क्रमगुणित मान से बदलें।
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((4-2)!) का मान क्या है?
What is the value of ((4-2)!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (1)
B (2)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
First (4-2=2), then (2!=2). Solve the bracket first.
Step 2
Why this answer is correct
The correct answer is B. (2). First (4-2=2), then (2!=2). Solve the bracket first.
Step 3
Exam Tip
पहले (4-2=2), फिर (2!=2)। कोष्ठक को पहले हल करें।
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(5!) को (3!) की सहायता से कैसे लिखेंगे?
How can (5!) be written using (3!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A \(5\times4\times3!\)
B \(5\times3!\)
C \(4\times3!\)
D (5+4+3!)
Explanation opens after your attempt
Correct Answer
A. \(5\times4\times3!\)
Step 1
Concept
\(5!=5\times4\times3!\). It is correct to stop the expansion and write a smaller factorial.
Step 2
Why this answer is correct
The correct answer is A. \(5\times4\times3!\). \(5!=5\times4\times3!\). It is correct to stop the expansion and write a smaller factorial.
Step 3
Exam Tip
\(5!=5\times4\times3!\) होता है। बीच में रुककर छोटा क्रमगुणित लिखना सही है।
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\(\frac{11!}{10!}\) का मान क्या है?
What is the value of \(\frac{11!}{10!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (10)
B (11)
C (21)
D (110)
Explanation opens after your attempt
Step 1
Concept
\(\frac{11!}{10!}=11\) because \(11!=11\times10!\). Consecutive factorials cancel easily.
Step 2
Why this answer is correct
The correct answer is B. (11). \(\frac{11!}{10!}=11\) because \(11!=11\times10!\). Consecutive factorials cancel easily.
Step 3
Exam Tip
\(\frac{11!}{10!}=11\) क्योंकि \(11!=11\times10!\)। लगातार क्रमगुणित आसानी से कटते हैं।
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(6!+0!-1!) का मान क्या है?
What is the value of (6!+0!-1!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (718)
B (719)
C (720)
D (721)
Explanation opens after your attempt
Step 1
Concept
(6!=720), (0!=1), and (1!=1), so the value is (720). (0!) and (1!) are equal.
Step 2
Why this answer is correct
The correct answer is C. (720). (6!=720), (0!=1), and (1!=1), so the value is (720). (0!) and (1!) are equal.
Step 3
Exam Tip
(6!=720), (0!=1) और (1!=1), इसलिए मान (720) है। (0!) और (1!) दोनों बराबर हैं।
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(\frac{(n+4)!}{(n+3)!}) का सरल रूप क्या है?
What is the simplified form of (\frac{(n+4)!}{(n+3)!})?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (n+2)
B (n+3)
C (n+4)
D (1)
Explanation opens after your attempt
Step 1
Concept
((n+4)!=(n+4)(n+3)!), so the simplified form is (n+4). Cancel the adjacent factorial part.
Step 2
Why this answer is correct
The correct answer is C. (n+4). ((n+4)!=(n+4)(n+3)!), so the simplified form is (n+4). Cancel the adjacent factorial part.
Step 3
Exam Tip
((n+4)!=(n+4)(n+3)!), इसलिए सरल रूप (n+4) है। क्रमगुणित में पास वाला भाग काटें।
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\(4!\div2!\) और (3!) का योग क्या है?
What is the sum of \(4!\div2!\) and (3!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (16)
B (18)
C (20)
D (24)
Explanation opens after your attempt
Step 1
Concept
\(4!\div2!=12\) and (3!=6), so the sum is (18). Simplify the division and factorial separately.
Step 2
Why this answer is correct
The correct answer is B. (18). \(4!\div2!=12\) and (3!=6), so the sum is (18). Simplify the division and factorial separately.
Step 3
Exam Tip
\(4!\div2!=12\) और (3!=6), इसलिए योग (18) है। भाग और क्रमगुणित दोनों को अलग-अलग सरल करें।
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\(9\times8!\) किसके बराबर है?
\(9\times8!\) is equal to which of the following?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (7!)
B (8!)
C (9!)
D (10!)
Explanation opens after your attempt
Step 1
Concept
\(9!=9\times8!\), so the given form is (9!). This is a basic factorial identity.
Step 2
Why this answer is correct
The correct answer is C. (9!). \(9!=9\times8!\), so the given form is (9!). This is a basic factorial identity.
Step 3
Exam Tip
\(9!=9\times8!\), इसलिए दिया गया रूप (9!) है। यह क्रमगुणित की मूल पहचान है।
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\(\frac{5!}{2!,3!}\) का मान क्या है?
What is the value of \(\frac{5!}{2!,3!}\)?
#factorial_notation
#permutations_combinations
#class_11
#easy
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A (6)
B (8)
C (10)
D (12)
Explanation opens after your attempt
Step 1
Concept
\(\frac{5!}{2!,3!}=\frac{120}{2\times6}=10\). Keep the correct value of each factorial.
Step 2
Why this answer is correct
The correct answer is C. (10). \(\frac{5!}{2!,3!}=\frac{120}{2\times6}=10\). Keep the correct value of each factorial.
Step 3
Exam Tip
\(\frac{5!}{2!,3!}=\frac{120}{2\times6}=10\)। हर क्रमगुणित का सही मान रखें।
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(2!+5!) का मान क्या है?
What is the value of (2!+5!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
50 50-50 2 wrong hide
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A (120)
B (121)
C (122)
D (124)
Explanation opens after your attempt
Step 1
Concept
(2!=2) and (5!=120), so the sum is (122). Do not forget the smaller factorial while adding.
Step 2
Why this answer is correct
The correct answer is C. (122). (2!=2) and (5!=120), so the sum is (122). Do not forget the smaller factorial while adding.
Step 3
Exam Tip
(2!=2) और (5!=120), इसलिए योग (122) है। जोड़ते समय छोटे क्रमगुणित को न भूलें।
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(7!) को (5!) की सहायता से कैसे लिखेंगे?
How can (7!) be written using (5!)?
#factorial_notation
#permutations_combinations
#class_11
#easy
50 50-50 2 wrong hide
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A \(7\times5!\)
B \(6\times5!\)
C (7+6+5!)
D \(7\times6\times5!\)
Explanation opens after your attempt
Correct Answer
D. \(7\times6\times5!\)
Step 1
Concept
\(7!=7\times6\times5!\). Expand the larger factorial down to the smaller factorial.
Step 2
Why this answer is correct
The correct answer is D. \(7\times6\times5!\). \(7!=7\times6\times5!\). Expand the larger factorial down to the smaller factorial.
Step 3
Exam Tip
\(7!=7\times6\times5!\) होता है। बड़े क्रमगुणित को छोटे क्रमगुणित तक विस्तार करें।
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\(12\times11\times10!\) किसके बराबर है?
\(12\times11\times10!\) is equal to which of the following?
#factorial_notation
#permutations_combinations
#class_11
#easy
50 50-50 2 wrong hide
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A (10!)
B (11!)
C (12!)
D (13!)
Explanation opens after your attempt
Step 1
Concept
\(12!=12\times11\times10!\), so the given form is (12!). Recognize factorial expansion down to a smaller factorial.
Step 2
Why this answer is correct
The correct answer is C. (12!). \(12!=12\times11\times10!\), so the given form is (12!). Recognize factorial expansion down to a smaller factorial.
Step 3
Exam Tip
\(12!=12\times11\times10!\), इसलिए दिया गया रूप (12!) है। क्रमगुणित विस्तार को छोटे क्रमगुणित तक पहचानें।
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