Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है • Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है • Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
In the permutation recurrence, the new factor is the choices for the (r)th position, (n-r+1). In exams connect the multiplier with the next position.
Step 2
Why this answer is correct
The correct answer is C. (n-r+1). In the permutation recurrence, the new factor is the choices for the (r)th position, (n-r+1). In exams connect the multiplier with the next position.
Step 3
Exam Tip
क्रमचय recurrence में नया गुणक (r)वें स्थान के विकल्प (n-r+1) होता है। परीक्षा में multiplier को next position से जोड़ें।
The ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\), so for (r=4), \(\frac{n-3}{4}=2\). In exams use the ratio formula for adjacent combinations.
Step 2
Why this answer is correct
The correct answer is B. \(n-3=2\times4\). The ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\), so for (r=4), \(\frac{n-3}{4}=2\). In exams use the ratio formula for adjacent combinations.
Step 3
Exam Tip
अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है, इसलिए (r=4) पर \(\frac{n-3}{4}=2\) मिलेगा। परीक्षा में adjacent combinations में ratio formula लगाएँ।
In equal combinations, different lower indices are complementary, so (5+8=n). In exams check the sum of indices in equal (C) terms.
Step 2
Why this answer is correct
The correct answer is A. (13). In equal combinations, different lower indices are complementary, so (5+8=n). In exams check the sum of indices in equal (C) terms.
Step 3
Exam Tip
समान combinations में अलग lower indices पूरक होते हैं, इसलिए (5+8=n) है। परीक्षा में equal (C) terms में indices का योग देखें।
The relation is \(^{n}P_r=^{n}C_r\times r!\), so here the factor is (4!). In exams add the arrangements of selected objects when going from (C) to (P).
Step 2
Why this answer is correct
The correct answer is C. (4!). The relation is \(^{n}P_r=^{n}C_r\times r!\), so here the factor is (4!). In exams add the arrangements of selected objects when going from (C) to (P).
Step 3
Exam Tip
संबंध \(^{n}P_r=^{n}C_r\times r!\) है, इसलिए यहाँ factor (4!) होगा। परीक्षा में (C) से (P) जाने पर चुनी वस्तुओं की arrangements जोड़ें।
The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=6) gives \(\frac{n-5}{6}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{n-5}{6}\). The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=6) gives \(\frac{n-5}{6}\).
Step 3
Exam Tip
सामान्य अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। (r=6) रखने पर \(\frac{n-5}{6}\) मिलता है।
The ratio is (n-6+1=n-5), and (n-5=9) gives (n=14). In exams the ratio of consecutive permutations gives the last factor.
Step 2
Why this answer is correct
The correct answer is C. (14). The ratio is (n-6+1=n-5), and (n-5=9) gives (n=14). In exams the ratio of consecutive permutations gives the last factor.
Step 3
Exam Tip
अनुपात (n-6+1=n-5) होता है और (n-5=9) से (n=14) है। परीक्षा में consecutive permutations का ratio last factor देता है।
\(^{14}C_{11}=^{14}C_3\), and \(^{14}C_3=\frac{^{14}P_3}{3!}\). In exams first use the complement to get a smaller index.
Step 2
Why this answer is correct
The correct answer is B. \(^{14}C_{11}=\frac{^{14}P_3}{3!}\). \(^{14}C_{11}=^{14}C_3\), and \(^{14}C_3=\frac{^{14}P_3}{3!}\). In exams first use the complement to get a smaller index.
Step 3
Exam Tip
\(^{14}C_{11}=^{14}C_3\) और \(^{14}C_3=\frac{^{14}P_3}{3!}\) है। परीक्षा में पहले complement से छोटा index लें।
Put (n=11) and (r=7) in \(^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}\). In exams the upper index of both terms decreases by (1).
Step 2
Why this answer is correct
The correct answer is D. \(^{10}C_7+^{10}C_6\). Put (n=11) and (r=7) in \(^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}\). In exams the upper index of both terms decreases by (1).
Step 3
Exam Tip
\(^{n}C_r=^{n-1}C_r+^{n-1}C_{r-1}\) में (n=11) और (r=7) रखें। परीक्षा में दोनों terms का upper index (1) कम होता है।
In Pascal's identity, the other case is not choosing the special object, \(^{n-1}C_r\). In exams separate included and excluded cases.
Step 2
Why this answer is correct
The correct answer is A. \(^{n-1}C_r\). In Pascal's identity, the other case is not choosing the special object, \(^{n-1}C_r\). In exams separate included and excluded cases.
Step 3
Exam Tip
पास्कल पहचान में दूसरा case विशेष वस्तु को न चुनने का \(^{n-1}C_r\) होता है। परीक्षा में included और excluded case अलग करें।
The coefficients are \(^{10}C_3\) and \(^{10}C_7\), and (3+7=10). In exams coefficients of complementary powers are equal.
Step 2
Why this answer is correct
The correct answer is C. दोनों बराबर हैं / Both are equal. The coefficients are \(^{10}C_3\) and \(^{10}C_7\), and (3+7=10). In exams coefficients of complementary powers are equal.
Step 3
Exam Tip
Coefficients \(^{10}C_3\) और \(^{10}C_7\) हैं और (3+7=10) है। परीक्षा में पूरक powers के coefficients बराबर होते हैं।
The coefficients are \(^{11}C_5\) and \(^{11}C_6\), which have complementary indices. In exams identify binomial symmetry.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (5+6=11) / Because (5+6=11). The coefficients are \(^{11}C_5\) and \(^{11}C_6\), which have complementary indices. In exams identify binomial symmetry.
Step 3
Exam Tip
Coefficients \(^{11}C_5\) और \(^{11}C_6\) हैं जो पूरक indices हैं। परीक्षा में binomial symmetry पहचानें।
The sum of odd indexed combinations equals the even indexed sum, \(2^{n-1}\). In exams remember the alternating identity from ((1-1)^n).
Step 2
Why this answer is correct
The correct answer is C. \(2^{n-1}\). The sum of odd indexed combinations equals the even indexed sum, \(2^{n-1}\). In exams remember the alternating identity from ((1-1)^n).
Step 3
Exam Tip
Odd indexed combinations का योग even indexed sum के बराबर \(2^{n-1}\) होता है। परीक्षा में ((1-1)^n) से alternating identity याद रखें।
First there are \(^{10}C_5\) ways to form the committee and (5) choices for coordinator. In exams multiply when a role follows selection.
Step 2
Why this answer is correct
The correct answer is C. \(^{10}C_5\times5\). First there are \(^{10}C_5\) ways to form the committee and (5) choices for coordinator. In exams multiply when a role follows selection.
Step 3
Exam Tip
पहले committee के \(^{10}C_5\) तरीके हैं और संयोजक के (5) विकल्प हैं। परीक्षा में selection के बाद role हो तो multiply करें।
One person has a distinct role and the order of the remaining (4) members is irrelevant. In exams remove overcounting of remaining members when there is a special role.
Step 2
Why this answer is correct
The correct answer is A. \(^{10}P_5\div4!\). One person has a distinct role and the order of the remaining (4) members is irrelevant. In exams remove overcounting of remaining members when there is a special role.
Step 3
Exam Tip
एक व्यक्ति का role अलग है और बाकी (4) सदस्यों का क्रम महत्वहीन है। परीक्षा में special role होने पर remaining members का overcount हटाएँ।
There is unordered selection in two categories, so combinations are multiplied. In exams count category-wise selections separately.
Step 2
Why this answer is correct
The correct answer is D. \(^{7}C_3\times{}^{6}C_2\). There is unordered selection in two categories, so combinations are multiplied. In exams count category-wise selections separately.
Step 3
Exam Tip
दो categories में बिना क्रम selection हो रहा है, इसलिए combinations multiply होंगे। परीक्षा में category-wise selection को अलग-अलग गिनें।
Exactly (3) boys and (2) girls must be selected. In exams take only that case for exact conditions.
Step 2
Why this answer is correct
The correct answer is A. \(^{7}C_3\times{}^{6}C_2\). Exactly (3) boys and (2) girls must be selected. In exams take only that case for exact conditions.
Step 3
Exam Tip
ठीक (3) लड़के और (2) लड़कियाँ चुननी हैं। परीक्षा में exact condition में केवल वही case लें।
C. \(^{6}C_3{}^{7}C_2+^{6}C_4{}^{7}C_1+^{6}C_5{}^{7}C_0\)
Step 1
Concept
The number of girls can be (3), (4), or (5). In exams add all valid cases in at least conditions.
Step 2
Why this answer is correct
The correct answer is C. \(^{6}C_3{}^{7}C_2+^{6}C_4{}^{7}C_1+^{6}C_5{}^{7}C_0\). The number of girls can be (3), (4), or (5). In exams add all valid cases in at least conditions.
Step 3
Exam Tip
लड़कियों की संख्या (3), (4) या (5) हो सकती है। परीक्षा में at least condition में सभी valid cases जोड़ें।
The first digit cannot be (0), so there are (7) choices and (8) choices for each of the remaining four places. In exams treat leading zero separately.
Step 2
Why this answer is correct
The correct answer is A. \(7\times8^4\). The first digit cannot be (0), so there are (7) choices and (8) choices for each of the remaining four places. In exams treat leading zero separately.
Step 3
Exam Tip
पहला digit (0) नहीं हो सकता, इसलिए (7) choices हैं और बाकी चार स्थानों पर (8) choices हैं। परीक्षा में leading zero को अलग देखें।
There are (7) non-zero choices for the first place, and then (4) places are filled from the remaining (7) digits. In exams handle the first place separately.
Step 2
Why this answer is correct
The correct answer is B. \(7\times{}^{7}P_4\). There are (7) non-zero choices for the first place, and then (4) places are filled from the remaining (7) digits. In exams handle the first place separately.
Step 3
Exam Tip
पहले स्थान के लिए (7) non-zero choices हैं और फिर बाकी (4) स्थान (7) बची digits से भरते हैं। परीक्षा में पहले स्थान को अलग handle करें।
Different lower indices are complementary, so (r+(r-3)=n). In exams solve equal combinations using the complement rule.
Step 2
Why this answer is correct
The correct answer is C. (n=2r-3). Different lower indices are complementary, so (r+(r-3)=n). In exams solve equal combinations using the complement rule.
Step 3
Exam Tip
अलग lower indices पूरक होंगे, इसलिए (r+(r-3)=n) है। परीक्षा में equal combination को complement rule से हल करें।
The ratio is \(\frac{n-r}{r+1}\), and it is set equal to (3). In exams form the equation directly from the ratio formula.
Step 2
Why this answer is correct
The correct answer is B. (n-r=3(r+1)). The ratio is \(\frac{n-r}{r+1}\), and it is set equal to (3). In exams form the equation directly from the ratio formula.
Step 3
Exam Tip
अनुपात \(\frac{n-r}{r+1}\) है और उसे (3) के बराबर रखा गया है। परीक्षा में ratio formula से सीधे equation बनाएं।
A pentagon needs an unordered selection of (5) points. In exams do not count the order of points when forming a shape.
Step 2
Why this answer is correct
The correct answer is C. \(^{12}C_5\). A pentagon needs an unordered selection of (5) points. In exams do not count the order of points when forming a shape.
Step 3
Exam Tip
Pentagon के लिए (5) points का unordered selection चाहिए। परीक्षा में shape बनाते समय points का order न गिनें।
In a directed segment, changing start and end changes the object. In exams use permutation when direction exists.
Step 2
Why this answer is correct
The correct answer is D. \(^{12}P_2\). In a directed segment, changing start and end changes the object. In exams use permutation when direction exists.
Step 3
Exam Tip
Directed segment में start और end बदलने से object बदलता है। परीक्षा में direction हो तो permutation लगाएँ।
Each circular arrangement is counted (9) times in the linear count due to rotations. In exams treat rotations as extra count in a circle.
Step 2
Why this answer is correct
The correct answer is A. Circular count \(=\frac{9!}{9}\). Each circular arrangement is counted (9) times in the linear count due to rotations. In exams treat rotations as extra count in a circle.
Step 3
Exam Tip
हर circular arrangement linear count में (9) rotations से गिनी जाती है। परीक्षा में circle में rotations को extra count मानें।
First removing rotations gives (7!), then mirror images being the same makes us divide by (2). In exams always check reflection in necklace problems.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{7!}{2}\). First removing rotations gives (7!), then mirror images being the same makes us divide by (2). In exams always check reflection in necklace problems.
Step 3
Exam Tip
पहले rotations हटाकर (7!) मिलता है, फिर mirror images same होने से (2) से भाग देते हैं। परीक्षा में necklace में reflection जरूर जाँचें।
Internal interchanges of repeated letters do not create new arrangements. In exams divide by the factorial of each repeated group.
Step 2
Why this answer is correct
The correct answer is A. (4!2!). Internal interchanges of repeated letters do not create new arrangements. In exams divide by the factorial of each repeated group.
Step 3
Exam Tip
Repeated letters की अंदरूनी अदला-बदली नई arrangement नहीं बनाती। परीक्षा में हर repeated group के factorial से भाग दें।
Internal orders of two identical groups are not different, so divide by (4!3!). In exams multiply the factorials in the denominator.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{10!}{4!3!}\). Internal orders of two identical groups are not different, so divide by (4!3!). In exams multiply the factorials in the denominator.
Step 3
Exam Tip
दो identical groups के internal orders अलग नहीं दिखते इसलिए (4!3!) से भाग देते हैं। परीक्षा में factorial denominator को multiply करें।
B. यह असंभव है क्योंकि (r!>1)/This is impossible because (r!>1)
Step 1
Concept
\(^{n}P_r=^{n}C_r r!\), and for (r>1), (r!>1). In exams verify equality of (P) and (C) using factorials.
Step 2
Why this answer is correct
The correct answer is B. यह असंभव है क्योंकि (r!>1) / This is impossible because (r!>1). \(^{n}P_r=^{n}C_r r!\), and for (r>1), (r!>1). In exams verify equality of (P) and (C) using factorials.
Step 3
Exam Tip
\(^{n}P_r=^{n}C_r r!\) और (r>1) पर (r!>1) होता है। परीक्षा में (P) और (C) की equality को factorial से जाँचें।
Different lower indices are complementary, so (x+2x=21) and (x=7). In exams solve equal (C) terms using the complement rule.
Step 2
Why this answer is correct
The correct answer is C. (7). Different lower indices are complementary, so (x+2x=21) and (x=7). In exams solve equal (C) terms using the complement rule.
Step 3
Exam Tip
अलग lower indices पूरक हैं इसलिए (x+2x=21) और (x=7) है। परीक्षा में equal (C) terms को complement rule से हल करें।
This is the sum of non-empty selections \(2^n-1\), and \(2^7-1=127\). In exams subtract (1) when the empty selection is removed.
Step 2
Why this answer is correct
The correct answer is B. (7). This is the sum of non-empty selections \(2^n-1\), and \(2^7-1=127\). In exams subtract (1) when the empty selection is removed.
Step 3
Exam Tip
यह non-empty selections का sum \(2^n-1\) है और \(2^7-1=127\) है। परीक्षा में empty selection हटाने पर (1) घटाएँ।
A. चुने और न चुने groups के internal orders हटाने के लिए/To remove internal orders of chosen and unchosen groups
Step 1
Concept
In the (n!) count, the orders inside both groups are counted extra. In exams understand both corrections (r!) and ((n-r)!).
Step 2
Why this answer is correct
The correct answer is A. चुने और न चुने groups के internal orders हटाने के लिए / To remove internal orders of chosen and unchosen groups. In the (n!) count, the orders inside both groups are counted extra. In exams understand both corrections (r!) and ((n-r)!).
Step 3
Exam Tip
(n!) वाली गिनती में दोनों groups के अंदर के क्रम extra गिने जाते हैं। परीक्षा में (r!) और ((n-r)!) दोनों corrections समझें।
Order inside each labelled group is irrelevant, so divide by (a!b!c!). In exams do not divide extra among labelled groups themselves.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n!}{a!b!c!}\). Order inside each labelled group is irrelevant, so divide by (a!b!c!). In exams do not divide extra among labelled groups themselves.
Step 3
Exam Tip
हर labelled group के अंदर order महत्वहीन है इसलिए (a!b!c!) से भाग दिया जाता है। परीक्षा में labelled groups को आपस में extra divide न करें।
After choosing the first (5)-group the second is fixed, and interchanging the two groups creates duplicate count. In exams divide by (2!) for equal groups.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{^{10}C_5}{2!}\). After choosing the first (5)-group the second is fixed, and interchanging the two groups creates duplicate count. In exams divide by (2!) for equal groups.
Step 3
Exam Tip
पहला (5)-group चुनने पर दूसरा तय है और दोनों groups interchange होने से duplicate count आता है। परीक्षा में equal groups हों तो (2!) से भाग दें।
A. गलत, क्योंकि सही गुणक (4!=24) है/False, because the correct multiplier is (4!=24)
Step 1
Concept
\(^{n}P_r=^{n}C_r r!\), so for (r=4) the multiplier is (24). In exams verify the multiplier using factorials.
Step 2
Why this answer is correct
The correct answer is A. गलत, क्योंकि सही गुणक (4!=24) है / False, because the correct multiplier is (4!=24). \(^{n}P_r=^{n}C_r r!\), so for (r=4) the multiplier is (24). In exams verify the multiplier using factorials.
Step 3
Exam Tip
\(^{n}P_r=^{n}C_r r!\) है, इसलिए (r=4) पर गुणक (24) होगा। परीक्षा में multiplier को factorial से verify करें।
A. \(^{n}P_r\) में चयन के साथ चुनी वस्तुओं की व्यवस्था भी गिनी जाती है/\(^{n}P_r\) counts selection along with arrangement of selected objects
Step 1
Concept
Permutation is (r!) times combination because the selected objects must be ordered. In exams remember \(P=C\times r!\) conceptually.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}P_r\) में चयन के साथ चुनी वस्तुओं की व्यवस्था भी गिनी जाती है / \(^{n}P_r\) counts selection along with arrangement of selected objects. Permutation is (r!) times combination because the selected objects must be ordered. In exams remember \(P=C\times r!\) conceptually.
Step 3
Exam Tip
Permutation combination से (r!) गुना होता है क्योंकि चुनी वस्तुओं को क्रम देना पड़ता है। परीक्षा में \(P=C\times r!\) को concept से याद रखें।
