(5) अलग-अलग चाबियों को एक की-रिंग में सीधी सूची के रूप में लिखने पर कितने क्रम होंगे?
If (5) different keys are written as a straight list, how many orders are possible?
#permutations
#class11
#easy
A (25)
B (60)
C (100)
D (120)
Explanation opens after your attempt
Step 1
Concept
In a straight list, the arrangement of (5) distinct objects is (5!=120). Do not use circular counting here.
Step 2
Why this answer is correct
The correct answer is D. (120). In a straight list, the arrangement of (5) distinct objects is (5!=120). Do not use circular counting here.
Step 3
Exam Tip
सीधी सूची में (5) अलग वस्तुओं की व्यवस्था (5!=120) है। यहाँ circular counting नहीं करनी है।
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\({}^{3}P_{2}\) का मान क्या है?
What is the value of \({}^{3}P_{2}\)?
#permutations
#class11
#easy
A (3)
B (9)
C (6)
D (2)
Explanation opens after your attempt
Step 1
Concept
\({}^{3}P_{2}=3\times2=6\). In small permutations, write decreasing factors directly.
Step 2
Why this answer is correct
The correct answer is C. (6). \({}^{3}P_{2}=3\times2=6\). In small permutations, write decreasing factors directly.
Step 3
Exam Tip
\({}^{3}P_{2}=3\times2=6\) है। छोटे permutation में सीधे घटते गुणनखंड लिखें।
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(9) बच्चों में से (3) बच्चों को मंच पर क्रम से खड़ा करने के तरीके कितने हैं?
How many ways are there to make (3) children from (9) children stand on a stage in order?
#permutations
#class11
#easy
A (81)
B (504)
C (84)
D (729)
Explanation opens after your attempt
Step 1
Concept
The answer is \({}^{9}P_{3}=9\times8\times7=504\). Left-right order on a stage is important.
Step 2
Why this answer is correct
The correct answer is B. (504). The answer is \({}^{9}P_{3}=9\times8\times7=504\). Left-right order on a stage is important.
Step 3
Exam Tip
उत्तर \({}^{9}P_{3}=9\times8\times7=504\) है। मंच पर बायाँ-दायाँ क्रम महत्त्वपूर्ण होता है।
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यदि (3) अलग-अलग खिलौनों को (3) बच्चों के सामने एक पंक्ति में रखा जाए तो व्यवस्थाएँ कितनी होंगी?
If (3) different toys are placed in a row before (3) children, how many arrangements are possible?
#permutations
#class11
#easy
A (6)
B (3)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
The arrangements of (3) distinct toys are (3!=6). If all objects are distinct, total arrangements are factorial.
Step 2
Why this answer is correct
The correct answer is A. (6). The arrangements of (3) distinct toys are (3!=6). If all objects are distinct, total arrangements are factorial.
Step 3
Exam Tip
(3) अलग खिलौनों की arrangements (3!=6) हैं। सभी objects अलग हों तो total arrangements factorial होती हैं।
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किसी परीक्षा में (5) विद्यार्थियों में से प्रथम, द्वितीय और तृतीय स्थान कितने तरीकों से मिल सकते हैं?
In an exam, in how many ways can first, second and third ranks be obtained among (5) students?
#permutations
#class11
#easy
A (15)
B (25)
C (30)
D (60)
Explanation opens after your attempt
Step 1
Concept
The three ranks are ordered, so \({}^{5}P_{3}=60\). In rank questions, use permutation, not combination.
Step 2
Why this answer is correct
The correct answer is D. (60). The three ranks are ordered, so \({}^{5}P_{3}=60\). In rank questions, use permutation, not combination.
Step 3
Exam Tip
तीन rank क्रम वाले हैं इसलिए \({}^{5}P_{3}=60\)। rank वाले प्रश्नों में combination नहीं, permutation लें।
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शब्द (LEVEL) में (5) अक्षर हैं, (L) दो बार और (E) दो बार आता है। अलग व्यवस्थाएँ कितनी होंगी?
The word (LEVEL) has (5) letters, with (L) twice and (E) twice. How many distinct arrangements are possible?
#permutations
#class11
#easy
A (60)
B (20)
C (30)
D (120)
Explanation opens after your attempt
Step 1
Concept
The arrangements are \(\frac{5!}{2!2!}=30\). Do not forget to divide by factorials of identical letters.
Step 2
Why this answer is correct
The correct answer is C. (30). The arrangements are \(\frac{5!}{2!2!}=30\). Do not forget to divide by factorials of identical letters.
Step 3
Exam Tip
व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। समान अक्षरों के factorial से भाग देना न भूलें।
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(7) अलग-अलग फाइलों में से (3) फाइलों को शेल्फ पर क्रम में रखने के तरीके कितने हैं?
How many ways are there to place (3) files from (7) different files on a shelf in order?
#permutations
#class11
#easy
A (35)
B (210)
C (343)
D (21)
Explanation opens after your attempt
Step 1
Concept
This is \({}^{7}P_{3}=7\times6\times5=210\). Changing order on a shelf changes the arrangement.
Step 2
Why this answer is correct
The correct answer is B. (210). This is \({}^{7}P_{3}=7\times6\times5=210\). Changing order on a shelf changes the arrangement.
Step 3
Exam Tip
यह \({}^{7}P_{3}=7\times6\times5=210\) है। शेल्फ पर क्रम बदलने से व्यवस्था बदलती है।
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अंकों (3,4,5,6) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बन सकती हैं?
How many (3)-digit numbers can be formed from digits (3,4,5,6) without repetition?
#permutations
#class11
#easy
A (24)
B (12)
C (64)
D (36)
Explanation opens after your attempt
Step 1
Concept
For three places, there are \(4\times3\times2=24\) ways. Without repetition, choices decrease at each next place.
Step 2
Why this answer is correct
The correct answer is A. (24). For three places, there are \(4\times3\times2=24\) ways. Without repetition, choices decrease at each next place.
Step 3
Exam Tip
तीन स्थानों के लिए \(4\times3\times2=24\) तरीके हैं। बिना पुनरावृत्ति में हर अगले स्थान पर विकल्प घटता है।
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\({}^{4}P_{4}\) का मान क्या है?
What is the value of \({}^{4}P_{4}\)?
#permutations
#class11
#easy
A (8)
B (12)
C (16)
D (24)
Explanation opens after your attempt
Step 1
Concept
\({}^{4}P_{4}=4!=24\). When (r=n), \({}^{n}P_{n}=n!\).
Step 2
Why this answer is correct
The correct answer is D. (24). \({}^{4}P_{4}=4!=24\). When (r=n), \({}^{n}P_{n}=n!\).
Step 3
Exam Tip
\({}^{4}P_{4}=4!=24\) होता है। जब (r=n) हो तो \({}^{n}P_{n}=n!\)।
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(6) विषयों में से (2) विषयों को पहले और दूसरे पीरियड में लगाने के तरीके कितने हैं?
In how many ways can (2) subjects from (6) subjects be placed in the first and second periods?
#permutations
#class11
#easy
A (12)
B (36)
C (30)
D (15)
Explanation opens after your attempt
Step 1
Concept
There are (6) choices for the first period and (5) for the second, so (30). Timetable order uses permutation.
Step 2
Why this answer is correct
The correct answer is C. (30). There are (6) choices for the first period and (5) for the second, so (30). Timetable order uses permutation.
Step 3
Exam Tip
पहले पीरियड के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए (30)। timetable order में permutation लगता है।
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शब्द (RAVI) के अक्षरों की कुल व्यवस्थाएँ कितनी हैं?
What is the total number of arrangements of the letters of the word (RAVI)?
#permutations
#class11
#easy
A (16)
B (24)
C (12)
D (4)
Explanation opens after your attempt
Step 1
Concept
The word (RAVI) has (4) distinct letters, so (4!=24). Apply factorial when all letters are distinct.
Step 2
Why this answer is correct
The correct answer is B. (24). The word (RAVI) has (4) distinct letters, so (4!=24). Apply factorial when all letters are distinct.
Step 3
Exam Tip
(RAVI) में (4) अलग अक्षर हैं इसलिए (4!=24)। सभी distinct letters हों तो factorial लगाएं।
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(10) विद्यार्थियों में से मॉनिटर और सह-मॉनिटर चुनने के तरीके कितने हैं?
How many ways are there to choose a monitor and assistant monitor from (10) students?
#permutations
#class11
#easy
A (90)
B (20)
C (45)
D (100)
Explanation opens after your attempt
Step 1
Concept
There are two distinct posts, so \({}^{10}P_{2}=10\times9=90\). Order matters for different designations.
Step 2
Why this answer is correct
The correct answer is A. (90). There are two distinct posts, so \({}^{10}P_{2}=10\times9=90\). Order matters for different designations.
Step 3
Exam Tip
दो अलग पद हैं इसलिए \({}^{10}P_{2}=10\times9=90\)। अलग designation में order महत्त्वपूर्ण होता है।
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(3) अलग-अलग सिक्कों को एक क्रम में रखने के तरीके कितने हैं?
How many ways are there to place (3) different coins in an order?
#permutations
#class11
#easy
A (3)
B (9)
C (12)
D (6)
Explanation opens after your attempt
Step 1
Concept
The arrangement of (3) distinct coins is (3!=6). Use factorial for full arrangement of distinct objects.
Step 2
Why this answer is correct
The correct answer is D. (6). The arrangement of (3) distinct coins is (3!=6). Use factorial for full arrangement of distinct objects.
Step 3
Exam Tip
(3) अलग सिक्कों की व्यवस्था (3!=6) है। अलग objects की पूरी arrangement factorial से करें।
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\({}^{8}P_{0}\) का मान क्या होगा?
What will be the value of \({}^{8}P_{0}\)?
#permutations
#class11
#easy
A (8)
B (0)
C (1)
D (64)
Explanation opens after your attempt
Step 1
Concept
\({}^{n}P_{0}=1\), so \({}^{8}P_{0}=1\). The arrangement of zero objects is counted as one.
Step 2
Why this answer is correct
The correct answer is C. (1). \({}^{n}P_{0}=1\), so \({}^{8}P_{0}=1\). The arrangement of zero objects is counted as one.
Step 3
Exam Tip
\({}^{n}P_{0}=1\) इसलिए \({}^{8}P_{0}=1\)। शून्य वस्तुओं की arrangement एक मानी जाती है।
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अंकों (0,1,2) से बिना पुनरावृत्ति कितनी (2)-अंकीय संख्याएँ बन सकती हैं?
How many (2)-digit numbers can be formed from digits (0,1,2) without repetition?
#permutations
#class11
#easy
A (6)
B (4)
C (3)
D (2)
Explanation opens after your attempt
Step 1
Concept
Zero cannot be in the tens place, so there are (2) choices for tens and (2) for units, total (4). In number formation, check zero in the first place.
Step 2
Why this answer is correct
The correct answer is B. (4). Zero cannot be in the tens place, so there are (2) choices for tens and (2) for units, total (4). In number formation, check zero in the first place.
Step 3
Exam Tip
दहाई स्थान पर (0) नहीं आ सकता इसलिए (2) विकल्प और इकाई पर (2) विकल्प हैं, कुल (4)। संख्या बनाते समय पहले स्थान पर (0) की जाँच करें।
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(6) अलग-अलग गेंदों को एक पंक्ति में रखने के तरीके कितने होंगे?
How many ways are possible to place (6) different balls in a row?
#permutations
#class11
#easy
A (720)
B (36)
C (120)
D (60)
Explanation opens after your attempt
Step 1
Concept
The arrangement of (6) distinct balls is (6!=720). If all objects are distinct, total arrangements are factorial.
Step 2
Why this answer is correct
The correct answer is A. (720). The arrangement of (6) distinct balls is (6!=720). If all objects are distinct, total arrangements are factorial.
Step 3
Exam Tip
(6) अलग गेंदों की व्यवस्था (6!=720) है। सभी वस्तुएँ अलग हों तो कुल व्यवस्था factorial होती है।
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(4) लोगों में से (3) लोगों को एक पंक्ति में कितने तरीकों से खड़ा किया जा सकता है?
In how many ways can (3) people be made to stand in a row from (4) people?
#permutations
#class11
#easy
A (12)
B (16)
C (4)
D (24)
Explanation opens after your attempt
Step 1
Concept
This is \({}^{4}P_{3}=4\times3\times2=24\). Standing in a row is an ordered arrangement.
Step 2
Why this answer is correct
The correct answer is D. (24). This is \({}^{4}P_{3}=4\times3\times2=24\). Standing in a row is an ordered arrangement.
Step 3
Exam Tip
यह \({}^{4}P_{3}=4\times3\times2=24\) है। पंक्ति में खड़ा करना ordered arrangement है।
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शब्द (BOOK) में (4) अक्षर हैं और (O) दो बार आता है। इसके अलग-अलग arrangements कितने होंगे?
The word (BOOK) has (4) letters and (O) occurs twice. How many distinct arrangements are possible?
#permutations
#class11
#easy
A (24)
B (8)
C (12)
D (6)
Explanation opens after your attempt
Step 1
Concept
Because two (O)'s are identical, arrangements are \(\frac{4!}{2!}=12\). For identical objects, divide by their factorial.
Step 2
Why this answer is correct
The correct answer is C. (12). Because two (O)'s are identical, arrangements are \(\frac{4!}{2!}=12\). For identical objects, divide by their factorial.
Step 3
Exam Tip
दो समान (O) होने से व्यवस्थाएँ \(\frac{4!}{2!}=12\) होंगी। समान वस्तुओं के लिए factorial से भाग दें।
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(5) अलग-अलग कार्डों में से (3) कार्डों को क्रम में लगाने के तरीके कितने हैं?
How many ways are there to arrange (3) cards selected from (5) different cards in order?
#permutations
#class11
#easy
A (15)
B (60)
C (10)
D (125)
Explanation opens after your attempt
Step 1
Concept
This is \({}^{5}P_{3}=5\times4\times3=60\). Changing the order of cards changes the arrangement.
Step 2
Why this answer is correct
The correct answer is B. (60). This is \({}^{5}P_{3}=5\times4\times3=60\). Changing the order of cards changes the arrangement.
Step 3
Exam Tip
यह \({}^{5}P_{3}=5\times4\times3=60\) है। कार्डों का क्रम बदलने से व्यवस्था बदलती है।
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यदि (n=5) और (r=2) हो तो \({}^{n}P_{r}\) का मान क्या है?
If (n=5) and (r=2), what is the value of \({}^{n}P_{r}\)?
#permutations
#class11
#easy
A (20)
B (10)
C (25)
D (5)
Explanation opens after your attempt
Step 1
Concept
\({}^{5}P_{2}=5\times4=20\). Start from (n) and take (r) decreasing factors.
Step 2
Why this answer is correct
The correct answer is A. (20). \({}^{5}P_{2}=5\times4=20\). Start from (n) and take (r) decreasing factors.
Step 3
Exam Tip
\({}^{5}P_{2}=5\times4=20\)। (n) से शुरू करके (r) तक घटते गुणनखंड लें।
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(6) चित्रों में से (2) चित्रों को पहले और दूसरे स्थान पर लगाने के तरीके कितने हैं?
In how many ways can (2) pictures from (6) pictures be placed in first and second positions?
#permutations
#class11
#easy
A (12)
B (15)
C (36)
D (30)
Explanation opens after your attempt
Step 1
Concept
There are (6) choices for the first position and (5) for the second, so \(6\times5=30\). Count order when positions are distinct.
Step 2
Why this answer is correct
The correct answer is D. (30). There are (6) choices for the first position and (5) for the second, so \(6\times5=30\). Count order when positions are distinct.
Step 3
Exam Tip
पहले स्थान के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए \(6\times5=30\)। स्थान अलग-अलग हों तो क्रम गिनें।
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\({}^{10}P_{1}\) का मान क्या है?
What is the value of \({}^{10}P_{1}\)?
#permutations
#class11
#easy
A (1)
B (100)
C (10)
D (90)
Explanation opens after your attempt
Step 1
Concept
Since \({}^{n}P_{1}=n\), \({}^{10}P_{1}=10\). For one position, write (n) directly.
Step 2
Why this answer is correct
The correct answer is C. (10). Since \({}^{n}P_{1}=n\), \({}^{10}P_{1}=10\). For one position, write (n) directly.
Step 3
Exam Tip
\({}^{n}P_{1}=n\) होता है इसलिए \({}^{10}P_{1}=10\)। एक पद की व्यवस्था में सीधे (n) लिखें।
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शब्द (SUN) के अक्षरों से कितनी व्यवस्थाएँ बनती हैं?
How many arrangements are formed from the letters of the word (SUN)?
#permutations
#class11
#easy
A (9)
B (6)
C (3)
D (18)
Explanation opens after your attempt
Step 1
Concept
The word (SUN) has (3) distinct letters, so (3!=6). Count the letters and apply factorial for small words.
Step 2
Why this answer is correct
The correct answer is B. (6). The word (SUN) has (3) distinct letters, so (3!=6). Count the letters and apply factorial for small words.
Step 3
Exam Tip
(SUN) में (3) अलग अक्षर हैं इसलिए (3!=6)। छोटे शब्दों में अक्षर गिनकर factorial लगाएं।
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(7) अलग-अलग पेन में से (4) पेन को क्रम से रखने के तरीके कितने हैं?
How many ways are there to arrange (4) pens selected from (7) different pens in order?
#permutations
#class11
#easy
A (840)
B (210)
C (28)
D (35)
Explanation opens after your attempt
Step 1
Concept
\({}^{7}P_{4}=7\times6\times5\times4=840\). For (r) places, multiply (r) decreasing factors.
Step 2
Why this answer is correct
The correct answer is A. (840). \({}^{7}P_{4}=7\times6\times5\times4=840\). For (r) places, multiply (r) decreasing factors.
Step 3
Exam Tip
\({}^{7}P_{4}=7\times6\times5\times4=840\)। (r) स्थानों के लिए (r) घटते गुणनखंड लें।
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अंकों (1,3,5,7,9) से बिना पुनरावृत्ति कितनी (1)-अंकीय संख्याएँ बनेंगी?
How many (1)-digit numbers can be formed from digits (1,3,5,7,9) without repetition?
#permutations
#class11
#easy
A (1)
B (10)
C (25)
D (5)
Explanation opens after your attempt
Step 1
Concept
There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.
Step 2
Why this answer is correct
The correct answer is D. (5). There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.
Step 3
Exam Tip
एक स्थान के लिए (5) विकल्प हैं इसलिए उत्तर (5) है। (1)-स्थान वाली व्यवस्था में उपलब्ध विकल्प ही उत्तर होते हैं।
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अक्षरों (A,B,C,D) में से (3) अक्षर लेकर कितने क्रम बनाए जा सकते हैं?
How many ordered arrangements can be made by taking (3) letters from (A,B,C,D)?
#permutations
#class11
#easy
A (12)
B (16)
C (24)
D (64)
Explanation opens after your attempt
Step 1
Concept
The answer is \({}^{4}P_{3}=4\times3\times2=24\). Choosing and arranging in order is permutation.
Step 2
Why this answer is correct
The correct answer is C. (24). The answer is \({}^{4}P_{3}=4\times3\times2=24\). Choosing and arranging in order is permutation.
Step 3
Exam Tip
उत्तर \({}^{4}P_{3}=4\times3\times2=24\) है। चुनना और क्रम में रखना permutation है।
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(4) मित्रों को (2) विशेष सीटों पर कितने तरीकों से बैठाया जा सकता है?
In how many ways can (4) friends be seated on (2) special seats?
#permutations
#class11
#easy
A (6)
B (12)
C (8)
D (16)
Explanation opens after your attempt
Step 1
Concept
Order matters on two special seats, so \({}^{4}P_{2}=12\). If seats are distinct, use permutation.
Step 2
Why this answer is correct
The correct answer is B. (12). Order matters on two special seats, so \({}^{4}P_{2}=12\). If seats are distinct, use permutation.
Step 3
Exam Tip
दो विशेष सीटों पर क्रम महत्त्वपूर्ण है इसलिए \({}^{4}P_{2}=12\)। सीटें अलग हों तो permutation लें।
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यदि \({}^{n}P_{0}\) पूछा जाए तो उसका मान क्या होता है?
If \({}^{n}P_{0}\) is asked, what is its value?
#permutations
#class11
#easy
A (1)
B (0)
C (n)
D (n!)
Explanation opens after your attempt
Step 1
Concept
\({}^{n}P_{0}=1\) because there is one way to choose nothing. Remember this standard result.
Step 2
Why this answer is correct
The correct answer is A. (1). \({}^{n}P_{0}=1\) because there is one way to choose nothing. Remember this standard result.
Step 3
Exam Tip
\({}^{n}P_{0}=1\) होता है क्योंकि कोई वस्तु न चुनने का एक तरीका होता है। यह standard result याद रखें।
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शब्द (MATH) के सभी अक्षरों को कितने तरीकों से क्रमबद्ध किया जा सकता है?
In how many ways can all letters of the word (MATH) be ordered?
#permutations
#class11
#easy
A (8)
B (12)
C (16)
D (24)
Explanation opens after your attempt
Step 1
Concept
The word (MATH) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.
Step 2
Why this answer is correct
The correct answer is D. (24). The word (MATH) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.
Step 3
Exam Tip
(MATH) में (4) अलग अक्षर हैं इसलिए (4!=24)। सभी अलग अक्षरों की पूरी व्यवस्था factorial से मिलती है।
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(3) अलग-अलग पुरस्कार (5) विद्यार्थियों में इस प्रकार बाँटने हैं कि एक विद्यार्थी को अधिकतम एक पुरस्कार मिले। कितने तरीके होंगे?
(3) different prizes are to be distributed among (5) students so that a student gets at most one prize. How many ways are possible?
#permutations
#class11
#easy
A (15)
B (30)
C (60)
D (125)
Explanation opens after your attempt
Step 1
Concept
The prizes are distinct and students cannot repeat, so \({}^{5}P_{3}=60\). Distribution of distinct prizes makes order important.
Step 2
Why this answer is correct
The correct answer is C. (60). The prizes are distinct and students cannot repeat, so \({}^{5}P_{3}=60\). Distribution of distinct prizes makes order important.
Step 3
Exam Tip
पुरस्कार अलग हैं और विद्यार्थी दोहराए नहीं जा सकते इसलिए \({}^{5}P_{3}=60\)। अलग पुरस्कारों के वितरण में क्रम महत्त्वपूर्ण होता है।
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