(6) विषयों में से (4) विषय चुनने के कितने तरीके हैं?
How many ways are there to choose (4) subjects from (6) subjects?
#combinations
#class11
#easy
A (10)
B (12)
C (15)
D (24)
Explanation opens after your attempt
Step 1
Concept
\(\binom{6}{4}=\binom{6}{2}=15\). Remember \(\binom{n}{r}=\binom{n}{n-r}\).
Step 2
Why this answer is correct
The correct answer is C. (15). \(\binom{6}{4}=\binom{6}{2}=15\). Remember \(\binom{n}{r}=\binom{n}{n-r}\).
Step 3
Exam Tip
\(\binom{6}{4}=\binom{6}{2}=15\) होता है। \(\binom{n}{r}=\binom{n}{n-r}\) याद रखें।
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(13) विद्यार्थियों में से कक्षा प्रतिनिधि समिति के लिए (2) विद्यार्थी कितने तरीकों से चुने जा सकते हैं?
In how many ways can (2) students be selected from (13) students for a class representative committee?
#combinations
#class11
#easy
A (26)
B (78)
C (156)
D (169)
Explanation opens after your attempt
Step 1
Concept
It is only selection for a committee, so \(\binom{13}{2}=78\). If posts are not different, order is not counted.
Step 2
Why this answer is correct
The correct answer is B. (78). It is only selection for a committee, so \(\binom{13}{2}=78\). If posts are not different, order is not counted.
Step 3
Exam Tip
समिति के लिए केवल चयन है इसलिए \(\binom{13}{2}=78\) होगा। पद अलग न हों तो क्रम नहीं गिना जाता।
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(7) कुर्सियों में से (3) कुर्सियां चुनने के कितने तरीके हैं?
How many ways are there to select (3) chairs from (7) chairs?
#combinations
#class11
#easy
A (21)
B (35)
C (42)
D (210)
Explanation opens after your attempt
Step 1
Concept
No arrangement is required, so \(\binom{7}{3}=35\). Choosing chairs is a combination.
Step 2
Why this answer is correct
The correct answer is B. (35). No arrangement is required, so \(\binom{7}{3}=35\). Choosing chairs is a combination.
Step 3
Exam Tip
चयन में व्यवस्था नहीं करनी है इसलिए \(\binom{7}{3}=35\) है। कुर्सियां चुनना संयोजन है।
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(5) लाल गेंदों और (3) काली गेंदों में से (2) लाल गेंदें चुनने के कितने तरीके हैं?
From (5) red balls and (3) black balls, how many ways are there to choose (2) red balls?
#combinations
#class11
#easy
A (8)
B (10)
C (15)
D (30)
Explanation opens after your attempt
Step 1
Concept
Only red balls are being selected. Hence \(\binom{5}{2}=10\) is correct.
Step 2
Why this answer is correct
The correct answer is B. (10). Only red balls are being selected. Hence \(\binom{5}{2}=10\) is correct.
Step 3
Exam Tip
केवल लाल गेंदों में से चयन है। इसलिए \(\binom{5}{2}=10\) सही है।
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(6) लाल गेंदों और (4) नीली गेंदों में से (1) लाल और (1) नीली गेंद कितने तरीकों से चुनी जा सकती है?
From (6) red balls and (4) blue balls, in how many ways can (1) red and (1) blue ball be selected?
#combinations
#class11
#easy
A (10)
B (20)
C (24)
D (30)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{6}{1}\binom{4}{1}=24\). Apply multiplication rule for selections from different colors.
Step 2
Why this answer is correct
The correct answer is C. (24). The ways are \(\binom{6}{1}\binom{4}{1}=24\). Apply multiplication rule for selections from different colors.
Step 3
Exam Tip
तरीके \(\binom{6}{1}\binom{4}{1}=24\) हैं। अलग रंगों के चयन में गुणा नियम लगाएं।
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(11) अलग-अलग कार्डों में से (3) कार्ड चुनने के कितने तरीके होंगे?
How many ways are there to choose (3) cards from (11) different cards?
#combinations
#class11
#easy
A (33)
B (99)
C (165)
D (990)
Explanation opens after your attempt
Step 1
Concept
This is a direct use of \(\binom{11}{3}=165\). Order is not considered while choosing cards.
Step 2
Why this answer is correct
The correct answer is C. (165). This is a direct use of \(\binom{11}{3}=165\). Order is not considered while choosing cards.
Step 3
Exam Tip
यह \(\binom{11}{3}=165\) का सीधा प्रयोग है। कार्ड चुनने में क्रम नहीं देखा जाता।
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(12) उम्मीदवारों में से (2) उम्मीदवारों को पुरस्कार के लिए चुनने के कितने तरीके हैं?
How many ways are there to select (2) candidates for an award from (12) candidates?
#combinations
#class11
#easy
A (24)
B (66)
C (132)
D (144)
Explanation opens after your attempt
Step 1
Concept
The order of the two selected candidates is not important. Thus \(\binom{12}{2}=66\).
Step 2
Why this answer is correct
The correct answer is B. (66). The order of the two selected candidates is not important. Thus \(\binom{12}{2}=66\).
Step 3
Exam Tip
दो समान चयनित उम्मीदवारों का क्रम महत्वपूर्ण नहीं है। अतः \(\binom{12}{2}=66\) होगा।
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(9) सदस्यों में से (3) सदस्यों की समिति कितने तरीकों से चुनी जा सकती है?
In how many ways can a committee of (3) members be selected from (9) members?
#combinations
#class11
#easy
A (27)
B (72)
C (84)
D (504)
Explanation opens after your attempt
Step 1
Concept
This is a committee selection question, so \(\binom{9}{3}=84\). If posts are not assigned, order is not counted.
Step 2
Why this answer is correct
The correct answer is C. (84). This is a committee selection question, so \(\binom{9}{3}=84\). If posts are not assigned, order is not counted.
Step 3
Exam Tip
समिति चयन का प्रश्न है इसलिए \(\binom{9}{3}=84\) है। समिति में पद न हों तो क्रम नहीं गिना जाता।
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(10) व्यक्तियों में से (2) व्यक्तियों के हाथ मिलाने की कुल संख्या क्या होगी?
What is the total number of handshakes among (10) persons taken (2) at a time?
#combinations
#class11
#easy
A (20)
B (45)
C (90)
D (100)
Explanation opens after your attempt
Step 1
Concept
Each handshake is a pair of (2) persons. Therefore the total number is \(\binom{10}{2}=45\).
Step 2
Why this answer is correct
The correct answer is B. (45). Each handshake is a pair of (2) persons. Therefore the total number is \(\binom{10}{2}=45\).
Step 3
Exam Tip
हर हाथ मिलाना (2) व्यक्तियों की एक जोड़ी है। इसलिए कुल संख्या \(\binom{10}{2}=45\) है।
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(6) बिंदुओं से कितने वृत्त बनाए जा सकते हैं यदि कोई (3) बिंदु एक सीध में नहीं हैं?
How many circles can be formed from (6) points if no (3) points are collinear?
#combinations
#class11
#easy
A (12)
B (15)
C (20)
D (36)
Explanation opens after your attempt
Step 1
Concept
A circle is determined by (3) points. Hence \(\binom{6}{3}=20\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (20). A circle is determined by (3) points. Hence \(\binom{6}{3}=20\) is correct.
Step 3
Exam Tip
एक वृत्त को (3) बिंदु निर्धारित करते हैं। इसलिए \(\binom{6}{3}=20\) सही है।
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(7) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। इनसे कितने त्रिभुज बन सकते हैं?
There are (7) points, no (3) of which are collinear. How many triangles can be formed?
#combinations
#class11
#easy
A (21)
B (35)
C (42)
D (70)
Explanation opens after your attempt
Step 1
Concept
A triangle needs (3) points, so \(\binom{7}{3}=35\). The non-collinear condition makes triangles possible.
Step 2
Why this answer is correct
The correct answer is B. (35). A triangle needs (3) points, so \(\binom{7}{3}=35\). The non-collinear condition makes triangles possible.
Step 3
Exam Tip
त्रिभुज के लिए (3) बिंदु चाहिए इसलिए \(\binom{7}{3}=35\) है। सीध में न होने की शर्त त्रिभुज को संभव बनाती है।
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(8) बिंदुओं में से (2) बिंदु चुनकर कितनी रेखाएं बनाई जा सकती हैं?
How many lines can be formed by selecting (2) points from (8) points?
#combinations
#class11
#easy
A (16)
B (28)
C (56)
D (64)
Explanation opens after your attempt
Step 1
Concept
Two points form one line, so \(\binom{8}{2}=28\). The order of points is not counted.
Step 2
Why this answer is correct
The correct answer is B. (28). Two points form one line, so \(\binom{8}{2}=28\). The order of points is not counted.
Step 3
Exam Tip
दो बिंदु एक रेखा बनाते हैं इसलिए \(\binom{8}{2}=28\) है। बिंदुओं का क्रम नहीं गिना जाता।
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(4) सेब और (5) आमों में से (1) सेब और (2) आम कितने तरीकों से चुने जा सकते हैं?
From (4) apples and (5) mangoes, in how many ways can (1) apple and (2) mangoes be selected?
#combinations
#class11
#easy
A (10)
B (20)
C (40)
D (80)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). Multiply selections from separate fruit groups.
Step 2
Why this answer is correct
The correct answer is C. (40). The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). Multiply selections from separate fruit groups.
Step 3
Exam Tip
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग फल वर्गों के चयन को गुणा करें।
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(5) लड़कों और (4) लड़कियों में से (2) लड़के चुनने के कितने तरीके हैं?
From (5) boys and (4) girls, how many ways are there to select (2) boys?
#combinations
#class11
#easy
A (10)
B (20)
C (9)
D (30)
Explanation opens after your attempt
Step 1
Concept
Only boys are being selected, so \(\binom{5}{2}=10\). Do not get confused by extra information.
Step 2
Why this answer is correct
The correct answer is A. (10). Only boys are being selected, so \(\binom{5}{2}=10\). Do not get confused by extra information.
Step 3
Exam Tip
केवल लड़कों में से चयन है इसलिए \(\binom{5}{2}=10\) होगा। अनावश्यक सूचना से भ्रमित न हों।
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(4) लड़कों और (3) लड़कियों में से (1) लड़का और (1) लड़की कितने तरीकों से चुने जा सकते हैं?
From (4) boys and (3) girls, in how many ways can (1) boy and (1) girl be selected?
#combinations
#class11
#easy
A (7)
B (12)
C (24)
D (1)
Explanation opens after your attempt
Step 1
Concept
The selection is \(\binom{4}{1}\binom{3}{1}=12\). Multiply choices when selecting from different groups.
Step 2
Why this answer is correct
The correct answer is B. (12). The selection is \(\binom{4}{1}\binom{3}{1}=12\). Multiply choices when selecting from different groups.
Step 3
Exam Tip
चयन \(\binom{4}{1}\binom{3}{1}=12\) है। अलग-अलग वर्गों से चयन में गुणा करते हैं।
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(6) दोस्तों में से (2) दोस्तों की जोड़ी कितने तरीकों से चुनी जा सकती है?
In how many ways can a pair of friends be selected from (6) friends?
#combinations
#class11
#easy
A (12)
B (15)
C (18)
D (30)
Explanation opens after your attempt
Step 1
Concept
For a pair, \(\binom{6}{2}=15\). In a pair, first and second are not counted separately.
Step 2
Why this answer is correct
The correct answer is B. (15). For a pair, \(\binom{6}{2}=15\). In a pair, first and second are not counted separately.
Step 3
Exam Tip
जोड़ी के लिए \(\binom{6}{2}=15\) होगा। जोड़ी में पहला और दूसरा अलग नहीं गिने जाते।
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(5) रंगों में से (5) रंग चुनने के कितने तरीके हैं?
How many ways are there to choose all (5) colors from (5) colors?
#combinations
#class11
#easy
A (0)
B (1)
C (5)
D (25)
Explanation opens after your attempt
Step 1
Concept
There is only one way to choose all objects. Hence \(\binom{5}{5}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is only one way to choose all objects. Hence \(\binom{5}{5}=1\).
Step 3
Exam Tip
सभी वस्तुएं चुनने का केवल एक तरीका होता है। इसलिए \(\binom{5}{5}=1\) है।
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(10) पेन में से (0) पेन चुनने का एक ही तरीका क्यों माना जाता है?
Why is selecting (0) pens from (10) pens considered one way?
#combinations
#class11
#easy
A क्योंकि \(\binom{10}{0}=1\) / Because \(\binom{10}{0}=1\)
B क्योंकि \(\binom{10}{0}=0\) / Because \(\binom{10}{0}=0\)
C क्योंकि \(\binom{10}{0}=10\) / Because \(\binom{10}{0}=10\)
D क्योंकि \(\binom{10}{0}=100\) / Because \(\binom{10}{0}=100\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि \(\binom{10}{0}=1\) / Because \(\binom{10}{0}=1\)
Step 1
Concept
Choosing nothing is also one definite selection. Therefore \(\binom{n}{0}=1\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(\binom{10}{0}=1\) / Because \(\binom{10}{0}=1\). Choosing nothing is also one definite selection. Therefore \(\binom{n}{0}=1\).
Step 3
Exam Tip
कुछ भी न चुनना भी एक निश्चित चयन है। इसलिए \(\binom{n}{0}=1\) होता है।
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(9) प्रश्नों में से (1) प्रश्न चुनने के कितने तरीके हैं?
How many ways are there to select (1) question from (9) questions?
#combinations
#class11
#easy
A (1)
B (8)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
Selecting one object gives \(\binom{9}{1}=9\). Remember \(\binom{n}{1}=n\).
Step 2
Why this answer is correct
The correct answer is C. (9). Selecting one object gives \(\binom{9}{1}=9\). Remember \(\binom{n}{1}=n\).
Step 3
Exam Tip
एक वस्तु चुनने पर \(\binom{9}{1}=9\) होता है। \(\binom{n}{1}=n\) याद रखें।
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(8) खिलाड़ियों में से (4) खिलाड़ियों का समूह कितने तरीकों से बनाया जा सकता है?
In how many ways can a group of (4) players be formed from (8) players?
#combinations
#class11
#easy
A (32)
B (56)
C (70)
D (1680)
Explanation opens after your attempt
Step 1
Concept
A group has no order, so \(\binom{8}{4}=70\). In exams, the word group usually indicates combination.
Step 2
Why this answer is correct
The correct answer is C. (70). A group has no order, so \(\binom{8}{4}=70\). In exams, the word group usually indicates combination.
Step 3
Exam Tip
समूह में क्रम नहीं होता इसलिए \(\binom{8}{4}=70\) होगा। परीक्षा में समूह शब्द दिखे तो संयोजन लगाएं।
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(7) अलग-अलग फूलों में से (2) फूल चुनने के कितने तरीके हैं?
How many ways are there to choose (2) flowers from (7) different flowers?
#combinations
#class11
#easy
A (14)
B (21)
C (35)
D (42)
Explanation opens after your attempt
Step 1
Concept
It is \(\binom{7}{2}=\frac{7\cdot6}{2}=21\). Such questions form pairs of objects.
Step 2
Why this answer is correct
The correct answer is B. (21). It is \(\binom{7}{2}=\frac{7\cdot6}{2}=21\). Such questions form pairs of objects.
Step 3
Exam Tip
यह \(\binom{7}{2}=\frac{7\cdot6}{2}=21\) है। ऐसे प्रश्नों में वस्तुओं की जोड़ी बनती है।
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(6) विद्यार्थियों में से (3) विद्यार्थियों की टीम कितने तरीकों से चुनी जा सकती है?
In how many ways can a team of (3) students be selected from (6) students?
#combinations
#class11
#easy
A (18)
B (15)
C (20)
D (12)
Explanation opens after your attempt
Step 1
Concept
Order is not important while selecting a team. Hence \(\binom{6}{3}=20\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (20). Order is not important while selecting a team. Hence \(\binom{6}{3}=20\) is correct.
Step 3
Exam Tip
टीम चुनने में क्रम महत्वपूर्ण नहीं होता। इसलिए \(\binom{6}{3}=20\) सही है।
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यदि (5) पुस्तकों में से (2) पुस्तकें चुननी हों तो कुल कितने चयन होंगे?
If (2) books are to be selected from (5) books, how many selections are possible?
#combinations
#class11
#easy
A (8)
B (10)
C (12)
D (20)
Explanation opens after your attempt
Step 1
Concept
This is a direct use of \(\binom{5}{2}=10\). In exams, order is not counted in selection.
Step 2
Why this answer is correct
The correct answer is B. (10). This is a direct use of \(\binom{5}{2}=10\). In exams, order is not counted in selection.
Step 3
Exam Tip
यह \(\binom{5}{2}=10\) का सरल प्रयोग है। परीक्षा में चयन का क्रम नहीं गिना जाता।
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शब्द (ENGINEERING) के अक्षरों से कितनी व्यवस्थाएँ बनेंगी जिनमें कोई भी दो स्वर साथ-साथ न आएँ?
How many arrangements of the letters of (ENGINEERING) can be formed such that no two vowels are adjacent?
#permutations
#class11
#expert
#gap_method
A (6300)
B (10500)
C (15120)
D (12600)
Explanation opens after your attempt
Correct Answer
D. (12600)
Step 1
Concept
Arrange the consonants in \(\frac{6!}{3!2!}\) ways and place the vowels in (7) gaps in \(\binom{7}{5}\frac{5!}{3!2!}\) ways. Use the gap method when vowels must not be adjacent.
Step 2
Why this answer is correct
The correct answer is D. (12600). Arrange the consonants in \(\frac{6!}{3!2!}\) ways and place the vowels in (7) gaps in \(\binom{7}{5}\frac{5!}{3!2!}\) ways. Use the gap method when vowels must not be adjacent.
Step 3
Exam Tip
पहले व्यंजनों को \(\frac{6!}{3!2!}\) तरीकों से सजाएँ और (7) खाली स्थानों में स्वरों को \(\binom{7}{5}\frac{5!}{3!2!}\) तरीकों से रखें। स्वर साथ न हों तो gap method लगाएँ।
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(10) अलग-अलग वस्तुओं को एक पंक्ति में सजाया जाता है। तीन निश्चित वस्तुओं का आपसी क्रम (A) पहले, फिर (B), फिर (C) होना चाहिए, लेकिन वे साथ-साथ होना जरूरी नहीं है। कुल व्यवस्थाएँ कितनी हैं?
(10) distinct objects are arranged in a row. Three particular objects must appear in the relative order (A) before (B) before (C), but they need not be adjacent. How many arrangements are possible?
#permutations
#class11
#expert
#relative_order
A (1209600)
B (604800)
C (1814400)
D (3628800)
Explanation opens after your attempt
Correct Answer
B. (604800)
Step 1
Concept
Among total (10!) arrangements, the (3!) relative orders of the three special objects are equally likely, so the count is \(\frac{10!}{3!}\). If relative order is fixed, divide by that group's factorial.
Step 2
Why this answer is correct
The correct answer is B. (604800). Among total (10!) arrangements, the (3!) relative orders of the three special objects are equally likely, so the count is \(\frac{10!}{3!}\). If relative order is fixed, divide by that group's factorial.
Step 3
Exam Tip
कुल (10!) क्रमों में तीन विशेष वस्तुओं के (3!) आपसी क्रम समान रूप से संभव हैं, इसलिए \(\frac{10!}{3!}\) है। relative order fixed हो तो कुल को उस group के factorial से बाँटें।
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शब्द (SELECTION) के अक्षरों से बनने वाली व्यवस्थाओं में सभी स्वर साथ-साथ हों, ऐसी कितनी व्यवस्थाएँ हैं?
In arrangements of the letters of (SELECTION), how many have all vowels together?
#permutations
#class11
#expert
#vowel_block
A (14400)
B (7200)
C (28800)
D (43200)
Explanation opens after your attempt
Correct Answer
A. (14400)
Step 1
Concept
The vowels (E,E,I,O) form one block, so (6) outside units arrange in (6!) ways and the vowels arrange internally in \(\frac{4!}{2!}\) ways. Do not forget repeated vowels inside the vowel block.
Step 2
Why this answer is correct
The correct answer is A. (14400). The vowels (E,E,I,O) form one block, so (6) outside units arrange in (6!) ways and the vowels arrange internally in \(\frac{4!}{2!}\) ways. Do not forget repeated vowels inside the vowel block.
Step 3
Exam Tip
स्वर (E,E,I,O) एक ब्लॉक हैं, इसलिए बाहरी (6) इकाइयाँ (6!) तरीकों से और अंदर स्वर \(\frac{4!}{2!}\) तरीकों से सजते हैं। vowel block में repeated vowels का भाग देना न भूलें।
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(8) अलग-अलग कार्डों को एक पंक्ति में रखना है। दो विशेष कार्डों के बीच ठीक (2) कार्ड हों, ऐसे कितने क्रम हैं?
(8) distinct cards are to be arranged in a row. How many arrangements have exactly (2) cards between two particular cards?
#permutations
#class11
#expert
#exact_gap
A (7200)
B (8640)
C (10080)
D (5760)
Explanation opens after your attempt
Step 1
Concept
The positions of the two particular cards differ by (3), giving (5) position pairs and (2) orders, then the remaining (6!) arrangements. Count position pairs for exact gaps.
Step 2
Why this answer is correct
The correct answer is B. (8640). The positions of the two particular cards differ by (3), giving (5) position pairs and (2) orders, then the remaining (6!) arrangements. Count position pairs for exact gaps.
Step 3
Exam Tip
दो विशेष कार्डों के स्थानों में अंतर (3) होगा, ऐसे (5) स्थान-जोड़े हैं और उनका क्रम (2) तरीकों से है, फिर (6!) क्रम हैं। exact gap में position pairs गिनें।
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(9) अलग-अलग खिलाड़ियों को एक पंक्ति में खड़ा करना है। कप्तान हमेशा गोलकीपर से बाएँ हो लेकिन उनके बीच लोगों की संख्या कोई भी हो सकती है। कितने क्रम संभव हैं?
(9) distinct players are to stand in a row. The captain must always be to the left of the goalkeeper, with any number of people between them. How many arrangements are possible?
#permutations
#class11
#expert
#relative_order
A (362880)
B (181440)
C (90720)
D (40320)
Explanation opens after your attempt
Correct Answer
B. (181440)
Step 1
Concept
In total (9!) orders, the two relative orders of captain and goalkeeper are equally likely, so half are valid. If relative order is fixed, divide by (2).
Step 2
Why this answer is correct
The correct answer is B. (181440). In total (9!) orders, the two relative orders of captain and goalkeeper are equally likely, so half are valid. If relative order is fixed, divide by (2).
Step 3
Exam Tip
कुल (9!) क्रमों में कप्तान और गोलकीपर के दो आपसी क्रम बराबर हैं, इसलिए आधे क्रम मान्य हैं। relative order fixed हो तो कुल को (2) से बाँटें।
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शब्द (INDEPENDENT) के अक्षरों से बनने वाली अलग-अलग व्यवस्थाओं की संख्या क्या है?
What is the number of distinct arrangements of the letters of (INDEPENDENT)?
#permutations
#class11
#expert
#repeated_letters
A (831600)
B (1663200)
C (277200)
D (554400)
Explanation opens after your attempt
Correct Answer
A. (831600)
Step 1
Concept
There are (11) letters with (N,E,D) repeated (3,3,2) times, so the count is \(\frac{11!}{3!3!2!}\). Write all repeated-letter frequencies.
Step 2
Why this answer is correct
The correct answer is A. (831600). There are (11) letters with (N,E,D) repeated (3,3,2) times, so the count is \(\frac{11!}{3!3!2!}\). Write all repeated-letter frequencies.
Step 3
Exam Tip
कुल (11) अक्षर हैं और (N,E,D) क्रमशः (3,3,2) बार आते हैं, इसलिए \(\frac{11!}{3!3!2!}\) है। repeated letters की सभी आवृत्तियाँ लिखें।
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अंकों (0,1,2,3,4,5,6,7,8,9) से बिना पुनरावृत्ति के कितने (6) अंकों के पासवर्ड बन सकते हैं जिनका पहला अंक शून्य नहीं है और अंतिम अंक सम है?
How many (6)-digit passwords can be formed from digits (0,1,2,3,4,5,6,7,8,9) without repetition such that the first digit is not zero and the last digit is even?
#permutations
#class11
#expert
#password_digits
A (60480)
B (67200)
C (80640)
D (54432)
Explanation opens after your attempt
Correct Answer
D. (54432)
Step 1
Concept
If the last digit is (0), count (9P5); if it is (2,4,6,8), count \(4\cdot8\cdot8P4\). Separate zero and nonzero last-digit cases.
Step 2
Why this answer is correct
The correct answer is D. (54432). If the last digit is (0), count (9P5); if it is (2,4,6,8), count \(4\cdot8\cdot8P4\). Separate zero and nonzero last-digit cases.
Step 3
Exam Tip
अंतिम अंक (0) होने पर (9P5) और अंतिम अंक (2,4,6,8) होने पर \(4\cdot8\cdot8P4\) मिलते हैं। शून्य और गैर-शून्य अंतिम अंक के cases अलग करें।
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