(7) पुरुषों और (3) महिलाओं को पंक्ति में इस प्रकार बैठाना है कि महिलाएँ लगातार तीन स्थानों पर बैठें। कुल व्यवस्थाएँ कितनी हैं?
(7) men and (3) women are to be seated in a row such that the women occupy three consecutive places. How many arrangements are possible?
#permutations
#class11
#expert
#block_method
A (241920)
B (30240)
C (120960)
D (60480)
Explanation opens after your attempt
Correct Answer
A. (241920)
Step 1
Concept
Treat the women as one block giving (8!) orders and (3!) internal orders. Use the block method for consecutive objects.
Step 2
Why this answer is correct
The correct answer is A. (241920). Treat the women as one block giving (8!) orders and (3!) internal orders. Use the block method for consecutive objects.
Step 3
Exam Tip
महिलाओं को एक ब्लॉक मानकर (8!) क्रम बनते हैं और ब्लॉक के अंदर (3!) क्रम हैं। लगातार रहने पर ब्लॉक विधि लगती है।
Login to save your score, XP, coins and progress. Login
शब्द (COMMITTEE) के अक्षरों से बनने वाली अलग-अलग व्यवस्थाओं की संख्या क्या है?
What is the number of distinct arrangements of the letters of the word (COMMITTEE)?
#permutations
#class11
#expert
#repeated_letters
A (45360)
B (90720)
C (22680)
D (15120)
Explanation opens after your attempt
Correct Answer
B. (90720)
Step 1
Concept
There are (9) letters and (M,T,E) each repeat (2) times, so the count is \(\frac{9!}{2!2!2!}\). Identical letters reduce permutations.
Step 2
Why this answer is correct
The correct answer is B. (90720). There are (9) letters and (M,T,E) each repeat (2) times, so the count is \(\frac{9!}{2!2!2!}\). Identical letters reduce permutations.
Step 3
Exam Tip
कुल (9) अक्षर हैं और (M,T,E) प्रत्येक (2) बार आते हैं, इसलिए \(\frac{9!}{2!2!2!}\) है। समान अक्षर permutation को घटाते हैं।
Login to save your score, XP, coins and progress. Login
(9) अलग-अलग लोगों को एक पंक्ति में बैठाना है। दो निश्चित लोग एक-दूसरे के ठीक बीच में (3) लोग छोड़कर बैठें, ऐसे कितने क्रम हैं?
(9) distinct people are to be seated in a row. Two particular people must have exactly (3) people between them. How many arrangements are possible?
#permutations
#class11
#expert
#position_constraint
A (10080)
B (20160)
C (40320)
D (80640)
Explanation opens after your attempt
Correct Answer
B. (20160)
Step 1
Concept
The two particular people occupy positions (4) apart, giving (5) position pairs and (2) orders, then arrange the remaining (7!). Count position pairs for distance constraints.
Step 2
Why this answer is correct
The correct answer is B. (20160). The two particular people occupy positions (4) apart, giving (5) position pairs and (2) orders, then arrange the remaining (7!). Count position pairs for distance constraints.
Step 3
Exam Tip
दो विशेष लोगों के स्थान (4) दूरी पर होंगे, ऐसे (5) जोड़े हैं और उनका क्रम (2) तरीकों से होगा, फिर बाकी (7!) हैं। दूरी वाले प्रश्नों में स्थान-जोड़े गिनें।
Login to save your score, XP, coins and progress. Login
अंकों (1,2,3,4,5,6,7,8,9) से बिना पुनरावृत्ति के (5) अंकों की कितनी संख्याएँ बनेंगी जिनमें अंक (1) और (2) दोनों हों?
How many (5)-digit numbers can be formed from digits (1,2,3,4,5,6,7,8,9) without repetition such that both digits (1) and (2) are included?
#permutations
#class11
#expert
#digits_required
A (2520)
B (5040)
C (3360)
D (1680)
Explanation opens after your attempt
Step 1
Concept
Choose the remaining (3) digits from (7) digits and arrange all (5) digits in (5!) ways. Include compulsory digits first.
Step 2
Why this answer is correct
The correct answer is A. (2520). Choose the remaining (3) digits from (7) digits and arrange all (5) digits in (5!) ways. Include compulsory digits first.
Step 3
Exam Tip
बाकी (3) अंक (7) अंकों से चुनें और कुल (5) अंकों को (5!) तरीकों से सजाएँ। आवश्यक अंकों को पहले शामिल करें।
Login to save your score, XP, coins and progress. Login
(7) अलग-अलग कुर्सियों पर (4) अलग-अलग विद्यार्थियों को बैठाने के कितने तरीके हैं?
In how many ways can (4) distinct students be seated on (7) distinct chairs?
#permutations
#class11
#expert
#npr
A (210)
B (5040)
C (840)
D (720)
Explanation opens after your attempt
Step 1
Concept
Choosing and ordering (4) chairs together gives (7P4). Use permutation when distinct people occupy distinct positions.
Step 2
Why this answer is correct
The correct answer is C. (840). Choosing and ordering (4) chairs together gives (7P4). Use permutation when distinct people occupy distinct positions.
Step 3
Exam Tip
पहले (4) कुर्सियाँ चुनने और क्रम देने का संयुक्त परिणाम (7P4) है। अलग स्थानों पर अलग व्यक्तियों के लिए permutation प्रयोग करें।
Login to save your score, XP, coins and progress. Login
शब्द (MATHEMATICS) के अक्षरों से कितनी अलग-अलग व्यवस्थाएँ बनेंगी?
How many distinct arrangements can be made from the letters of the word (MATHEMATICS)?
#permutations
#class11
#expert
#repeated_letters
A (4989600)
B (9979200)
C (2494800)
D (3628800)
Explanation opens after your attempt
Correct Answer
A. (4989600)
Step 1
Concept
There are (11) letters and (M,A,T) each repeat (2) times, so the count is \(\frac{11!}{2!2!2!}\). Correctly identify repeated letters.
Step 2
Why this answer is correct
The correct answer is A. (4989600). There are (11) letters and (M,A,T) each repeat (2) times, so the count is \(\frac{11!}{2!2!2!}\). Correctly identify repeated letters.
Step 3
Exam Tip
कुल (11) अक्षर हैं और (M,A,T) प्रत्येक (2) बार आते हैं, इसलिए \(\frac{11!}{2!2!2!}\) है। दोहराव की संख्या सही पहचानना जरूरी है।
Login to save your score, XP, coins and progress. Login
अंकों (0,1,2,3,4,5,6) से बिना पुनरावृत्ति के (4) अंकों की कितनी विषम संख्याएँ बन सकती हैं?
How many (4)-digit odd numbers can be formed from digits (0,1,2,3,4,5,6) without repetition?
#permutations
#class11
#expert
#digits
A (240)
B (300)
C (360)
D (420)
Explanation opens after your attempt
Step 1
Concept
The last digit is one of (1,3,5), and the first digit cannot be (0), giving \(3\cdot5\cdot5\cdot4\). Handle the leading zero condition separately.
Step 2
Why this answer is correct
The correct answer is B. (300). The last digit is one of (1,3,5), and the first digit cannot be (0), giving \(3\cdot5\cdot5\cdot4\). Handle the leading zero condition separately.
Step 3
Exam Tip
अंतिम अंक (1,3,5) में से होगा और पहले अंक में (0) नहीं आ सकता, इसलिए कुल \(3\cdot5\cdot5\cdot4\) है। शून्य वाले प्रश्नों में पहला स्थान अलग देखें।
Login to save your score, XP, coins and progress. Login
(5) लड़कों और (4) लड़कियों को एक पंक्ति में इस प्रकार बैठाना है कि कोई भी दो लड़कियाँ साथ-साथ न बैठें। कुल व्यवस्थाएँ कितनी होंगी?
(5) boys and (4) girls are to be seated in a row so that no two girls sit together. How many arrangements are possible?
#permutations
#class11
#expert
#no_adjacent
A (17280)
B (2880)
C (14400)
D (8640)
Explanation opens after your attempt
Correct Answer
A. (17280)
Step 1
Concept
Arrange boys in (5!) ways and place (4) girls in (6) gaps in (6P4) ways. The gap method is most useful here.
Step 2
Why this answer is correct
The correct answer is A. (17280). Arrange boys in (5!) ways and place (4) girls in (6) gaps in (6P4) ways. The gap method is most useful here.
Step 3
Exam Tip
पहले लड़कों को (5!) तरीकों से रखें और (6) खाली स्थानों में (4) लड़कियाँ (6P4) तरीकों से रखें। ऐसी समस्याओं में खाली स्थान विधि सबसे उपयोगी है।
Login to save your score, XP, coins and progress. Login
शब्द (MISSISSIPPI) के अक्षरों से बनने वाली अलग-अलग व्यवस्थाओं की संख्या क्या है?
What is the number of distinct arrangements of the letters of the word (MISSISSIPPI)?
#permutations
#class11
#expert
#repeated_letters
A (34650)
B (27720)
C (11550)
D (69300)
Explanation opens after your attempt
Correct Answer
A. (34650)
Step 1
Concept
There are (11) letters with (I,S,P) repeated (4,4,2) times, so the count is \(\frac{11!}{4!4!2!}\). Do not count identical letters separately.
Step 2
Why this answer is correct
The correct answer is A. (34650). There are (11) letters with (I,S,P) repeated (4,4,2) times, so the count is \(\frac{11!}{4!4!2!}\). Do not count identical letters separately.
Step 3
Exam Tip
कुल (11) अक्षर हैं और (I,S,P) क्रमशः (4,4,2) बार आते हैं, इसलिए \(\frac{11!}{4!4!2!}\) है। समान अक्षरों की पुनरावृत्ति को अलग न गिनें।
Login to save your score, XP, coins and progress. Login
अंकों (1,2,3,4,5,6,7) से बिना पुनरावृत्ति के (4) अंकों की कितनी संख्याएँ बनेंगी जो (5) से विभाज्य हों?
How many (4)-digit numbers can be formed from digits (1,2,3,4,5,6,7) without repetition that are divisible by (5)?
#permutations
#class11
#expert
#digits
A (120)
B (60)
C (180)
D (90)
Explanation opens after your attempt
Step 1
Concept
The last digit must be (5), and the remaining (3) places have (6P3) arrangements. Apply divisibility conditions first.
Step 2
Why this answer is correct
The correct answer is A. (120). The last digit must be (5), and the remaining (3) places have (6P3) arrangements. Apply divisibility conditions first.
Step 3
Exam Tip
अंतिम अंक केवल (5) होगा और बाकी (3) स्थानों पर (6P3) व्यवस्थाएँ होंगी। विभाज्यता की शर्त पहले लगाएं।
Login to save your score, XP, coins and progress. Login
(8) व्यक्तियों को एक वृत्ताकार मेज के चारों ओर बैठाना है और दो विशेष व्यक्ति साथ-साथ नहीं बैठने चाहिए। कितने तरीकों से बैठाया जा सकता है?
(8) people are to be seated around a circular table and two particular people must not sit together. In how many ways can this be done?
#permutations
#class11
#expert
#circular_permutation
A (3600)
B (4320)
C (2880)
D (5040)
Explanation opens after your attempt
Step 1
Concept
Total circular arrangements are (7!) and adjacent cases are \(2\cdot6!\), so the answer is \(7!-2\cdot6!\). In circular order, fix one position.
Step 2
Why this answer is correct
The correct answer is A. (3600). Total circular arrangements are (7!) and adjacent cases are \(2\cdot6!\), so the answer is \(7!-2\cdot6!\). In circular order, fix one position.
Step 3
Exam Tip
कुल वृत्तीय व्यवस्थाएँ (7!) हैं और साथ बैठने वाली \(2\cdot6!\) हैं, इसलिए उत्तर \(7!-2\cdot6!\) है। वृत्तीय क्रम में एक स्थान स्थिर मानें।
Login to save your score, XP, coins and progress. Login
शब्द (ARRANGE) के सभी अक्षरों से बनने वाली अलग-अलग व्यवस्थाओं की संख्या क्या है?
What is the number of distinct arrangements formed using all letters of the word (ARRANGE)?
#permutations
#class11
#expert
#repeated_letters
A (1260)
B (2520)
C (840)
D (5040)
Explanation opens after your attempt
Step 1
Concept
There are (7) letters and (R) repeats twice, so the count is \(\frac{7!}{2!}\). Divide by factorials of repeated letters.
Step 2
Why this answer is correct
The correct answer is B. (2520). There are (7) letters and (R) repeats twice, so the count is \(\frac{7!}{2!}\). Divide by factorials of repeated letters.
Step 3
Exam Tip
कुल (7) अक्षर हैं और (R) दो बार आता है, इसलिए संख्या \(\frac{7!}{2!}\) है। दोहराए अक्षरों के लिए फैक्टरियल से भाग दें।
Login to save your score, XP, coins and progress. Login
(6) अलग-अलग पुस्तकों को एक शेल्फ पर इस प्रकार सजाना है कि (2) निश्चित पुस्तकें साथ-साथ रहें लेकिन उनका आपसी क्रम बदला जा सके। कुल व्यवस्थाओं की संख्या क्या होगी?
(6) distinct books are to be arranged on a shelf so that (2) particular books always remain together but their internal order may change. What is the number of arrangements?
#permutations
#class11
#expert
#arrangement
A (240)
B (120)
C (360)
D (480)
Explanation opens after your attempt
Step 1
Concept
Treat the two books as one block giving \(5!\times2!\) arrangements. In exams, make a block for objects that must stay together.
Step 2
Why this answer is correct
The correct answer is A. (240). Treat the two books as one block giving \(5!\times2!\) arrangements. In exams, make a block for objects that must stay together.
Step 3
Exam Tip
दो पुस्तकों को एक ब्लॉक मानकर \(5!\times2!\) व्यवस्थाएँ मिलती हैं। परीक्षा में साथ रहने वाली वस्तुओं को ब्लॉक बनाएं।
Login to save your score, XP, coins and progress. Login
शब्द (CIVIC) में (C) दो बार और (I) दो बार आता है। अलग arrangements कितने होंगे?
In the word (CIVIC), (C) occurs twice and (I) occurs twice. How many distinct arrangements are possible?
#permutations
#class11
#easy
A (60)
B (30)
C (20)
D (120)
Explanation opens after your attempt
Step 1
Concept
The arrangements are \(\frac{5!}{2!2!}=30\). It is necessary to divide by factorials of repeated letters.
Step 2
Why this answer is correct
The correct answer is B. (30). The arrangements are \(\frac{5!}{2!2!}=30\). It is necessary to divide by factorials of repeated letters.
Step 3
Exam Tip
व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। repeated letters के factorial से भाग देना जरूरी है।
Login to save your score, XP, coins and progress. Login
(11) अलग-अलग चार्टों में से (3) चार्टों को क्रम से लगाने के तरीके कितने हैं?
How many ways are there to arrange (3) charts from (11) different charts in order?
#permutations
#class11
#easy
A (990)
B (165)
C (1331)
D (33)
Explanation opens after your attempt
Step 1
Concept
\({}^{11}P_{3}=11\times10\times9=990\). Use \({}^{n}P_{r}\) in ordered selection.
Step 2
Why this answer is correct
The correct answer is A. (990). \({}^{11}P_{3}=11\times10\times9=990\). Use \({}^{n}P_{r}\) in ordered selection.
Step 3
Exam Tip
\({}^{11}P_{3}=11\times10\times9=990\) है। क्रम वाले चयन में \({}^{n}P_{r}\) लगाएं।
Login to save your score, XP, coins and progress. Login
अंकों (1,2,3,4,6) से बिना पुनरावृत्ति कितनी (2)-अंकीय सम संख्याएँ बनेंगी?
How many (2)-digit even numbers can be formed from digits (1,2,3,4,6) without repetition?
#permutations
#class11
#easy
A (15)
B (20)
C (10)
D (12)
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 D. (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 और दहाई पर (4) विकल्प हैं, कुल (12)। even number में इकाई स्थान पहले देखें।
Login to save your score, XP, coins and progress. Login
(7) अलग-अलग पोस्टकार्डों को बाएँ से दाएँ सजाने के तरीके कितने हैं?
In how many ways can (7) different postcards be arranged from left to right?
#permutations
#class11
#easy
A (720)
B (49)
C (5040)
D (120)
Explanation opens after your attempt
Step 1
Concept
The left-to-right arrangement of (7) distinct postcards is (7!=5040). Use factorial in linear arrangement.
Step 2
Why this answer is correct
The correct answer is C. (5040). The left-to-right arrangement of (7) distinct postcards is (7!=5040). Use factorial in linear arrangement.
Step 3
Exam Tip
बाएँ से दाएँ (7) अलग पोस्टकार्डों की व्यवस्था (7!=5040) है। linear arrangement में factorial लें।
Login to save your score, XP, coins and progress. Login
\({}^{12}P_{0}+{}^{12}P_{1}\) का मान क्या होगा?
What will be the value of \({}^{12}P_{0}+{}^{12}P_{1}\)?
#permutations
#class11
#easy
A (12)
B (13)
C (144)
D (1)
Explanation opens after your attempt
Step 1
Concept
\({}^{12}P_{0}=1\) and \({}^{12}P_{1}=12\), so the sum is (13). Remember standard values.
Step 2
Why this answer is correct
The correct answer is B. (13). \({}^{12}P_{0}=1\) and \({}^{12}P_{1}=12\), so the sum is (13). Remember standard values.
Step 3
Exam Tip
\({}^{12}P_{0}=1\) और \({}^{12}P_{1}=12\), इसलिए योग (13) है। standard values याद रखें।
Login to save your score, XP, coins and progress. Login
शब्द (REFER) में (R) दो बार और (E) दो बार आता है। अलग व्यवस्थाएँ कितनी होंगी?
In the word (REFER), (R) occurs twice and (E) occurs twice. How many distinct arrangements are possible?
#permutations
#class11
#easy
A (30)
B (60)
C (120)
D (20)
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 A. (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 न गिनें।
Login to save your score, XP, coins and progress. Login
(9) गायकों में से प्रथम, द्वितीय और तृतीय प्रदर्शन क्रम चुनने के तरीके कितने हैं?
In how many ways can first, second and third performance order be chosen from (9) singers?
#permutations
#class11
#easy
A (84)
B (729)
C (504)
D (504)
Explanation opens after your attempt
Step 1
Concept
Performance order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Changing order changes the programme.
Step 2
Why this answer is correct
The correct answer is D. (504). Performance order matters, so \({}^{9}P_{3}=9\times8\times7=504\). Changing order changes the programme.
Step 3
Exam Tip
प्रदर्शन क्रम महत्त्वपूर्ण है इसलिए \({}^{9}P_{3}=9\times8\times7=504\)। order बदलने से कार्यक्रम बदल जाता है।
Login to save your score, XP, coins and progress. Login
अंकों (4,5,6,7,8,9) से बिना पुनरावृत्ति कितनी (4)-अंकीय संख्याएँ बनेंगी?
How many (4)-digit numbers can be formed from digits (4,5,6,7,8,9) without repetition?
#permutations
#class11
#easy
A (360)
B (720)
C (360)
D (1296)
Explanation opens after your attempt
Step 1
Concept
For four places, there are \(6\times5\times4\times3=360\) ways. Without repetition, choices decrease.
Step 2
Why this answer is correct
The correct answer is C. (360). For four places, there are \(6\times5\times4\times3=360\) ways. Without repetition, choices decrease.
Step 3
Exam Tip
चार स्थानों के लिए \(6\times5\times4\times3=360\) तरीके हैं। बिना पुनरावृत्ति में विकल्प घटते हैं।
Login to save your score, XP, coins and progress. Login
(5) अलग-अलग घंटियों को (5) निर्धारित स्थानों पर लगाने के तरीके कितने हैं?
In how many ways can (5) different bells be placed at (5) fixed positions?
#permutations
#class11
#easy
A (25)
B (120)
C (60)
D (10)
Explanation opens after your attempt
Step 1
Concept
(5) different bells can be placed at (5) positions in (5!=120) ways. Arrangements at fixed positions use factorial.
Step 2
Why this answer is correct
The correct answer is B. (120). (5) different bells can be placed at (5) positions in (5!=120) ways. Arrangements at fixed positions use factorial.
Step 3
Exam Tip
(5) अलग घंटियाँ (5) स्थानों पर (5!=120) तरीकों से लगेंगी। निर्धारित स्थानों की arrangement factorial से होती है।
Login to save your score, XP, coins and progress. Login
शब्द (MOUSE) के सभी अक्षरों से कितने अलग शब्द बन सकते हैं?
How many distinct words can be formed using all letters of (MOUSE)?
#permutations
#class11
#easy
A (120)
B (60)
C (25)
D (240)
Explanation opens after your attempt
Step 1
Concept
The word (MOUSE) has (5) distinct letters, so (5!=120). For distinct letters, factorial gives total arrangements.
Step 2
Why this answer is correct
The correct answer is A. (120). The word (MOUSE) has (5) distinct letters, so (5!=120). For distinct letters, factorial gives total arrangements.
Step 3
Exam Tip
(MOUSE) में (5) अलग अक्षर हैं इसलिए (5!=120)। अलग अक्षरों में factorial ही कुल व्यवस्था देता है।
Login to save your score, XP, coins and progress. Login
(9) अलग-अलग पेनड्राइव में से (2) पेनड्राइव को पहले और दूसरे स्लॉट में लगाने के तरीके कितने हैं?
How many ways are there to place (2) pen drives from (9) different pen drives in the first and second slots?
#permutations
#class11
#easy
A (18)
B (81)
C (36)
D (72)
Explanation opens after your attempt
Step 1
Concept
There are (9) choices for the first slot and (8) for the second, so (72). Count permutation when slots are distinct.
Step 2
Why this answer is correct
The correct answer is D. (72). There are (9) choices for the first slot and (8) for the second, so (72). Count permutation when slots are distinct.
Step 3
Exam Tip
पहले स्लॉट के लिए (9) और दूसरे के लिए (8) विकल्प हैं इसलिए (72)। स्लॉट अलग हों तो permutation गिनें।
Login to save your score, XP, coins and progress. Login
\({}^{5}P_{0}+{}^{5}P_{2}\) का मान क्या है?
What is the value of \({}^{5}P_{0}+{}^{5}P_{2}\)?
#permutations
#class11
#easy
A (20)
B (25)
C (21)
D (15)
Explanation opens after your attempt
Step 1
Concept
\({}^{5}P_{0}=1\) and \({}^{5}P_{2}=20\), so the sum is (21). Find the values separately first.
Step 2
Why this answer is correct
The correct answer is C. (21). \({}^{5}P_{0}=1\) and \({}^{5}P_{2}=20\), so the sum is (21). Find the values separately first.
Step 3
Exam Tip
\({}^{5}P_{0}=1\) और \({}^{5}P_{2}=20\), इसलिए योग (21) है। पहले अलग-अलग values निकालें।
Login to save your score, XP, coins and progress. Login
(6) अलग-अलग मोतियों को सीधी कढ़ाई में लगाने के तरीके कितने हैं?
How many ways are there to arrange (6) different beads in a straight embroidery line?
#permutations
#class11
#easy
A (360)
B (720)
C (120)
D (36)
Explanation opens after your attempt
Step 1
Concept
In a straight line, the arrangement of (6) distinct beads is (6!=720). Do not treat it as a circular arrangement.
Step 2
Why this answer is correct
The correct answer is B. (720). In a straight line, the arrangement of (6) distinct beads is (6!=720). Do not treat it as a circular arrangement.
Step 3
Exam Tip
सीधी रेखा में (6) अलग मोतियों की व्यवस्था (6!=720) है। इसे circular arrangement न मानें।
Login to save your score, XP, coins and progress. Login
अंकों (3,6,9) से पुनरावृत्ति की अनुमति होने पर कितनी (2)-अंकीय संख्याएँ बनेंगी?
How many (2)-digit numbers can be formed from digits (3,6,9) if repetition is allowed?
#permutations
#class11
#easy
A (9)
B (6)
C (12)
D (3)
Explanation opens after your attempt
Step 1
Concept
There are (3) choices at both places, so \(3^2=9\). With repetition allowed, choices remain the same.
Step 2
Why this answer is correct
The correct answer is A. (9). There are (3) choices at both places, so \(3^2=9\). With repetition allowed, choices remain the same.
Step 3
Exam Tip
दोनों स्थानों पर (3) विकल्प हैं इसलिए \(3^2=9\)। repetition allowed हो तो विकल्प वही रहते हैं।
Login to save your score, XP, coins and progress. Login
(15) विद्यार्थियों में से प्रथम, द्वितीय और तृतीय rank देने के तरीके कितने हैं?
In how many ways can first, second and third ranks be given among (15) students?
#permutations
#class11
#easy
A (455)
B (3375)
C (2730)
D (2730)
Explanation opens after your attempt
Step 1
Concept
The three ranks are ordered, so \({}^{15}P_{3}=15\times14\times13=2730\). Rank questions use permutation.
Step 2
Why this answer is correct
The correct answer is D. (2730). The three ranks are ordered, so \({}^{15}P_{3}=15\times14\times13=2730\). Rank questions use permutation.
Step 3
Exam Tip
तीन rank क्रम वाले हैं इसलिए \({}^{15}P_{3}=15\times14\times13=2730\)। rank वाले प्रश्न permutation के होते हैं।
Login to save your score, XP, coins and progress. Login
शब्द (COCO) के अलग-अलग arrangements कितने होंगे?
How many distinct arrangements are possible for the word (COCO)?
#permutations
#class11
#easy
A (12)
B (24)
C (6)
D (4)
Explanation opens after your attempt
Step 1
Concept
(C) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated letters.
Step 2
Why this answer is correct
The correct answer is C. (6). (C) occurs twice and (O) occurs twice, so \(\frac{4!}{2!2!}=6\). Divide by both repeated letters.
Step 3
Exam Tip
(C) दो बार और (O) दो बार हैं इसलिए \(\frac{4!}{2!2!}=6\)। दो repeated letters हों तो दोनों से भाग दें।
Login to save your score, XP, coins and progress. Login
अक्षरों (P,Q,R,S,T,U) से बिना पुनरावृत्ति कितने (2)-अक्षरी कोड बनेंगे?
How many (2)-letter codes can be formed from (P,Q,R,S,T,U) without repetition?
#permutations
#class11
#easy
A (12)
B (30)
C (36)
D (15)
Explanation opens after your attempt
Step 1
Concept
\({}^{6}P_{2}=6\times5=30\) codes are possible. In a code, changing the position of letters changes the code.
Step 2
Why this answer is correct
The correct answer is B. (30). \({}^{6}P_{2}=6\times5=30\) codes are possible. In a code, changing the position of letters changes the code.
Step 3
Exam Tip
\({}^{6}P_{2}=6\times5=30\) कोड बनेंगे। कोड में अक्षर का स्थान बदलने से कोड बदलता है।
Login to save your score, XP, coins and progress. Login
