Class 11 Mathematics - Permutations And Combinations - Permutations Easy Quiz

Mathematics Permutations And Combinations Permutations Easy Class 11 Level 51

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Question 1/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक पहचान कोड में (3) अक्षर और (1) अंक है। अक्षरों के लिए (5) विकल्प और अंक के लिए (8) विकल्प हैं। पुनरावृत्ति सहित कितने कोड बनेंगे?

An identity code has (3) letters and (1) digit. There are (5) choices for each letter and (8) choices for the digit. With repetition allowed, how many codes are possible?

Explanation opens after your attempt
Correct Answer

D. (1000)

Step 1

Concept

Total choices are \(5\times5\times5\times8=1000\). In mixed codes, write choices for each position separately.

Step 2

Why this answer is correct

The correct answer is D. (1000). Total choices are \(5\times5\times5\times8=1000\). In mixed codes, write choices for each position separately.

Step 3

Exam Tip

कुल विकल्प \(5\times5\times5\times8=1000\) हैं। मिश्रित कोड में हर स्थान का विकल्प अलग लिखें।

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Question 2/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक सिक्का उछालने और एक पासा फेंकने पर कुल परिणाम कितने होंगे?

How many total outcomes are possible when a coin is tossed and a die is rolled?

Explanation opens after your attempt
Correct Answer

A. (12)

Step 1

Concept

A coin has (2) outcomes and a die has (6), so \(2\times6=12\). Multiply when two experiments are performed together.

Step 2

Why this answer is correct

The correct answer is A. (12). A coin has (2) outcomes and a die has (6), so \(2\times6=12\). Multiply when two experiments are performed together.

Step 3

Exam Tip

सिक्के के (2) और पासे के (6) परिणाम हैं इसलिए \(2\times6=12\)। दो प्रयोग साथ हों तो गुणन करें।

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Question 3/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

दो अलग-अलग सिक्के उछाले जाते हैं। कुल परिणाम कितने होंगे?

Two distinct coins are tossed. How many total outcomes are possible?

Explanation opens after your attempt
Correct Answer

B. (4)

Step 1

Concept

Each coin has (2) outcomes, so \(2\times2=4\). Treat distinct coins as separate stages.

Step 2

Why this answer is correct

The correct answer is B. (4). Each coin has (2) outcomes, so \(2\times2=4\). Treat distinct coins as separate stages.

Step 3

Exam Tip

हर सिक्के के (2) परिणाम हैं इसलिए \(2\times2=4\)। अलग-अलग सिक्कों को अलग चरण मानें।

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Question 4/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

दो पासे एक साथ फेंके जाते हैं। कुल क्रमित परिणाम कितने होंगे?

Two dice are rolled together. How many ordered outcomes are possible?

Explanation opens after your attempt
Correct Answer

C. (36)

Step 1

Concept

The first die has (6) outcomes and the second has (6), so \(6\times6=36\). In ordered outcomes, the roles of both dice are distinct.

Step 2

Why this answer is correct

The correct answer is C. (36). The first die has (6) outcomes and the second has (6), so \(6\times6=36\). In ordered outcomes, the roles of both dice are distinct.

Step 3

Exam Tip

पहले पासे के (6) और दूसरे के (6) परिणाम हैं इसलिए \(6\times6=36\)। क्रमित परिणाम में दोनों पासों की भूमिका अलग रहती है।

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Question 5/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक चक्र में (4) रंग और (3) आकार उपलब्ध हैं। एक रंग और एक आकार चुनकर डिज़ाइन कितने तरीकों से बनेगा?

A wheel has (4) colors and (3) shapes available. In how many ways can a design be made by choosing one color and one shape?

Explanation opens after your attempt
Correct Answer

D. (12)

Step 1

Concept

One color and one shape are needed, so \(4\times3=12\). Multiply two independent attributes.

Step 2

Why this answer is correct

The correct answer is D. (12). One color and one shape are needed, so \(4\times3=12\). Multiply two independent attributes.

Step 3

Exam Tip

एक रंग और एक आकार चाहिए इसलिए \(4\times3=12\)। दो स्वतंत्र विशेषताओं को गुणा करें।

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Question 6/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक पासवर्ड में पहले (3) अक्षरों में से एक और फिर (4) अंकों में से एक चुनना है। कितने पासवर्ड बनेंगे?

A password is made by choosing one of (3) letters first and then one of (4) digits. How many passwords are possible?

Explanation opens after your attempt
Correct Answer

A. (12)

Step 1

Concept

The first stage has (3) choices and the second has (4), so \(3\times4=12\). Fill password positions in order.

Step 2

Why this answer is correct

The correct answer is A. (12). The first stage has (3) choices and the second has (4), so \(3\times4=12\). Fill password positions in order.

Step 3

Exam Tip

पहले चरण में (3) और दूसरे में (4) विकल्प हैं इसलिए \(3\times4=12\)। पासवर्ड के स्थानों को क्रम से भरें।

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Question 7/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक प्रतियोगिता में (5) कविता विषय और (6) चित्रकला विषय हैं। विद्यार्थी केवल एक विषय चुन सकता है। चयन के कितने तरीके हैं?

In a competition, there are (5) poetry topics and (6) painting topics. A student can choose only one topic. How many choices are possible?

Explanation opens after your attempt
Correct Answer

B. (11)

Step 1

Concept

Only one topic is to be chosen, so (5+6=11). Add when option groups are separate and not chosen together.

Step 2

Why this answer is correct

The correct answer is B. (11). Only one topic is to be chosen, so (5+6=11). Add when option groups are separate and not chosen together.

Step 3

Exam Tip

केवल एक विषय चुनना है इसलिए (5+6=11)। विकल्प समूह अलग हों और साथ नहीं चुने जाते हों तो जोड़ें।

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Question 8/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक स्कूल यात्रा के लिए (3) संग्रहालय या (4) पार्क में से एक स्थान चुनना है। कुल विकल्प कितने हैं?

For a school trip, one place is to be chosen from (3) museums or (4) parks. How many total options are there?

Explanation opens after your attempt
Correct Answer

C. (7)

Step 1

Concept

The place will be a museum or a park, so (3+4=7). Use the addition rule for separate “or” choices.

Step 2

Why this answer is correct

The correct answer is C. (7). The place will be a museum or a park, so (3+4=7). Use the addition rule for separate “or” choices.

Step 3

Exam Tip

स्थान संग्रहालय या पार्क में से होगा इसलिए (3+4=7)। “या” वाले अलग चयन में योग का नियम लगाएँ।

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Question 9/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक ऐप में (4) भाषा विकल्प और (2) थीम विकल्प हैं। एक भाषा और एक थीम चुनने के कितने तरीके हैं?

An app has (4) language options and (2) theme options. In how many ways can one language and one theme be selected?

Explanation opens after your attempt
Correct Answer

D. (8)

Step 1

Concept

Both selections are made together, so \(4\times2=8\). Multiply independent setting options.

Step 2

Why this answer is correct

The correct answer is D. (8). Both selections are made together, so \(4\times2=8\). Multiply independent setting options.

Step 3

Exam Tip

दोनों चयन साथ करने हैं इसलिए \(4\times2=8\)। सेटिंग्स के स्वतंत्र विकल्पों को गुणा करें।

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Question 10/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक छात्र (2) बैग, (5) पानी की बोतल और (3) टिफिन में से एक-एक चीज़ चुनता है। कुल तरीके कितने होंगे?

A student chooses one item each from (2) bags, (5) water bottles, and (3) lunch boxes. How many total ways are possible?

Explanation opens after your attempt
Correct Answer

A. (30)

Step 1

Concept

Total ways are \(2\times5\times3=30\). Multiply when one item is chosen from each category.

Step 2

Why this answer is correct

The correct answer is A. (30). Total ways are \(2\times5\times3=30\). Multiply when one item is chosen from each category.

Step 3

Exam Tip

कुल तरीके \(2\times5\times3=30\) हैं। एक-एक चीज़ अलग श्रेणियों से हो तो गुणा करें।

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Question 11/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक परीक्षा कोड में (2) अंकों के बाद (1) अक्षर है। अंक (0) से (9) तक और अक्षर (5) दिए गए हैं। पुनरावृत्ति सहित कितने कोड बनेंगे?

An exam code has (2) digits followed by (1) letter. Digits are from (0) to (9) and (5) letters are given. With repetition allowed, how many codes can be formed?

Explanation opens after your attempt
Correct Answer

B. (500)

Step 1

Concept

The two digit positions have (10) and (10) choices, and the letter has (5) choices. Total ways are \(10\times10\times5=500\).

Step 2

Why this answer is correct

The correct answer is B. (500). The two digit positions have (10) and (10) choices, and the letter has (5) choices. Total ways are \(10\times10\times5=500\).

Step 3

Exam Tip

दो अंक स्थानों के (10) और (10) विकल्प तथा अक्षर के (5) विकल्प हैं। कुल \(10\times10\times5=500\) हैं।

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Question 12/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक (4)-अंकीय पिन में प्रत्येक स्थान पर (1) से (6) तक अंक आ सकते हैं। पुनरावृत्ति की अनुमति है तो कुल पिन कितने हैं?

In a (4)-digit PIN, each position can have digits from (1) to (6). If repetition is allowed, how many PINs are possible?

Explanation opens after your attempt
Correct Answer

B. (1296)

Step 1

Concept

Each position has (6) choices, so \(6^4=1296\). When the same choices repeat, using powers is faster.

Step 2

Why this answer is correct

The correct answer is B. (1296). Each position has (6) choices, so \(6^4=1296\). When the same choices repeat, using powers is faster.

Step 3

Exam Tip

प्रत्येक स्थान के (6) विकल्प हैं इसलिए \(6^4=1296\)। समान विकल्प बार-बार हों तो घात का प्रयोग तेज होता है।

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Question 13/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

अंकों (1,2,3,4) से पुनरावृत्ति बिना (4)-अंकीय संख्याएँ कितनी बनेंगी?

How many (4)-digit numbers can be formed from digits (1,2,3,4) without repetition?

Explanation opens after your attempt
Correct Answer

C. (24)

Step 1

Concept

The four positions have (4,3,2,1) choices, so \(4\times3\times2\times1=24\). If all digits are used, choices decrease step by step.

Step 2

Why this answer is correct

The correct answer is C. (24). The four positions have (4,3,2,1) choices, so \(4\times3\times2\times1=24\). If all digits are used, choices decrease step by step.

Step 3

Exam Tip

चार स्थानों के विकल्प (4,3,2,1) हैं इसलिए \(4\times3\times2\times1=24\)। सभी अंक उपयोग हों तो विकल्प क्रम से घटते हैं।

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Question 14/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

अंकों (0,1,2,3) से पुनरावृत्ति बिना (3)-अंकीय सम संख्याएँ कितनी बनेंगी?

How many (3)-digit even numbers can be formed from digits (0,1,2,3) without repetition?

Explanation opens after your attempt
Correct Answer

B. (8)

Step 1

Concept

If the units digit is (0), there are \(3\times2=6\) ways, and if it is (2), there are \(2\times2=4\) ways. Total ways are (6+4=10).

Step 2

Why this answer is correct

The correct answer is B. (8). If the units digit is (0), there are \(3\times2=6\) ways, and if it is (2), there are \(2\times2=4\) ways. Total ways are (6+4=10).

Step 3

Exam Tip

यदि इकाई (0) है तो \(3\times2=6\) तरीके हैं और यदि इकाई (2) है तो सैकड़े के (2) तथा दहाई के (2) तरीके हैं। कुल (6+4=10) हैं।

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Question 15/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक कैफे में (3) प्रकार की चाय, (4) प्रकार की कॉफी और (2) प्रकार का जूस है। केवल एक पेय चुनने के कितने तरीके हैं?

A cafe has (3) types of tea, (4) types of coffee, and (2) types of juice. In how many ways can only one drink be chosen?

Explanation opens after your attempt
Correct Answer

C. (9)

Step 1

Concept

Only one drink is chosen, so (3+4+2=9). Add when the groups are alternatives, not simultaneous choices.

Step 2

Why this answer is correct

The correct answer is C. (9). Only one drink is chosen, so (3+4+2=9). Add when the groups are alternatives, not simultaneous choices.

Step 3

Exam Tip

केवल एक पेय चुनना है इसलिए (3+4+2=9)। एक ही श्रेणी नहीं बल्कि वैकल्पिक समूह हों तो जोड़ें।

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Question 16/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक टीम जर्सी के लिए (5) रंग और (4) आकार उपलब्ध हैं। एक रंग और एक आकार चुनने के कितने तरीके हैं?

For a team jersey, (5) colors and (4) sizes are available. In how many ways can one color and one size be chosen?

Explanation opens after your attempt
Correct Answer

D. (20)

Step 1

Concept

Both color and size are to be selected, so \(5\times4=20\). Multiply when product features are independent.

Step 2

Why this answer is correct

The correct answer is D. (20). Both color and size are to be selected, so \(5\times4=20\). Multiply when product features are independent.

Step 3

Exam Tip

रंग और आकार दोनों चुनने हैं इसलिए \(5\times4=20\)। उत्पाद की अलग विशेषताएँ स्वतंत्र हों तो गुणा करें।

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Question 17/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक बैज में (3) पृष्ठभूमि रंग, (2) प्रतीक और (4) बॉर्डर विकल्प हैं। एक बैज कितने तरीकों से बनेगा?

A badge has (3) background colors, (2) symbols, and (4) border options. In how many ways can one badge be made?

Explanation opens after your attempt
Correct Answer

A. (24)

Step 1

Concept

The three stages are independent, so \(3\times2\times4=24\). In design questions, treat each feature as one stage.

Step 2

Why this answer is correct

The correct answer is A. (24). The three stages are independent, so \(3\times2\times4=24\). In design questions, treat each feature as one stage.

Step 3

Exam Tip

तीनों चरण स्वतंत्र हैं इसलिए \(3\times2\times4=24\)। डिज़ाइन प्रश्नों में हर फीचर को एक चरण मानें।

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Question 18/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक छात्र (6) ऑनलाइन कोर्स या (3) ऑफलाइन कोर्स में से केवल एक कोर्स चुनता है। कुल चयन कितने हैं?

A student chooses only one course from (6) online courses or (3) offline courses. How many total choices are there?

Explanation opens after your attempt
Correct Answer

B. (9)

Step 1

Concept

Only one course is to be chosen, so (6+3=9). Add when the options are not taken together.

Step 2

Why this answer is correct

The correct answer is B. (9). Only one course is to be chosen, so (6+3=9). Add when the options are not taken together.

Step 3

Exam Tip

एक ही कोर्स चुनना है इसलिए (6+3=9)। विकल्पों को एक साथ नहीं लेना हो तो योग करें।

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Question 19/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक कार्ड पर (4) चित्र विकल्प और (5) संदेश विकल्प हैं। एक चित्र और एक संदेश चुनकर कार्ड कितने तरीकों से बनेगा?

A card has (4) picture options and (5) message options. In how many ways can a card be made by choosing one picture and one message?

Explanation opens after your attempt
Correct Answer

C. (20)

Step 1

Concept

One picture and one message are both needed, so \(4\times5=20\). The multiplication principle applies to combined selection.

Step 2

Why this answer is correct

The correct answer is C. (20). One picture and one message are both needed, so \(4\times5=20\). The multiplication principle applies to combined selection.

Step 3

Exam Tip

एक चित्र और एक संदेश दोनों चाहिए इसलिए \(4\times5=20\)। संयुक्त चयन में गुणा सिद्धांत लागू होता है।

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Question 20/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक पेन ड्राइव का रंग (3) विकल्पों में से और क्षमता (4) विकल्पों में से चुनी जाती है। कुल मॉडल कितने बनेंगे?

A pen drive color is chosen from (3) options and capacity from (4) options. How many total models can be formed?

Explanation opens after your attempt
Correct Answer

D. (12)

Step 1

Concept

Total models are \(3\times4=12\). In product models, multiply each independent feature.

Step 2

Why this answer is correct

The correct answer is D. (12). Total models are \(3\times4=12\). In product models, multiply each independent feature.

Step 3

Exam Tip

कुल मॉडल \(3\times4=12\) होंगे। उत्पाद मॉडल में हर स्वतंत्र विशेषता को गुणा करें।

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Question 21/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक कक्षा में (4) मॉनिटर उम्मीदवार और (3) सह-मॉनिटर उम्मीदवार हैं। एक मॉनिटर और एक सह-मॉनिटर चुनने के कितने तरीके हैं?

In a class, there are (4) monitor candidates and (3) assistant monitor candidates. In how many ways can one monitor and one assistant monitor be chosen?

Explanation opens after your attempt
Correct Answer

A. (12)

Step 1

Concept

Two different posts are to be filled, so \(4\times3=12\). Treat selections for different posts as ordered stages.

Step 2

Why this answer is correct

The correct answer is A. (12). Two different posts are to be filled, so \(4\times3=12\). Treat selections for different posts as ordered stages.

Step 3

Exam Tip

दो अलग पद भरने हैं इसलिए \(4\times3=12\)। अलग पदों के चयन को क्रमबद्ध चरण मानें।

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Question 22/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक व्यक्ति (5) बसों में से एक से स्टेशन जाता है और फिर (2) ऑटो में से एक से घर जाता है। कुल यात्रा तरीके कितने हैं?

A person takes one of (5) buses to the station and then one of (2) autos to home. How many total travel ways are possible?

Explanation opens after your attempt
Correct Answer

B. (10)

Step 1

Concept

There are (5) choices first and (2) choices next, so \(5\times2=10\). Multiply the stages in a sequential journey.

Step 2

Why this answer is correct

The correct answer is B. (10). There are (5) choices first and (2) choices next, so \(5\times2=10\). Multiply the stages in a sequential journey.

Step 3

Exam Tip

पहले (5) और फिर (2) विकल्प हैं इसलिए \(5\times2=10\)। क्रम से होने वाली यात्रा में चरणों को गुणा करें।

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Question 23/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक वेबसाइट साइनअप में (3) प्रोफाइल फोटो विकल्प और (6) यूज़रनेम स्टाइल विकल्प हैं। एक प्रोफाइल बनाने के कितने तरीके हैं?

A website signup has (3) profile photo options and (6) username style options. In how many ways can one profile be made?

Explanation opens after your attempt
Correct Answer

C. (18)

Step 1

Concept

Both choices are independent, so \(3\times6=18\). In profile setup, multiply every required selection.

Step 2

Why this answer is correct

The correct answer is C. (18). Both choices are independent, so \(3\times6=18\). In profile setup, multiply every required selection.

Step 3

Exam Tip

दोनों चुनाव स्वतंत्र हैं इसलिए \(3\times6=18\)। प्रोफाइल सेटअप में हर अनिवार्य चयन को गुणा करें।

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Question 24/24 Easy Mathematics Permutations and Combinations Fundamental principle of counting Class 11 Level 51

एक गेम अवतार में (2) चेहरे, (4) टोपी और (3) पोशाक विकल्प हैं। एक अवतार कितने तरीकों से बनाया जा सकता है?

A game avatar has (2) face options, (4) hat options, and (3) costume options. In how many ways can one avatar be created?

Explanation opens after your attempt
Correct Answer

D. (24)

Step 1

Concept

Total avatars are \(2\times4\times3=24\). When there are three independent features, multiply all choices.

Step 2

Why this answer is correct

The correct answer is D. (24). Total avatars are \(2\times4\times3=24\). When there are three independent features, multiply all choices.

Step 3

Exam Tip

कुल अवतार \(2\times4\times3=24\) होंगे। तीन स्वतंत्र फीचर हों तो सभी विकल्पों को गुणा करें।

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