Class 11 Mathematics - Permutations And Combinations - Permutations Easy Quiz

Concept

Here the posts are different, so order matters and the answer is \({}^{6}P_{3}=120\). In exams, use permutation when posts are distinct.

Why this answer is correct

The correct answer is B. (120). Here the posts are different, so order matters and the answer is \({}^{6}P_{3}=120\). In exams, use permutation when posts are distinct.

Exam Tip

यहाँ पद अलग हैं इसलिए क्रम महत्त्वपूर्ण है और उत्तर \({}^{6}P_{3}=120\) है। परीक्षा में पद अलग हों तो permutation लें।

Concept

The word (CAT) has (3) distinct letters, so (3!=6) words are possible. For small words, arrange all letters using factorial.

Why this answer is correct

The correct answer is C. (6). The word (CAT) has (3) distinct letters, so (3!=6) words are possible. For small words, arrange all letters using factorial.

Exam Tip

(CAT) में (3) अलग अक्षर हैं इसलिए (3!=6) शब्द बनेंगे। छोटे शब्दों में सभी अक्षरों की व्यवस्था सीधे factorial से करें।

Concept

The arrangement of (4) distinct flags is (4!=24). Placing in a straight line is a linear permutation.

Why this answer is correct

The correct answer is D. (24). The arrangement of (4) distinct flags is (4!=24). Placing in a straight line is a linear permutation.

Exam Tip

(4) अलग झंडों की व्यवस्था (4!=24) होती है। सीधी रेखा में लगाना linear permutation है।

Concept

\({}^{7}P_{2}=7\times6=42\). For small (r), multiply (r) decreasing factors.

Why this answer is correct

The correct answer is A. (42). \({}^{7}P_{2}=7\times6=42\). For small (r), multiply (r) decreasing factors.

Exam Tip

\({}^{7}P_{2}=7\times6=42\) होता है। छोटे (r) के लिए घटते हुए (r) गुणनखंड लें।

Concept

Captain and vice-captain are distinct posts, so \({}^{8}P_{2}=56\). When roles differ, order matters.

Why this answer is correct

The correct answer is B. (56). Captain and vice-captain are distinct posts, so \({}^{8}P_{2}=56\). When roles differ, order matters.

Exam Tip

कप्तान और उपकप्तान अलग पद हैं इसलिए \({}^{8}P_{2}=56\) होगा। जब जिम्मेदारी अलग हो तो क्रम महत्त्वपूर्ण होता है।

Concept

The arrangement of three distinct digits is (3!=6). Without repetition, choices decrease place by place.

Why this answer is correct

The correct answer is C. (6). The arrangement of three distinct digits is (3!=6). Without repetition, choices decrease place by place.

Exam Tip

तीनों अलग अंकों की व्यवस्था (3!=6) है। बिना पुनरावृत्ति में हर स्थान के विकल्प घटते हैं।

Concept

The number of ways to seat (5) people in (5) places is (5!=120). Remember factorial for seating in a row.

Why this answer is correct

The correct answer is D. (120). The number of ways to seat (5) people in (5) places is (5!=120). Remember factorial for seating in a row.

Exam Tip

(5) लोगों को (5) स्थानों पर बैठाने के तरीके (5!=120) हैं। seating in a row में factorial याद रखें।

Concept

Since \({}^{n}P_{1}=n\), \({}^{5}P_{1}=5\). Choosing and arranging one object gives (n) ways.

Why this answer is correct

The correct answer is A. (5). Since \({}^{n}P_{1}=n\), \({}^{5}P_{1}=5\). Choosing and arranging one object gives (n) ways.

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{5}P_{1}=5\)। एक वस्तु चुनकर व्यवस्थित करने में (n) तरीके होते हैं।

Concept

The word (DOG) has (3) distinct letters, so there are (3!=6) arrangements. Use factorial when letters are distinct.

Why this answer is correct

The correct answer is B. (6). The word (DOG) has (3) distinct letters, so there are (3!=6) arrangements. Use factorial when letters are distinct.

Exam Tip

(DOG) में (3) अलग अक्षर हैं इसलिए (3!=6) व्यवस्थाएँ हैं। अक्षर अलग हों तो सीधे factorial प्रयोग करें।

Concept

The answer is \({}^{9}P_{2}=9\times8=72\). After the first place, the second choice decreases by one.

Why this answer is correct

The correct answer is C. (72). The answer is \({}^{9}P_{2}=9\times8=72\). After the first place, the second choice decreases by one.

Exam Tip

उत्तर \({}^{9}P_{2}=9\times8=72\) है। पहले स्थान के बाद दूसरा विकल्प एक कम हो जाता है।

Concept

First and second places are ordered, so \({}^{7}P_{2}=42\). Ranking questions use permutation.

Why this answer is correct

The correct answer is D. (42). First and second places are ordered, so \({}^{7}P_{2}=42\). Ranking questions use permutation.

Exam Tip

प्रथम और द्वितीय स्थान क्रम वाले हैं इसलिए \({}^{7}P_{2}=42\) होगा। रैंकिंग में permutation लगता है।

Concept

\({}^{6}P_{3}=6\times5\times4=120\). When (r=3), take three decreasing factors.

Why this answer is correct

The correct answer is B. (120). \({}^{6}P_{3}=6\times5\times4=120\). When (r=3), take three decreasing factors.

Exam Tip

\({}^{6}P_{3}=6\times5\times4=120\) है। (r=3) हो तो तीन घटते गुणनखंड लें।

Concept

The prizes are distinct and students cannot repeat, so \({}^{5}P_{3}=60\). Distribution of distinct prizes makes order important.

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.

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।

Concept

The word (MATH) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.

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.

Exam Tip

(MATH) में (4) अलग अक्षर हैं इसलिए (4!=24)। सभी अलग अक्षरों की पूरी व्यवस्था factorial से मिलती है।

Concept

\({}^{n}P_{0}=1\) because there is one way to choose nothing. Remember this standard result.

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.

Exam Tip

\({}^{n}P_{0}=1\) होता है क्योंकि कोई वस्तु न चुनने का एक तरीका होता है। यह standard result याद रखें।

Concept

Order matters on two special seats, so \({}^{4}P_{2}=12\). If seats are distinct, use permutation.

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.

Exam Tip

दो विशेष सीटों पर क्रम महत्त्वपूर्ण है इसलिए \({}^{4}P_{2}=12\)। सीटें अलग हों तो permutation लें।

Concept

The answer is \({}^{4}P_{3}=4\times3\times2=24\). Choosing and arranging in order is permutation.

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.

Exam Tip

उत्तर \({}^{4}P_{3}=4\times3\times2=24\) है। चुनना और क्रम में रखना permutation है।

Concept

There are (5) choices for one place, so the answer is (5). For a one-place arrangement, available choices are the answer.

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.

Exam Tip

एक स्थान के लिए (5) विकल्प हैं इसलिए उत्तर (5) है। (1)-स्थान वाली व्यवस्था में उपलब्ध विकल्प ही उत्तर होते हैं।

Concept

\({}^{7}P_{4}=7\times6\times5\times4=840\). For (r) places, multiply (r) decreasing factors.

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.

Exam Tip

\({}^{7}P_{4}=7\times6\times5\times4=840\)। (r) स्थानों के लिए (r) घटते गुणनखंड लें।

Concept

The word (SUN) has (3) distinct letters, so (3!=6). Count the letters and apply factorial for small words.

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.

Exam Tip

(SUN) में (3) अलग अक्षर हैं इसलिए (3!=6)। छोटे शब्दों में अक्षर गिनकर factorial लगाएं।

Concept

Since \({}^{n}P_{1}=n\), \({}^{10}P_{1}=10\). For one position, write (n) directly.

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.

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{10}P_{1}=10\)। एक पद की व्यवस्था में सीधे (n) लिखें।

Concept

There are (6) choices for the first position and (5) for the second, so \(6\times5=30\). Count order when positions are distinct.

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.

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।

Concept

\({}^{5}P_{2}=5\times4=20\). Start from (n) and take (r) decreasing factors.

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.

Exam Tip

\({}^{5}P_{2}=5\times4=20\)। (n) से शुरू करके (r) तक घटते गुणनखंड लें।

Concept

This is \({}^{5}P_{3}=5\times4\times3=60\). Changing the order of cards changes the arrangement.

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.

Exam Tip

यह \({}^{5}P_{3}=5\times4\times3=60\) है। कार्डों का क्रम बदलने से व्यवस्था बदलती है।

Concept

Because two (O)'s are identical, arrangements are \(\frac{4!}{2!}=12\). For identical objects, divide by their factorial.

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.

Exam Tip

दो समान (O) होने से व्यवस्थाएँ \(\frac{4!}{2!}=12\) होंगी। समान वस्तुओं के लिए factorial से भाग दें।

Concept

This is \({}^{4}P_{3}=4\times3\times2=24\). Standing in a row is an ordered arrangement.

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.

Exam Tip

यह \({}^{4}P_{3}=4\times3\times2=24\) है। पंक्ति में खड़ा करना ordered arrangement है।

Concept

The arrangement of (6) distinct balls is (6!=720). If all objects are distinct, total arrangements are factorial.

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.

Exam Tip

(6) अलग गेंदों की व्यवस्था (6!=720) है। सभी वस्तुएँ अलग हों तो कुल व्यवस्था factorial होती है।

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.

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.

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।

Concept

\({}^{n}P_{0}=1\), so \({}^{8}P_{0}=1\). The arrangement of zero objects is counted as one.

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.

Exam Tip

\({}^{n}P_{0}=1\) इसलिए \({}^{8}P_{0}=1\)। शून्य वस्तुओं की arrangement एक मानी जाती है।

Concept

The arrangement of (3) distinct coins is (3!=6). Use factorial for full arrangement of distinct objects.

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.

Exam Tip

(3) अलग सिक्कों की व्यवस्था (3!=6) है। अलग objects की पूरी arrangement factorial से करें।

Concept

There are two distinct posts, so \({}^{10}P_{2}=10\times9=90\). Order matters for different designations.

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.

Exam Tip

दो अलग पद हैं इसलिए \({}^{10}P_{2}=10\times9=90\)। अलग designation में order महत्त्वपूर्ण होता है।

Concept

The word (RAVI) has (4) distinct letters, so (4!=24). Apply factorial when all letters are distinct.

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.

Exam Tip

(RAVI) में (4) अलग अक्षर हैं इसलिए (4!=24)। सभी distinct letters हों तो factorial लगाएं।

Concept

There are (6) choices for the first period and (5) for the second, so (30). Timetable order uses permutation.

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.

Exam Tip

पहले पीरियड के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए (30)। timetable order में permutation लगता है।

Concept

\({}^{4}P_{4}=4!=24\). When (r=n), \({}^{n}P_{n}=n!\).

Why this answer is correct

The correct answer is D. (24). \({}^{4}P_{4}=4!=24\). When (r=n), \({}^{n}P_{n}=n!\).

Exam Tip

\({}^{4}P_{4}=4!=24\) होता है। जब (r=n) हो तो \({}^{n}P_{n}=n!\)।

Concept

For three places, there are \(4\times3\times2=24\) ways. Without repetition, choices decrease at each next place.

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.

Exam Tip

तीन स्थानों के लिए \(4\times3\times2=24\) तरीके हैं। बिना पुनरावृत्ति में हर अगले स्थान पर विकल्प घटता है।

Concept

This is \({}^{7}P_{3}=7\times6\times5=210\). Changing order on a shelf changes the arrangement.

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.

Exam Tip

यह \({}^{7}P_{3}=7\times6\times5=210\) है। शेल्फ पर क्रम बदलने से व्यवस्था बदलती है।

Concept

The arrangements are \(\frac{5!}{2!2!}=30\). Do not forget to divide by factorials of identical letters.

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.

Exam Tip

व्यवस्थाएँ \(\frac{5!}{2!2!}=30\) होंगी। समान अक्षरों के factorial से भाग देना न भूलें।

Concept

The three ranks are ordered, so \({}^{5}P_{3}=60\). In rank questions, use permutation, not combination.

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.

Exam Tip

तीन rank क्रम वाले हैं इसलिए \({}^{5}P_{3}=60\)। rank वाले प्रश्नों में combination नहीं, permutation लें।

Concept

The arrangements of (3) distinct toys are (3!=6). If all objects are distinct, total arrangements are factorial.

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.

Exam Tip

(3) अलग खिलौनों की arrangements (3!=6) हैं। सभी objects अलग हों तो total arrangements factorial होती हैं।

Concept

The answer is \({}^{9}P_{3}=9\times8\times7=504\). Left-right order on a stage is important.

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.

Exam Tip

उत्तर \({}^{9}P_{3}=9\times8\times7=504\) है। मंच पर बायाँ-दायाँ क्रम महत्त्वपूर्ण होता है।

Concept

\({}^{3}P_{2}=3\times2=6\). In small permutations, write decreasing factors directly.

Why this answer is correct

The correct answer is C. (6). \({}^{3}P_{2}=3\times2=6\). In small permutations, write decreasing factors directly.

Exam Tip

\({}^{3}P_{2}=3\times2=6\) है। छोटे permutation में सीधे घटते गुणनखंड लिखें।

Concept

In a straight list, the arrangement of (5) distinct objects is (5!=120). Do not use circular counting here.

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.

Exam Tip

सीधी सूची में (5) अलग वस्तुओं की व्यवस्था (5!=120) है। यहाँ circular counting नहीं करनी है।

Concept

Left-to-right order matters, so (4!=24). Changing the order changes the arrangement.

Why this answer is correct

The correct answer is B. (24). Left-to-right order matters, so (4!=24). Changing the order changes the arrangement.

Exam Tip

बाएँ से दाएँ क्रम महत्त्वपूर्ण है इसलिए (4!=24)। क्रम बदलते ही arrangement बदल जाती है।

Concept

For (2) ordered positions from (5) letters, \({}^{5}P_{2}=20\). In ordered pairs, ((P,Q)) and ((Q,P)) are different.

Why this answer is correct

The correct answer is C. (20). For (2) ordered positions from (5) letters, \({}^{5}P_{2}=20\). In ordered pairs, ((P,Q)) and ((Q,P)) are different.

Exam Tip

(5) अक्षरों से (2) ordered positions के लिए \({}^{5}P_{2}=20\)। ordered pairs में ((P,Q)) और ((Q,P)) अलग होते हैं।

Concept

There are (4) letters and two identical (O)'s, so \(\frac{4!}{2!}=12\). Use a denominator for repeated letters.

Why this answer is correct

The correct answer is D. (12). There are (4) letters and two identical (O)'s, so \(\frac{4!}{2!}=12\). Use a denominator for repeated letters.

Exam Tip

कुल अक्षर (4) हैं और (O) दो समान हैं इसलिए \(\frac{4!}{2!}=12\)। repeated letters में denominator लगाएं।

Concept

\({}^{6}P_{2}=6\times5=30\). For two places, take (n) and then (n-1).

Why this answer is correct

The correct answer is A. (30). \({}^{6}P_{2}=6\times5=30\). For two places, take (n) and then (n-1).

Exam Tip

\({}^{6}P_{2}=6\times5=30\) है। दो स्थान हों तो पहले (n), फिर (n-1) लें।

Concept

The answer is \({}^{8}P_{3}=8\times7\times6=336\). Changing the order of selected books changes the way.

Why this answer is correct

The correct answer is B. (336). The answer is \({}^{8}P_{3}=8\times7\times6=336\). Changing the order of selected books changes the way.

Exam Tip

उत्तर \({}^{8}P_{3}=8\times7\times6=336\) है। चुनी गई पुस्तकों का क्रम बदलने से तरीका बदलता है।

Concept

\({}^{6}P_{3}=6\times5\times4=120\). Use permutation for ordered arrangements.

Why this answer is correct

The correct answer is A. (120). \({}^{6}P_{3}=6\times5\times4=120\). Use permutation for ordered arrangements.

Exam Tip

\({}^{6}P_{3}=6\times5\times4=120\) होता है। क्रम वाली व्यवस्था में permutation लगाएं।

Concept

There are (4) choices for the first place and (3) for the second, so \(4\times3=12\). In passwords, changing positions changes the arrangement.

Why this answer is correct

The correct answer is B. (12). There are (4) choices for the first place and (3) for the second, so \(4\times3=12\). In passwords, changing positions changes the arrangement.

Exam Tip

पहले स्थान के लिए (4) और दूसरे के लिए (3) विकल्प हैं, इसलिए \(4\times3=12\)। पासवर्ड में स्थान बदलने से arrangement बदलती है।

Concept

The arrangement of (7) distinct objects is (7!=5040). Use factorial when all objects are distinct.

Why this answer is correct

The correct answer is A. (5040). The arrangement of (7) distinct objects is (7!=5040). Use factorial when all objects are distinct.

Exam Tip

(7) अलग वस्तुओं की व्यवस्था (7!=5040) होती है। सभी वस्तुएँ अलग हों तो factorial लगाएं।

Concept

The positions are ordered, so \({}^{9}P_{2}=9\times8=72\). Use permutation in ranking questions.

Why this answer is correct

The correct answer is B. (72). The positions are ordered, so \({}^{9}P_{2}=9\times8=72\). Use permutation in ranking questions.

Exam Tip

स्थान क्रम वाले हैं इसलिए \({}^{9}P_{2}=9\times8=72\)। रैंकिंग में क्रमचय का प्रयोग करें।

Concept

The word (LAMP) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.

Why this answer is correct

The correct answer is C. (24). The word (LAMP) has (4) distinct letters, so (4!=24). Full arrangement of distinct letters is found by factorial.

Exam Tip

(LAMP) में (4) अलग अक्षर हैं इसलिए (4!=24)। अलग अक्षरों की पूरी व्यवस्था factorial से मिलती है।

Concept

There are (4) choices for tens and (3) for units, so \(4\times3=12\). Changing places changes the number.

Why this answer is correct

The correct answer is D. (12). There are (4) choices for tens and (3) for units, so \(4\times3=12\). Changing places changes the number.

Exam Tip

दहाई के लिए (4) और इकाई के लिए (3) विकल्प हैं इसलिए \(4\times3=12\)। स्थान बदलने से संख्या बदलती है।

Concept

\({}^{8}P_{2}=8\times7=56\). When (r=2), take two decreasing factors.

Why this answer is correct

The correct answer is A. (56). \({}^{8}P_{2}=8\times7=56\). When (r=2), take two decreasing factors.

Exam Tip

\({}^{8}P_{2}=8\times7=56\) होता है। (r=2) हो तो दो घटते गुणनखंड लें।

Concept

There are (6) choices for the first position and (5) for the second, so (30). Order matters in distinct positions.

Why this answer is correct

The correct answer is B. (30). There are (6) choices for the first position and (5) for the second, so (30). Order matters in distinct positions.

Exam Tip

पहले स्थान के लिए (6) और दूसरे के लिए (5) विकल्प हैं इसलिए (30)। अलग स्थानों में क्रम महत्त्वपूर्ण होता है।

Concept

Since (P) occurs twice, arrangements are \(\frac{5!}{2!}=60\). Divide by the factorial of repeated letters.

Why this answer is correct

The correct answer is C. (60). Since (P) occurs twice, arrangements are \(\frac{5!}{2!}=60\). Divide by the factorial of repeated letters.

Exam Tip

(P) दो बार है इसलिए व्यवस्थाएँ \(\frac{5!}{2!}=60\) होंगी। समान अक्षरों के factorial से भाग दें।

Concept

The chairs are distinct, so \({}^{5}P_{3}=5\times4\times3=60\). Changing seats changes the arrangement.

Why this answer is correct

The correct answer is D. (60). The chairs are distinct, so \({}^{5}P_{3}=5\times4\times3=60\). Changing seats changes the arrangement.

Exam Tip

कुर्सियाँ अलग हैं इसलिए \({}^{5}P_{3}=5\times4\times3=60\)। सीटों का क्रम बदलने से व्यवस्था बदलती है।

Concept

When (r=n), \({}^{n}P_{n}=n!\), so \({}^{6}P_{6}=720\). Use factorial when all are selected.

Why this answer is correct

The correct answer is A. (720). When (r=n), \({}^{n}P_{n}=n!\), so \({}^{6}P_{6}=720\). Use factorial when all are selected.

Exam Tip

जब (r=n) हो तो \({}^{n}P_{n}=n!\), इसलिए \({}^{6}P_{6}=720\)। पूरा चयन होने पर factorial लें।

Concept

Zero cannot be in the tens place, so \(3\times3=9\) numbers are possible. Check the first-place condition in number formation.

Why this answer is correct

The correct answer is B. (9). Zero cannot be in the tens place, so \(3\times3=9\) numbers are possible. Check the first-place condition in number formation.

Exam Tip

दहाई स्थान पर (0) नहीं आ सकता, इसलिए \(3\times3=9\) संख्याएँ बनेंगी। संख्या बनाते समय पहले स्थान की शर्त देखें।

Concept

(4) distinct flowers can be placed in (4) distinct positions in (4!=24) ways. Placing one each counts arrangements.

Why this answer is correct

The correct answer is C. (24). (4) distinct flowers can be placed in (4) distinct positions in (4!=24) ways. Placing one each counts arrangements.

Exam Tip

(4) अलग फूल (4) अलग स्थानों पर (4!=24) तरीकों से रखे जा सकते हैं। एक-एक वस्तु रखने में व्यवस्था गिनी जाती है।

Concept

The three positions are distinct, so \({}^{7}P_{3}=7\times6\times5=210\). Use permutation when positions have different names.

Why this answer is correct

The correct answer is D. (210). The three positions are distinct, so \({}^{7}P_{3}=7\times6\times5=210\). Use permutation when positions have different names.

Exam Tip

तीनों स्थान अलग हैं इसलिए \({}^{7}P_{3}=7\times6\times5=210\)। स्थानों के नाम अलग हों तो क्रमचय लें।

Concept

The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate fixed positions first.

Why this answer is correct

The correct answer is A. (6). The first position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Separate fixed positions first.

Exam Tip

पहला स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed position को पहले अलग करें।

Concept

The three posts are distinct, so \({}^{10}P_{3}=10\times9\times8=720\). Order matters for different posts.

Why this answer is correct

The correct answer is B. (720). The three posts are distinct, so \({}^{10}P_{3}=10\times9\times8=720\). Order matters for different posts.

Exam Tip

तीन पद अलग हैं इसलिए \({}^{10}P_{3}=10\times9\times8=720\)। अलग पदों में order महत्त्वपूर्ण होता है।

Concept

There are (3) choices for each place, so \(3\times3=9\). With repetition, choices do not decrease.

Why this answer is correct

The correct answer is C. (9). There are (3) choices for each place, so \(3\times3=9\). With repetition, choices do not decrease.

Exam Tip

दोनों स्थानों पर (3) विकल्प हैं इसलिए \(3\times3=9\)। पुनरावृत्ति हो तो विकल्प कम नहीं होते।

Concept

The first position is fixed and the remaining (4) bags can be arranged in (4!=24) ways. Arrange the remaining items after fixing the item.

Why this answer is correct

The correct answer is D. (24). The first position is fixed and the remaining (4) bags can be arranged in (4!=24) ways. Arrange the remaining items after fixing the item.

Exam Tip

पहला स्थान निश्चित है और शेष (4) बैग (4!=24) तरीकों से लगेंगे। fixed item के बाद बचे items की व्यवस्था करें।

Concept

Since \({}^{n}P_{1}=n\), \({}^{9}P_{1}=9\). For one position, count the available objects directly.

Why this answer is correct

The correct answer is A. (9). Since \({}^{n}P_{1}=n\), \({}^{9}P_{1}=9\). For one position, count the available objects directly.

Exam Tip

\({}^{n}P_{1}=n\) होता है इसलिए \({}^{9}P_{1}=9\)। एक स्थान के लिए सीधे उपलब्ध वस्तुएँ गिनें।

Concept

The word (DELHI) has (5) distinct letters, so (5!=120). The number of distinct letters is the base of the factorial.

Why this answer is correct

The correct answer is B. (120). The word (DELHI) has (5) distinct letters, so (5!=120). The number of distinct letters is the base of the factorial.

Exam Tip

(DELHI) में (5) अलग अक्षर हैं इसलिए (5!=120)। अलग अक्षरों की संख्या ही factorial का आधार होती है।

Concept

This is \({}^{8}P_{4}=8\times7\times6\times5=1680\). Take (4) decreasing factors for (4) positions.

Why this answer is correct

The correct answer is A. (1680). This is \({}^{8}P_{4}=8\times7\times6\times5=1680\). Take (4) decreasing factors for (4) positions.

Exam Tip

यह \({}^{8}P_{4}=8\times7\times6\times5=1680\) है। (4) स्थानों के लिए (4) घटते गुणनखंड लें।

Concept

There will be \({}^{5}P_{3}=5\times4\times3=60\) ways. Changing the order of letters changes the arrangement.

Why this answer is correct

The correct answer is B. (60). There will be \({}^{5}P_{3}=5\times4\times3=60\) ways. Changing the order of letters changes the arrangement.

Exam Tip

\({}^{5}P_{3}=5\times4\times3=60\) तरीके होंगे। अक्षरों का क्रम बदलने से arrangement बदलती है।

Concept

(M) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Repeated letters reduce total arrangements.

Why this answer is correct

The correct answer is A. (3). (M) occurs twice, so there are \(\frac{3!}{2!}=3\) arrangements. Repeated letters reduce total arrangements.

Exam Tip

(M) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements हैं। समान अक्षर होने पर कुल arrangements घट जाती हैं।

Concept

The answer is \({}^{6}P_{4}=6\times5\times4\times3=360\). Order matters in a row.

Why this answer is correct

The correct answer is C. (360). The answer is \({}^{6}P_{4}=6\times5\times4\times3=360\). Order matters in a row.

Exam Tip

उत्तर \({}^{6}P_{4}=6\times5\times4\times3=360\) है। पंक्ति में क्रम महत्त्वपूर्ण होता है।

Concept

For three places, there are \(5\times4\times3=60\) ways. Without repetition, each next choice decreases.

Why this answer is correct

The correct answer is D. (60). For three places, there are \(5\times4\times3=60\) ways. Without repetition, each next choice decreases.

Exam Tip

तीन स्थानों के लिए \(5\times4\times3=60\) तरीके हैं। बिना पुनरावृत्ति में हर अगला विकल्प कम होता है।

Concept

\({}^{n}P_{0}=1\), so \({}^{7}P_{0}=1\). Choosing zero objects is counted in one way.

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.

Exam Tip

\({}^{n}P_{0}=1\) होता है इसलिए \({}^{7}P_{0}=1\)। शून्य वस्तु चुनने का एक ही तरीका माना जाता है।

Concept

(4) different labels can be pasted on (4) different boxes in (4!=24) ways. Factorial is useful in matching arrangements.

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.

Exam Tip

(4) अलग लेबल (4) अलग डिब्बों पर (4!=24) तरीकों से लगेंगे। matching arrangements में factorial काम आता है।

Concept

There are (6) letters and two identical (O)'s, so \(\frac{6!}{2!}=360\). Use a denominator for a repeated letter.

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.

Exam Tip

कुल (6) अक्षर हैं और (O) दो समान हैं इसलिए \(\frac{6!}{2!}=360\)। repeated letter के लिए denominator लगाएं।

Concept

There are (5) choices for the first city and (4) for the second, so (20). Changing trip order changes the plan.

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.

Exam Tip

पहले शहर के लिए (5) और दूसरे के लिए (4) विकल्प हैं इसलिए (20)। यात्रा क्रम बदलने से योजना बदलती है।

Concept

There are (3) choices for the hundreds place, then (3) and (2) choices, total (18). Do not put (0) in the first place.

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.

Exam Tip

सैकड़ा स्थान पर (3) विकल्प हैं, फिर (3) और (2) विकल्प हैं, कुल (18)। पहले स्थान पर (0) न रखें।

Concept

The ordered arrangement of (3) distinct flags is (3!=6). Changing top-bottom order changes the arrangement.

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.

Exam Tip

(3) अलग झंडों की क्रमबद्ध व्यवस्था (3!=6) है। ऊपर-नीचे का क्रम बदलने से arrangement बदलती है।

Concept

The two posts are distinct, so \({}^{11}P_{2}=11\times10=110\). Use permutation for distinct posts.

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.

Exam Tip

दो पद अलग हैं इसलिए \({}^{11}P_{2}=11\times10=110\)। अलग पदों के लिए क्रमचय लगाएं।

Concept

The last position is fixed and the remaining (3) letters can be arranged in (3!=6) ways. Apply the fixed condition first.

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.

Exam Tip

अंतिम स्थान निश्चित है और शेष (3) अक्षर (3!=6) तरीकों से लगेंगे। fixed condition को पहले लागू करें।

Concept

\({}^{5}P_{4}=5\times4\times3\times2=120\). When (r=4), take (4) decreasing factors.

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.

Exam Tip

\({}^{5}P_{4}=5\times4\times3\times2=120\) है। (r=4) हो तो (4) घटते गुणनखंड लें।

Concept

Right and left positions are distinct, so \({}^{6}P_{2}=30\). When positions are identified, order is counted.

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.

Exam Tip

दायाँ और बायाँ स्थान अलग हैं इसलिए \({}^{6}P_{2}=30\)। स्थानों की पहचान अलग हो तो क्रम गिना जाता है।

Concept

\({}^{4}P_{3}=4\times3\times2=24\) codes are possible. In a code, changing positions changes the code.

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.

Exam Tip

\({}^{4}P_{3}=4\times3\times2=24\) कोड बनेंगे। कोड में अक्षरों का स्थान बदलने से कोड बदल जाता है।

Concept

(E) occurs twice, so \(\frac{3!}{2!}=3\) arrangements are possible. Do not count repeated letters as distinct.

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.

Exam Tip

(E) दो बार है इसलिए \(\frac{3!}{2!}=3\) arrangements होंगे। repeated letters को अलग-अलग न गिनें।

Concept

First and second speakers are ordered positions, so \({}^{12}P_{2}=132\). Speaking order uses permutation.

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.

Exam Tip

पहला और दूसरा वक्ता क्रम वाले पद हैं इसलिए \({}^{12}P_{2}=132\)। बोलने के क्रम में permutation लगता है।

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.

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.

Exam Tip

तीनों स्थानों पर (3) विकल्प हैं इसलिए \(3^3=27\)। repetition allowed हो तो हर स्थान पर समान विकल्प रहते हैं।

Concept

On a straight string, the arrangement of (5) distinct beads is (5!=120). Do not treat it as a circular arrangement.

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.

Exam Tip

सीधी डोरी पर (5) अलग मोतियों की व्यवस्था (5!=120) है। इसे circular arrangement न मानें।

Concept

\({}^{4}P_{0}=1\) and \({}^{4}P_{1}=4\), so the sum is (5). Remember standard values.

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.

Exam Tip

\({}^{4}P_{0}=1\) और \({}^{4}P_{1}=4\), इसलिए योग (5) है। standard values याद रखें।

Concept

There are (7) choices for the first box and (6) for the second, so (42). Permutation is counted for distinct boxes.

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.

Exam Tip

पहले डिब्बे के लिए (7) और दूसरे के लिए (6) विकल्प हैं इसलिए (42)। अलग डिब्बों में क्रमचय गिना जाता है।

Concept

The word (BIHAR) has (5) distinct letters, so (5!=120). No denominator is needed when all letters are distinct.

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.

Exam Tip

(BIHAR) में (5) अलग अक्षर हैं इसलिए (5!=120)। सभी अक्षर अलग हों तो denominator नहीं लगता।

Concept

(4) different subjects can be arranged in (4) different periods in (4!=24) ways. Order matters in a timetable.

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.

Exam Tip

(4) अलग विषय (4) अलग पीरियड में (4!=24) तरीकों से लगते हैं। timetable में order महत्त्वपूर्ण होता है।

Concept

For four places, there are \(5\times4\times3\times2=120\) ways. Without repetition, choices keep decreasing.

Why this answer is correct