A. पहले स्थान के लिए (n) विकल्प होते हैं/There are (n) choices for the first position
Step 1
Concept
Any one of the (n) distinct objects can occupy the first position. In exams think by filling positions one by one.
Step 2
Why this answer is correct
The correct answer is A. पहले स्थान के लिए (n) विकल्प होते हैं / There are (n) choices for the first position. Any one of the (n) distinct objects can occupy the first position. In exams think by filling positions one by one.
Step 3
Exam Tip
पहले स्थान पर (n) अलग वस्तुओं में से कोई भी आ सकती है। परीक्षा में स्थानों को क्रम से भरकर सोचें।
Since (r) positions are to be filled there are (r) factors. In exams connect the number of factors with the number of positions.
Step 2
Why this answer is correct
The correct answer is B. (r). Since (r) positions are to be filled there are (r) factors. In exams connect the number of factors with the number of positions.
Step 3
Exam Tip
क्योंकि (r) स्थान भरने हैं इसलिए (r) गुणनखंड आते हैं। परीक्षा में गुणनखंडों की संख्या को स्थानों की संख्या से जोड़ें।
In ordered selection the remaining ((n-r)!) part is removed from (n!). In exams remember ((n-r)!) in the denominator for permutation.
Step 2
Why this answer is correct
The correct answer is B. (^{n}P_r=\frac{n!}{(n-r)!}). In ordered selection the remaining ((n-r)!) part is removed from (n!). In exams remember ((n-r)!) in the denominator for permutation.
Step 3
Exam Tip
क्रम वाले चयन में (n!) से बाकी ((n-r)!) भाग हटता है। परीक्षा में permutation के हर में ((n-r)!) याद रखें।
D. क्योंकि क्रम को हटाना होता है/Because order has to be removed
Step 1
Concept
In combination order is not counted so (r!) orders of selected objects are removed. In exams treat combination as selection without order.
Step 2
Why this answer is correct
The correct answer is D. क्योंकि क्रम को हटाना होता है / Because order has to be removed. In combination order is not counted so (r!) orders of selected objects are removed. In exams treat combination as selection without order.
Step 3
Exam Tip
Combination में क्रम नहीं गिना जाता इसलिए चुनी गई (r) वस्तुओं के (r!) क्रम हटते हैं। परीक्षा में combination को बिना क्रम का चयन मानें।
First (r) objects are chosen and then arranged in (r!) ways. In exams keep the idea of choose first and arrange later.
Step 2
Why this answer is correct
The correct answer is C. \(^{n}P_r=^{n}C_r\times r!\). First (r) objects are chosen and then arranged in (r!) ways. In exams keep the idea of choose first and arrange later.
Step 3
Exam Tip
पहले (r) वस्तुएँ चुनी जाती हैं और फिर उन्हें (r!) तरीकों से जमाया जाता है। परीक्षा में पहले चयन फिर व्यवस्था का विचार रखें।
B. \(^{n}P_r\) को (r!) से भाग देना/Dividing \(^{n}P_r\) by (r!)
Step 1
Concept
In combination extra orders are removed from permutation. In exams write \(^{n}P_r\) and divide by (r!).
Step 2
Why this answer is correct
The correct answer is B. \(^{n}P_r\) को (r!) से भाग देना / Dividing \(^{n}P_r\) by (r!). In combination extra orders are removed from permutation. In exams write \(^{n}P_r\) and divide by (r!).
Step 3
Exam Tip
Combination में permutation से अतिरिक्त क्रम हटाए जाते हैं। परीक्षा में \(^{n}P_r\) लिखकर (r!) से भाग दें।
D. (r) वस्तुएँ चुनना उतना ही है जितना (n-r) वस्तुएँ छोड़ना/Choosing (r) objects is same as leaving (n-r) objects
Step 1
Concept
When (r) are chosen the (n-r) objects are automatically left. In exams remember the connection between choosing and leaving.
Step 2
Why this answer is correct
The correct answer is D. (r) वस्तुएँ चुनना उतना ही है जितना (n-r) वस्तुएँ छोड़ना / Choosing (r) objects is same as leaving (n-r) objects. When (r) are chosen the (n-r) objects are automatically left. In exams remember the connection between choosing and leaving.
Step 3
Exam Tip
जब (r) चुने जाते हैं तब (n-r) वस्तुएँ अपने आप छूटती हैं। परीक्षा में चयन और त्याग का संबंध याद रखें।
A. क्योंकि कोई वस्तु न चुनने का केवल (1) तरीका है/Because there is only (1) way to choose no object
Step 1
Concept
Selecting zero objects is an empty selection and it happens in one way. In exams count empty selection as one way.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि कोई वस्तु न चुनने का केवल (1) तरीका है / Because there is only (1) way to choose no object. Selecting zero objects is an empty selection and it happens in one way. In exams count empty selection as one way.
Step 3
Exam Tip
शून्य वस्तुओं का चयन खाली चयन है और वह एक ही तरह से होता है। परीक्षा में खाली चयन को भी एक तरीका मानें।
D. सभी वस्तुएँ चुनने का केवल (1) तरीका है/There is only (1) way to choose all objects
Step 1
Concept
Choosing all (n) objects is one fixed selection. In exams understand the all selected case as (1) way.
Step 2
Why this answer is correct
The correct answer is D. सभी वस्तुएँ चुनने का केवल (1) तरीका है / There is only (1) way to choose all objects. Choosing all (n) objects is one fixed selection. In exams understand the all selected case as (1) way.
Step 3
Exam Tip
सभी (n) वस्तुओं को चुनना एक निश्चित चयन है। परीक्षा में सभी चुनी गई स्थिति को (1) तरीका समझें।
D. शून्य वस्तुओं की व्यवस्था का एक खाली तरीका/One empty way to arrange zero objects
Step 1
Concept
Filling zero positions is counted as one empty arrangement. In exams remember the use of (0!=1).
Step 2
Why this answer is correct
The correct answer is D. शून्य वस्तुओं की व्यवस्था का एक खाली तरीका / One empty way to arrange zero objects. Filling zero positions is counted as one empty arrangement. In exams remember the use of (0!=1).
Step 3
Exam Tip
शून्य स्थान भरने का एक खाली arrangement माना जाता है। परीक्षा में (0!=1) का उपयोग ध्यान रखें।
D. \(^{n}C_n=\frac{n!}{n!0!}=1\) सही रखने के लिए (0!=1) चाहिए/To keep \(^{n}C_n=\frac{n!}{n!0!}=1\) we need (0!=1)
Step 1
Concept
The formula gives \(^{n}C_n=1\) only when (0!=1). In exams understand factorial using boundary cases.
Step 2
Why this answer is correct
The correct answer is D. \(^{n}C_n=\frac{n!}{n!0!}=1\) सही रखने के लिए (0!=1) चाहिए / To keep \(^{n}C_n=\frac{n!}{n!0!}=1\) we need (0!=1). The formula gives \(^{n}C_n=1\) only when (0!=1). In exams understand factorial using boundary cases.
Step 3
Exam Tip
सूत्र में \(^{n}C_n=1\) तभी मिलता है जब (0!=1) हो। परीक्षा में सीमा मामलों से factorial समझें।
A. क्योंकि (n) अलग वस्तुओं को सभी (n) स्थानों पर जमाना है/Because (n) distinct objects are arranged in all (n) places
Step 1
Concept
Arranging all objects gives (n(n-1)\cdots1=n!). In exams connect complete arrangement with factorial.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (n) अलग वस्तुओं को सभी (n) स्थानों पर जमाना है / Because (n) distinct objects are arranged in all (n) places. Arranging all objects gives (n(n-1)\cdots1=n!). In exams connect complete arrangement with factorial.
Step 3
Exam Tip
सभी वस्तुओं की व्यवस्था (n(n-1)\cdots1=n!) होती है। परीक्षा में पूरी व्यवस्था को factorial से जोड़ें।
There are (5) choices for the first place and (4) for the second. In exams choices decrease when repetition is not allowed.
Step 2
Why this answer is correct
The correct answer is B. \(5\times4\). There are (5) choices for the first place and (4) for the second. In exams choices decrease when repetition is not allowed.
Step 3
Exam Tip
पहले स्थान के लिए (5) और दूसरे के लिए (4) विकल्प होंगे। परीक्षा में पुनरावृत्ति न हो तो विकल्प घटते हैं।
Seating in positions depends on order so permutation is used. In exams when position seat or rank appears check order.
Step 2
Why this answer is correct
The correct answer is D. \(^{6}P_3\). Seating in positions depends on order so permutation is used. In exams when position seat or rank appears check order.
Step 3
Exam Tip
पदों पर बैठाना क्रम पर निर्भर करता है इसलिए permutation लगेगा। परीक्षा में पद सीट या rank आने पर order देखें।
First choose (2) objects and then arrange them in (2!) ways. In exams understand permutation as combination times arrangement.
Step 2
Why this answer is correct
The correct answer is D. \(^{7}P_2=^{7}C_2\times2!\). First choose (2) objects and then arrange them in (2!) ways. In exams understand permutation as combination times arrangement.
Step 3
Exam Tip
पहले (2) वस्तुएँ चुनें और फिर उन्हें (2!) तरीकों से जमाएँ। परीक्षा में permutation को combination गुणा व्यवस्था समझें।
To get combination ordered arrangements are divided by (2!). In exams divide to remove internal orders.
Step 2
Why this answer is correct
The correct answer is B. \(^{8}C_2=\frac{^{8}P_2}{2!}\). To get combination ordered arrangements are divided by (2!). In exams divide to remove internal orders.
Step 3
Exam Tip
Combination पाने के लिए ordered arrangements को (2!) से भाग दिया जाता है। परीक्षा में internal orders हटाने के लिए भाग दें।
Only one position is to be filled so there are (n) choices. In exams for (r=1) permutation and combination both equal (n).
Step 2
Why this answer is correct
The correct answer is A. \(^{n}P_1=n\). Only one position is to be filled so there are (n) choices. In exams for (r=1) permutation and combination both equal (n).
Step 3
Exam Tip
एक स्थान भरना है इसलिए (n) विकल्प मिलते हैं। परीक्षा में (r=1) पर permutation और combination दोनों (n) होते हैं।
In the combination formula after cancelling ((n-2)!) the term (n(n-1)) remains. In exams remember \(^{n}C_2\) as the pair-making formula.
Step 2
Why this answer is correct
The correct answer is D. (^{n}C_2=\frac{n!}{2!(n-2)!}). In the combination formula after cancelling ((n-2)!) the term (n(n-1)) remains. In exams remember \(^{n}C_2\) as the pair-making formula.
Step 3
Exam Tip
Combination formula में ((n-2)!) कटने पर (n(n-1)) बचता है। परीक्षा में \(^{n}C_2\) को जोड़े बनाने का सूत्र मानें।
C. दो क्रमबद्ध स्थान भरना/Filling two ordered positions
Step 1
Concept
There are (n) choices for the first place and (n-1) for the second. In exams use the multiplication principle for ordered pairs.
Step 2
Why this answer is correct
The correct answer is C. दो क्रमबद्ध स्थान भरना / Filling two ordered positions. There are (n) choices for the first place and (n-1) for the second. In exams use the multiplication principle for ordered pairs.
Step 3
Exam Tip
पहले स्थान के लिए (n) और दूसरे के लिए (n-1) विकल्प होते हैं। परीक्षा में ordered pair में multiplication principle लगाएँ।
A. विशेष वस्तु को न चुनना/Not choosing the special object
Step 1
Concept
Choosing (r) from remaining (n-1) after excluding one fixed object is \(^{n-1}C_r\). In exams split cases into included and excluded.
Step 2
Why this answer is correct
The correct answer is A. विशेष वस्तु को न चुनना / Not choosing the special object. Choosing (r) from remaining (n-1) after excluding one fixed object is \(^{n-1}C_r\). In exams split cases into included and excluded.
Step 3
Exam Tip
एक निश्चित वस्तु को छोड़कर बाकी (n-1) में से (r) चुनना \(^{n-1}C_r\) है। परीक्षा में मामलों को शामिल और अलग में बाँटें।
C. विशेष वस्तु को चुनकर बाकी (r-1) चुनना/Choosing the special object and then choosing remaining (r-1)
Step 1
Concept
If the special object is chosen the remaining (r-1) objects must be chosen from (n-1). In exams write the included case separately.
Step 2
Why this answer is correct
The correct answer is C. विशेष वस्तु को चुनकर बाकी (r-1) चुनना / Choosing the special object and then choosing remaining (r-1). If the special object is chosen the remaining (r-1) objects must be chosen from (n-1). In exams write the included case separately.
Step 3
Exam Tip
यदि विशेष वस्तु चुनी गई है तो बाकी (r-1) वस्तुएँ (n-1) में से चुननी होंगी। परीक्षा में included case को अलग लिखें।
B. हर वस्तु को लेना या न लेना दो विकल्प होते हैं/Each object has two choices take or not take
Step 1
Concept
Each object has two choices so total subsets are \(2^n\). In exams think of subsets using include or exclude.
Step 2
Why this answer is correct
The correct answer is B. हर वस्तु को लेना या न लेना दो विकल्प होते हैं / Each object has two choices take or not take. Each object has two choices so total subsets are \(2^n\). In exams think of subsets using include or exclude.
Step 3
Exam Tip
हर वस्तु के लिए दो विकल्प होते हैं इसलिए कुल उपसमुच्चय \(2^n\) होते हैं। परीक्षा में subsets को include या exclude से सोचें।
D. क्योंकि कुल \(2^n\) selections में से खाली selection हटता है/Because the empty selection is removed from total \(2^n\) selections
Step 1
Concept
All selections include the empty selection too. In exams subtract the empty case when at least one appears.
Step 2
Why this answer is correct
The correct answer is D. क्योंकि कुल \(2^n\) selections में से खाली selection हटता है / Because the empty selection is removed from total \(2^n\) selections. All selections include the empty selection too. In exams subtract the empty case when at least one appears.
Step 3
Exam Tip
सभी selections में खाली selection भी शामिल होता है। परीक्षा में at least one आने पर empty case घटाएँ।
Two independent selections are needed together so the multiplication principle applies. In exams the word and often means multiply.
Step 2
Why this answer is correct
The correct answer is B. \(^{4}C_2\times{}^{5}C_1\). Two independent selections are needed together so the multiplication principle applies. In exams the word and often means multiply.
Step 3
Exam Tip
दो स्वतंत्र selections साथ-साथ चाहिए इसलिए गुणन सिद्धांत लगेगा। परीक्षा में and शब्द आने पर अक्सर multiply करें।
First choose (3) students and then arrange them in (3!) ways. In exams multiply when arrangement follows selection.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_3\times3!\). First choose (3) students and then arrange them in (3!) ways. In exams multiply when arrangement follows selection.
Step 3
Exam Tip
पहले (3) विद्यार्थी चुनें और फिर (3!) तरीकों से जमाएँ। परीक्षा में चयन के बाद व्यवस्था हो तो multiplication करें।
Rotations are considered same in a circle so fixing one object gives ((n-1)!) arrangements. In exams fix (1) object in circular arrangement.
Step 2
Why this answer is correct
The correct answer is D. ((n-1)!). Rotations are considered same in a circle so fixing one object gives ((n-1)!) arrangements. In exams fix (1) object in circular arrangement.
Step 3
Exam Tip
वृत्त में घुमाव समान माना जाता है इसलिए एक वस्तु स्थिर करके ((n-1)!) व्यवस्थाएँ मिलती हैं। परीक्षा में circular arrangement में (1) वस्तु स्थिर करें।
A. क्योंकि हर वृत्तीय व्यवस्था (n) घुमावों में गिनी जाती है/Because each circular arrangement is counted in (n) rotations
Step 1
Concept
Rotating a circle does not create a new arrangement so (n!) is divided by (n). In exams treat rotation as duplicate counting.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि हर वृत्तीय व्यवस्था (n) घुमावों में गिनी जाती है / Because each circular arrangement is counted in (n) rotations. Rotating a circle does not create a new arrangement so (n!) is divided by (n). In exams treat rotation as duplicate counting.
Step 3
Exam Tip
वृत्त में घुमाने से व्यवस्था नई नहीं मानी जाती इसलिए (n!) को (n) से भाग देते हैं। परीक्षा में rotation को duplicate count समझें।
A. क्योंकि समान अक्षरों के अंदरूनी क्रम अलग नहीं दिखते/Because internal orders of identical letters are not different
Step 1
Concept
Interchanging identical letters does not create a new word. In exams divide by the factorial of identical letters.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि समान अक्षरों के अंदरूनी क्रम अलग नहीं दिखते / Because internal orders of identical letters are not different. Interchanging identical letters does not create a new word. In exams divide by the factorial of identical letters.
Step 3
Exam Tip
समान अक्षरों की अदला-बदली से नया शब्द नहीं बनता। परीक्षा में identical letters के factorial से भाग दें।
B. जब दो प्रकार की वस्तुएँ क्रमशः (p) और (q) बार समान हों/When two types of objects are identical (p) and (q) times respectively
Step 1
Concept
Interchanges inside two identical groups are not counted separately so we divide by (p!q!). In exams place the factorial of each repeated group in the denominator.
Step 2
Why this answer is correct
The correct answer is B. जब दो प्रकार की वस्तुएँ क्रमशः (p) और (q) बार समान हों / When two types of objects are identical (p) and (q) times respectively. Interchanges inside two identical groups are not counted separately so we divide by (p!q!). In exams place the factorial of each repeated group in the denominator.
Step 3
Exam Tip
दो समानता समूहों की अदला-बदली अलग नहीं गिनी जाती इसलिए (p!q!) से भाग देते हैं। परीक्षा में हर repeated group का factorial हर में रखें।
Each position independently has (n) choices so the product is \(n\times n\times\cdots\times n=n^r\). In exams choices do not decrease when repetition is allowed.
Step 2
Why this answer is correct
The correct answer is C. \(n^r\). Each position independently has (n) choices so the product is \(n\times n\times\cdots\times n=n^r\). In exams choices do not decrease when repetition is allowed.
Step 3
Exam Tip
हर स्थान पर (n) विकल्प स्वतंत्र रूप से मिलते हैं इसलिए गुणन \(n\times n\times\cdots\times n=n^r\) होता है। परीक्षा में repetition allowed हो तो विकल्प घटते नहीं हैं।
A. क्योंकि हर अगले स्थान पर विकल्प घटते हैं/Because choices decrease at each next position
Step 1
Concept
Without repetition one object cannot be taken again. In exams pay attention to allowed and not allowed words.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि हर अगले स्थान पर विकल्प घटते हैं / Because choices decrease at each next position. Without repetition one object cannot be taken again. In exams pay attention to allowed and not allowed words.
Step 3
Exam Tip
बिना पुनरावृत्ति एक वस्तु दोबारा नहीं ली जा सकती। परीक्षा में allowed और not allowed शब्दों पर ध्यान दें।
C. पहले क्रमबद्ध गिनती फिर अतिरिक्त (r!) क्रम हटाना/Count ordered arrangements first then remove extra (r!) orders
Step 1
Concept
Combination is obtained by removing (r!) internal orders from permutation. In exams make the ordered count first and then divide.
Step 2
Why this answer is correct
The correct answer is C. पहले क्रमबद्ध गिनती फिर अतिरिक्त (r!) क्रम हटाना / Count ordered arrangements first then remove extra (r!) orders. Combination is obtained by removing (r!) internal orders from permutation. In exams make the ordered count first and then divide.
Step 3
Exam Tip
Combination को permutation से (r!) आंतरिक क्रम हटाकर निकाला जाता है। परीक्षा में पहले ordered count बनाकर फिर divide करें।
B. \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\)
Step 1
Concept
Simplifying the factorial formula gives the ratio \(\frac{n-r+1}{r}\). In exams ratio method is useful for adjacent combinations.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Simplifying the factorial formula gives the ratio \(\frac{n-r+1}{r}\). In exams ratio method is useful for adjacent combinations.
Step 3
Exam Tip
Factorial formula से सरल करने पर अनुपात \(\frac{n-r+1}{r}\) मिलता है। परीक्षा में adjacent combinations में ratio method उपयोगी है।
For the (r)th position (n-r+1) choices are added as a factor. In exams identify the factor of the next position.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}P_r=^{n}P_{r-1}\times(n-r+1)\). For the (r)th position (n-r+1) choices are added as a factor. In exams identify the factor of the next position.
Step 3
Exam Tip
(r)वें स्थान के लिए (n-r+1) विकल्प जुड़ते हैं। परीक्षा में next position का factor अलग पहचानें।
A. एक chosen object को चिह्नित करके double counting करना/Double counting by marking one chosen object
Step 1
Concept
In each (r)-selection one of the (r) chosen objects can be marked. In exams understand this relation through marked object counting.
Step 2
Why this answer is correct
The correct answer is A. एक chosen object को चिह्नित करके double counting करना / Double counting by marking one chosen object. In each (r)-selection one of the (r) chosen objects can be marked. In exams understand this relation through marked object counting.
Step 3
Exam Tip
हर (r)-selection में (r) chosen objects में से एक चिह्नित हो सकता है। परीक्षा में marked object वाली counting से यह relation समझें।
B. Adjacent combination ratio से/From adjacent combination ratio
Step 1
Concept
It is obtained by simplifying the ratio of \(^{n}C_r\) and \(^{n}C_{r-1}\). In exams use ratio for consecutive (C) terms.
Step 2
Why this answer is correct
The correct answer is B. Adjacent combination ratio से / From adjacent combination ratio. It is obtained by simplifying the ratio of \(^{n}C_r\) and \(^{n}C_{r-1}\). In exams use ratio for consecutive (C) terms.
Step 3
Exam Tip
यह \(^{n}C_r\) और \(^{n}C_{r-1}\) के अनुपात को सरल करके मिलता है। परीक्षा में consecutive (C) terms पर ratio लगाएँ।
There are \(^{n}C_r\) ways to choose (r) objects and then (r!) arrangements. In exams write ordered selection as choose and arrange.
Step 2
Why this answer is correct
The correct answer is C. \(^{n}C_r\times r!\). There are \(^{n}C_r\) ways to choose (r) objects and then (r!) arrangements. In exams write ordered selection as choose and arrange.
Step 3
Exam Tip
पहले (r) वस्तुएँ चुनने के \(^{n}C_r\) तरीके हैं और फिर (r!) व्यवस्थाएँ हैं। परीक्षा में ordered selection को choose and arrange लिखें।
The row arrangement of all (6) distinct books is \(^{6}P_6=6!\). In exams connect all distinct arrangement with factorial.
Step 2
Why this answer is correct
The correct answer is C. \(^{6}P_6=6!\). The row arrangement of all (6) distinct books is \(^{6}P_6=6!\). In exams connect all distinct arrangement with factorial.
Step 3
Exam Tip
सभी (6) अलग पुस्तकों की पंक्ति व्यवस्था \(^{6}P_6=6!\) होती है। परीक्षा में all distinct arrangement को factorial से जोड़ें।
Order is important when placing from top to bottom so the chosen (3) colours are arranged in (3!) ways. In exams add arrangement when positions appear.
Step 2
Why this answer is correct
The correct answer is B. \(^{5}C_3\times3!\). Order is important when placing from top to bottom so the chosen (3) colours are arranged in (3!) ways. In exams add arrangement when positions appear.
Step 3
Exam Tip
ऊपर से नीचे लगाने में क्रम महत्वपूर्ण है इसलिए चुने हुए (3) रंगों को (3!) तरीकों से लगाया जाएगा। परीक्षा में positions दिखें तो arrangement जोड़ें।
Chosen and not chosen objects together make (n) so the remaining number is (n-r). In exams this idea connects to \(^{n}C_r=^{n}C_{n-r}\).
Step 2
Why this answer is correct
The correct answer is C. (n-r). Chosen and not chosen objects together make (n) so the remaining number is (n-r). In exams this idea connects to \(^{n}C_r=^{n}C_{n-r}\).
Step 3
Exam Tip
चुनी गई और न चुनी गई वस्तुओं की कुल संख्या (n) होती है इसलिए बाकी (n-r) हैं। परीक्षा में यही विचार \(^{n}C_r=^{n}C_{n-r}\) से जुड़ता है।
