शब्द (TRAIN) के सभी अक्षरों से कितने अलग शब्द बनाए जा सकते हैं?
How many distinct words can be formed using all letters of the word (TRAIN)?
#permutations
#class11
#easy
A (60)
B (25)
C (120)
D (10)
Explanation opens after your attempt
Step 1
Concept
The word (TRAIN) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.
Step 2
Why this answer is correct
The correct answer is C. (120). The word (TRAIN) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.
Step 3
Exam Tip
(TRAIN) में (5) अलग अक्षर हैं इसलिए (5!=120)। सभी अक्षर अलग हों तो denominator नहीं लगता।
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(12) विद्यार्थियों में से कक्षा-प्रतिनिधि और सह-प्रतिनिधि चुनने के तरीके कितने हैं?
In how many ways can a class representative and assistant representative be chosen from (12) students?
#permutations
#class11
#easy
A (24)
B (132)
C (66)
D (144)
Explanation opens after your attempt
Step 1
Concept
The two posts are distinct, so \({}^{12}P_{2}=12\times11=132\). Order matters for different posts.
Step 2
Why this answer is correct
The correct answer is B. (132). The two posts are distinct, so \({}^{12}P_{2}=12\times11=132\). Order matters for different posts.
Step 3
Exam Tip
दो पद अलग हैं इसलिए \({}^{12}P_{2}=12\times11=132\)। अलग पदों में क्रम महत्त्वपूर्ण होता है।
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(8) अलग-अलग ट्रॉफियों को एक शोकेस में सीधी पंक्ति में कितने तरीकों से सजाया जा सकता है?
In how many ways can (8) different trophies be arranged in a straight row in a showcase?
#permutations
#class11
#easy
A (40320)
B (5040)
C (64)
D (720)
Explanation opens after your attempt
Correct Answer
A. (40320)
Step 1
Concept
The arrangement of (8) distinct objects is (8!=40320). Use factorial for all distinct objects.
Step 2
Why this answer is correct
The correct answer is A. (40320). The arrangement of (8) distinct objects is (8!=40320). Use factorial for all distinct objects.
Step 3
Exam Tip
(8) अलग वस्तुओं की व्यवस्था (8!=40320) होती है। सभी अलग वस्तुओं के लिए factorial लगाएं।
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शब्द (RADAR) में (R) दो बार और (A) दो बार आता है। अलग व्यवस्थाएँ कितनी होंगी?
In the word (RADAR), (R) occurs twice and (A) occurs twice. How many distinct arrangements are possible?
#permutations
#class11
#easy
A (120)
B (20)
C (30)
D (60)
Explanation opens after your attempt
Step 1
Concept
The arrangements are \(\frac{5!}{2!2!}=30\). Do not count identical letters as different.
Step 2
Why this answer is correct
The correct answer is C. (30). The arrangements are \(\frac{5!}{2!2!}=30\). Do not count identical letters as different.
Step 3
Exam Tip
व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। समान अक्षरों को अलग-अलग मानकर गिनती न करें।
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(9) अलग-अलग फाइलों में से (4) फाइलों को क्रम से रखने के तरीके कितने हैं?
How many ways are there to arrange (4) files from (9) different files in order?
#permutations
#class11
#easy
A (126)
B (3024)
C (504)
D (6561)
Explanation opens after your attempt
Step 1
Concept
\({}^{9}P_{4}=9\times8\times7\times6=3024\). Use \({}^{n}P_{r}\) for ordered selection.
Step 2
Why this answer is correct
The correct answer is B. (3024). \({}^{9}P_{4}=9\times8\times7\times6=3024\). Use \({}^{n}P_{r}\) for ordered selection.
Step 3
Exam Tip
\({}^{9}P_{4}=9\times8\times7\times6=3024\) है। क्रम वाले चयन में \({}^{n}P_{r}\) लगाएं।
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अंकों (2,3,4,5,6) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बनेंगी?
How many (2)-digit even numbers can be formed from digits (2,3,4,5,6) without repetition?
#permutations
#class11
#easy
A (12)
B (15)
C (10)
D (20)
Explanation opens after your attempt
Step 1
Concept
The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.
Step 2
Why this answer is correct
The correct answer is A. (12). The units place has (3) choices from (2,4,6), and the tens place has (4) choices, total (12). For even numbers, check the units place first.
Step 3
Exam Tip
इकाई स्थान पर (2,4,6) में से (3) विकल्प हैं और दहाई पर (4) विकल्प, कुल (12)। even number में इकाई स्थान पहले देखें।
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(6) अलग-अलग पोस्टरों को दीवार पर बाएँ से दाएँ लगाने के तरीके कितने हैं?
In how many ways can (6) different posters be arranged on a wall from left to right?
#permutations
#class11
#easy
A (120)
B (36)
C (60)
D (720)
Explanation opens after your attempt
Step 1
Concept
The left-to-right arrangement of (6) distinct posters is (6!=720). Use factorial in linear arrangement.
Step 2
Why this answer is correct
The correct answer is D. (720). The left-to-right arrangement of (6) distinct posters is (6!=720). Use factorial in linear arrangement.
Step 3
Exam Tip
बाएँ से दाएँ (6) अलग पोस्टरों की व्यवस्था (6!=720) है। linear arrangement में factorial लें।
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\({}^{10}P_{0}+{}^{10}P_{1}\) का मान क्या होगा?
What will be the value of \({}^{10}P_{0}+{}^{10}P_{1}\)?
#permutations
#class11
#easy
A (10)
B (100)
C (11)
D (1)
Explanation opens after your attempt
Step 1
Concept
\({}^{10}P_{0}=1\) and \({}^{10}P_{1}=10\), so the sum is (11). In simple sums, find the values separately first.
Step 2
Why this answer is correct
The correct answer is C. (11). \({}^{10}P_{0}=1\) and \({}^{10}P_{1}=10\), so the sum is (11). In simple sums, find the values separately first.
Step 3
Exam Tip
\({}^{10}P_{0}=1\) और \({}^{10}P_{1}=10\), इसलिए योग (11) है। आसान योग में पहले values अलग निकालें।
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शब्द (NOON) के अलग-अलग arrangements कितने हैं?
How many distinct arrangements are there for the word (NOON)?
#permutations
#class11
#easy
A (12)
B (6)
C (4)
D (24)
Explanation opens after your attempt
Step 1
Concept
(N) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated groups.
Step 2
Why this answer is correct
The correct answer is B. (6). (N) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated groups.
Step 3
Exam Tip
(N) दो बार और (O) दो बार हैं इसलिए \(\frac{4!}{2!2!}=6\)। दो repeated groups हों तो दोनों से भाग दें।
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(8) चित्रकारों में से (3) को प्रथम, द्वितीय और तृतीय पुरस्कार देने के तरीके कितने हैं?
In how many ways can first, second and third prizes be given to (3) painters from (8) painters?
#permutations
#class11
#easy
A (336)
B (56)
C (24)
D (512)
Explanation opens after your attempt
Step 1
Concept
The prizes are distinct, so \({}^{8}P_{3}=8\times7\times6=336\). Order is counted for distinct prizes.
Step 2
Why this answer is correct
The correct answer is A. (336). The prizes are distinct, so \({}^{8}P_{3}=8\times7\times6=336\). Order is counted for distinct prizes.
Step 3
Exam Tip
पुरस्कार अलग हैं इसलिए \({}^{8}P_{3}=8\times7\times6=336\)। अलग पुरस्कारों में order गिना जाता है।
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अंकों (1,2,3,4,5) से बिना पुनरावृत्ति कितनी (4)-अंकीय संख्याएँ बनेंगी?
How many (4)-digit numbers can be formed from digits (1,2,3,4,5) without repetition?
#permutations
#class11
#easy
A (24)
B (625)
C (100)
D (120)
Explanation opens after your attempt
Step 1
Concept
For four places, there are \(5\times4\times3\times2=120\) ways. Without repetition, choices keep decreasing.
Step 2
Why this answer is correct
The correct answer is D. (120). For four places, there are \(5\times4\times3\times2=120\) ways. Without repetition, choices keep decreasing.
Step 3
Exam Tip
चार स्थानों के लिए \(5\times4\times3\times2=120\) तरीके हैं। बिना पुनरावृत्ति में विकल्प घटते जाते हैं।
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(4) अलग-अलग विषयों को सोमवार के (4) पीरियड में लगाने के तरीके कितने हैं?
In how many ways can (4) different subjects be arranged in (4) periods on Monday?
#permutations
#class11
#easy
A (16)
B (8)
C (24)
D (12)
Explanation opens after your attempt
Step 1
Concept
(4) different subjects can be arranged in (4) different periods in (4!=24) ways. Order matters in a timetable.
Step 2
Why this answer is correct
The correct answer is C. (24). (4) different subjects can be arranged in (4) different periods in (4!=24) ways. Order matters in a timetable.
Step 3
Exam Tip
(4) अलग विषय (4) अलग पीरियड में (4!=24) तरीकों से लगते हैं। timetable में order महत्त्वपूर्ण होता है।
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शब्द (BIHAR) के सभी अक्षरों से कितने अलग शब्द बन सकते हैं?
How many distinct words can be formed using all letters of (BIHAR)?
#permutations
#class11
#easy
A (60)
B (120)
C (25)
D (10)
Explanation opens after your attempt
Step 1
Concept
The word (BIHAR) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.
Step 2
Why this answer is correct
The correct answer is B. (120). The word (BIHAR) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.
Step 3
Exam Tip
(BIHAR) में (5) अलग अक्षर हैं इसलिए (5!=120)। सभी अक्षर अलग हों तो denominator नहीं लगता।
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(7) अलग-अलग पेंसिलों में से (2) पेंसिलों को पहले और दूसरे डिब्बे में रखने के तरीके कितने हैं?
How many ways are there to place (2) pencils from (7) different pencils in the first and second boxes?
#permutations
#class11
#easy
A (42)
B (14)
C (21)
D (49)
Explanation opens after your attempt
Step 1
Concept
There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.
Step 2
Why this answer is correct
The correct answer is A. (42). There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.
Step 3
Exam Tip
पहले डिब्बे के लिए (7) और दूसरे के लिए (6) विकल्प हैं इसलिए (42)। अलग डिब्बों में क्रमचय गिना जाता है।
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\({}^{4}P_{0}+{}^{4}P_{1}\) का मान क्या है?
What is the value of \({}^{4}P_{0}+{}^{4}P_{1}\)?
#permutations
#class11
#easy
A (4)
B (1)
C (8)
D (5)
Explanation opens after your attempt
Step 1
Concept
\({}^{4}P_{0}=1\) and \({}^{4}P_{1}=4\), so the sum is (5). Remember standard values.
Step 2
Why this answer is correct
The correct answer is D. (5). \({}^{4}P_{0}=1\) and \({}^{4}P_{1}=4\), so the sum is (5). Remember standard values.
Step 3
Exam Tip
\({}^{4}P_{0}=1\) और \({}^{4}P_{1}=4\), इसलिए योग (5) है। standard values याद रखें।
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(5) अलग-अलग मोतियों को एक सीधी डोरी पर लगाने के तरीके कितने हैं?
How many ways are there to arrange (5) different beads on a straight string?
#permutations
#class11
#easy
A (60)
B (25)
C (120)
D (10)
Explanation opens after your attempt
Step 1
Concept
On a straight string, the arrangement of (5) distinct beads is (5!=120). Do not treat it as a circular arrangement.
Step 2
Why this answer is correct
The correct answer is C. (120). On a straight string, the arrangement of (5) distinct beads is (5!=120). Do not treat it as a circular arrangement.
Step 3
Exam Tip
सीधी डोरी पर (5) अलग मोतियों की व्यवस्था (5!=120) है। इसे circular arrangement न मानें।
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अंकों (4,5,6) से पुनरावृत्ति की अनुमति होने पर कितनी (3)-अंकीय संख्याएँ बनेंगी?
How many (3)-digit numbers can be formed from digits (4,5,6) if repetition is allowed?
#permutations
#class11
#easy
A (9)
B (27)
C (18)
D (6)
Explanation opens after your attempt
Step 1
Concept
There are (3) choices for each of the three places, so \(3^3=27\). With repetition allowed, choices remain the same at every place.
Step 2
Why this answer is correct
The correct answer is B. (27). There are (3) choices for each of the three places, so \(3^3=27\). With repetition allowed, choices remain the same at every place.
Step 3
Exam Tip
तीनों स्थानों पर (3) विकल्प हैं इसलिए \(3^3=27\)। repetition allowed हो तो हर स्थान पर समान विकल्प रहते हैं।
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(12) व्यक्तियों में से प्रथम वक्ता और द्वितीय वक्ता चुनने के तरीके कितने हैं?
How many ways are there to choose the first speaker and second speaker from (12) persons?
#permutations
#class11
#easy
A (132)
B (24)
C (66)
D (144)
Explanation opens after your attempt
Step 1
Concept
First and second speakers are ordered positions, so \({}^{12}P_{2}=132\). Speaking order uses permutation.
Step 2
Why this answer is correct
The correct answer is A. (132). First and second speakers are ordered positions, so \({}^{12}P_{2}=132\). Speaking order uses permutation.
Step 3
Exam Tip
पहला और दूसरा वक्ता क्रम वाले पद हैं इसलिए \({}^{12}P_{2}=132\)। बोलने के क्रम में permutation लगता है।
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शब्द (EYE) के अलग-अलग arrangements कितने होंगे?
How many distinct arrangements are possible for the word (EYE)?
#permutations
#class11
#easy
A (6)
B (2)
C (1)
D (3)
Explanation opens after your attempt
Step 1
Concept
(E) occurs twice, so \(\frac{3!}{2!}=3\) arrangements are possible. Do not count repeated letters as distinct.
Step 2
Why this answer is correct
The correct answer is D. (3). (E) occurs twice, so \(\frac{3!}{2!}=3\) arrangements are possible. Do not count repeated letters as distinct.
Step 3
Exam Tip
(E) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements होंगे। repeated letters को अलग-अलग न गिनें।
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अक्षरों (K,L,M,N) से बिना पुनरावृत्ति कितने (3)-अक्षरी कोड बनेंगे?
How many (3)-letter codes can be formed from letters (K,L,M,N) without repetition?
#permutations
#class11
#easy
A (16)
B (12)
C (24)
D (64)
Explanation opens after your attempt
Step 1
Concept
\({}^{4}P_{3}=4\times3\times2=24\) codes are possible. In a code, changing positions changes the code.
Step 2
Why this answer is correct
The correct answer is C. (24). \({}^{4}P_{3}=4\times3\times2=24\) codes are possible. In a code, changing positions changes the code.
Step 3
Exam Tip
\({}^{4}P_{3}=4\times3\times2=24\) कोड बनेंगे। कोड में अक्षरों का स्थान बदलने से कोड बदल जाता है।
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(6) अलग-अलग किताबों में से (2) किताबों को दाएँ और बाएँ स्थान पर रखने के तरीके कितने हैं?
How many ways are there to place (2) books from (6) different books in right and left positions?
#permutations
#class11
#easy
A (15)
B (30)
C (12)
D (36)
Explanation opens after your attempt
Step 1
Concept
Right and left positions are distinct, so \({}^{6}P_{2}=30\). When positions are identified, order is counted.
Step 2
Why this answer is correct
The correct answer is B. (30). Right and left positions are distinct, so \({}^{6}P_{2}=30\). When positions are identified, order is counted.
Step 3
Exam Tip
दायाँ और बायाँ स्थान अलग हैं इसलिए \({}^{6}P_{2}=30\)। स्थानों की पहचान अलग हो तो क्रम गिना जाता है।
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\({}^{5}P_{4}\) का मान क्या है?
What is the value of \({}^{5}P_{4}\)?
#permutations
#class11
#easy
A (120)
B (20)
C (60)
D (24)
Explanation opens after your attempt
Step 1
Concept
\({}^{5}P_{4}=5\times4\times3\times2=120\). When (r=4), take (4) decreasing factors.
Step 2
Why this answer is correct
The correct answer is A. (120). \({}^{5}P_{4}=5\times4\times3\times2=120\). When (r=4), take (4) decreasing factors.
Step 3
Exam Tip
\({}^{5}P_{4}=5\times4\times3\times2=120\) है। (r=4) हो तो (4) घटते गुणनखंड लें।
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शब्द (TEAM) के अक्षरों से कितने क्रम बनेंगे यदि अंतिम अक्षर (M) हो?
How many arrangements of the letters of (TEAM) are possible if the last letter is (M)?
#permutations
#class11
#easy
A (24)
B (12)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.
Step 2
Why this answer is correct
The correct answer is D. (6). The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.
Step 3
Exam Tip
अंतिम स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed condition को पहले लागू करें।
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(11) अभ्यर्थियों में से अध्यक्ष और सचिव चुनने के तरीके कितने हैं?
In how many ways can a president and secretary be chosen from (11) candidates?
#permutations
#class11
#easy
A (22)
B (121)
C (110)
D (55)
Explanation opens after your attempt
Step 1
Concept
The two posts are distinct, so \({}^{11}P_{2}=11\times10=110\). Use permutation for distinct posts.
Step 2
Why this answer is correct
The correct answer is C. (110). The two posts are distinct, so \({}^{11}P_{2}=11\times10=110\). Use permutation for distinct posts.
Step 3
Exam Tip
दो पद अलग हैं इसलिए \({}^{11}P_{2}=11\times10=110\)। अलग पदों के लिए क्रमचय लगाएं।
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(3) लाल, नीले और हरे अलग झंडों को एक डंडे पर ऊपर से नीचे लगाने के तरीके कितने हैं?
In how many ways can (3) distinct red, blue and green flags be placed on a pole from top to bottom?
#permutations
#class11
#easy
A (3)
B (6)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
The ordered arrangement of (3) distinct flags is (3!=6). Changing top-bottom order changes the arrangement.
Step 2
Why this answer is correct
The correct answer is B. (6). The ordered arrangement of (3) distinct flags is (3!=6). Changing top-bottom order changes the arrangement.
Step 3
Exam Tip
(3) अलग झंडों की क्रमबद्ध व्यवस्था (3!=6) है। ऊपर-नीचे का क्रम बदलने से arrangement बदलती है।
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अंकों (0,1,2,3) से बिना पुनरावृत्ति कितनी (3)-अंकीय संख्याएँ बन सकती हैं?
How many (3)-digit numbers can be formed from digits (0,1,2,3) without repetition?
#permutations
#class11
#easy
A (18)
B (24)
C (12)
D (9)
Explanation opens after your attempt
Step 1
Concept
There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Do not put (0) in the first place.
Step 2
Why this answer is correct
The correct answer is A. (18). There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Do not put (0) in the first place.
Step 3
Exam Tip
सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प हैं, कुल (18)। पहले स्थान पर (0) न रखें।
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(5) अलग-अलग शहरों में से (2) शहरों की यात्रा का क्रम चुनने के तरीके कितने हैं?
How many ways are there to choose an ordered trip of (2) cities from (5) different cities?
#permutations
#class11
#easy
A (10)
B (25)
C (5)
D (20)
Explanation opens after your attempt
Step 1
Concept
There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.
Step 2
Why this answer is correct
The correct answer is D. (20). There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.
Step 3
Exam Tip
पहले शहर के लिए (5) और दूसरे के लिए (4) विकल्प हैं इसलिए (20)। यात्रा क्रम बदलने से योजना बदलती है।
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शब्द (SCHOOL) में (O) दो बार आता है। इसके अलग अक्षर-क्रम कितने होंगे?
The word (SCHOOL) has (O) twice. How many distinct letter arrangements are possible?
#permutations
#class11
#easy
A (720)
B (60)
C (360)
D (180)
Explanation opens after your attempt
Step 1
Concept
There are (6) letters and two identical (O)'s, so \(\frac{6!}{2!}=360\). Use a denominator for a repeated letter.
Step 2
Why this answer is correct
The correct answer is C. (360). There are (6) letters and two identical (O)'s, so \(\frac{6!}{2!}=360\). Use a denominator for a repeated letter.
Step 3
Exam Tip
कुल (6) अक्षर हैं और (O) दो समान हैं इसलिए \(\frac{6!}{2!}=360\)। repeated letter के लिए denominator लगाएं।
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(4) अलग-अलग डिब्बों पर (4) अलग-अलग लेबल चिपकाने के तरीके कितने हैं?
In how many ways can (4) different labels be pasted on (4) different boxes?
#permutations
#class11
#easy
A (16)
B (24)
C (8)
D (12)
Explanation opens after your attempt
Step 1
Concept
(4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.
Step 2
Why this answer is correct
The correct answer is B. (24). (4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.
Step 3
Exam Tip
(4) अलग लेबल (4) अलग डिब्बों पर (4!=24) तरीकों से लगेंगे। matching arrangements में factorial काम आता है।
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\({}^{7}P_{0}\) का मान क्या है?
What is the value of \({}^{7}P_{0}\)?
#permutations
#class11
#easy
A (1)
B (0)
C (7)
D (49)
Explanation opens after your attempt
Step 1
Concept
\({}^{n}P_{0}=1\), so \({}^{7}P_{0}=1\). Choosing zero objects is counted in one way.
Step 2
Why this answer is correct
The correct answer is A. (1). \({}^{n}P_{0}=1\), so \({}^{7}P_{0}=1\). Choosing zero objects is counted in one way.
Step 3
Exam Tip
\({}^{n}P_{0}=1\) होता है इसलिए \({}^{7}P_{0}=1\)। शून्य वस्तु चुनने का एक ही तरीका माना जाता है।
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