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The relation is \(^{n}P_r=^{n}C_r r!\), and (20) is not equal to any (r!) for these options. With these options the question would be invalid.
Step 2
Why this answer is correct
The correct answer is B. (4). The relation is \(^{n}P_r=^{n}C_r r!\), and (20) is not equal to any (r!) for these options. With these options the question would be invalid.
Step 3
Exam Tip
संबंध \(^{n}P_r=^{n}C_r r!\) है और (4!=24) नहीं बल्कि (20) किसी (r!) के बराबर नहीं है। दिए गए विकल्पों में कोई सटीक नहीं है इसलिए यह प्रश्न अमान्य होता यदि विकल्प यही हों।
In equal combinations different lower indices are complementary, so (3+7=n). In exams check the sum of indices in equal (C) terms.
Step 2
Why this answer is correct
The correct answer is B. (10). In equal combinations different lower indices are complementary, so (3+7=n). In exams check the sum of indices in equal (C) terms.
Step 3
Exam Tip
बराबर संचयों में अलग lower indices पूरक होते हैं इसलिए (3+7=n) है। परीक्षा में समान (C) terms में indices का योग जाँचें।
The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=5) gives \(\frac{n-4}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{n-4}{5}\). The general ratio is \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\). Putting (r=5) gives \(\frac{n-4}{5}\).
Step 3
Exam Tip
सामान्य अनुपात \(\frac{^{n}C_r}{^{n}C_{r-1}}=\frac{n-r+1}{r}\) है। यहाँ (r=5) रखने पर \(\frac{n-4}{5}\) मिलता है।
The ratio is (n-5+1=n-4), and (n-4=6) gives (n=10). In exams the ratio of consecutive permutations gives the last factor.
Step 2
Why this answer is correct
The correct answer is B. (10). The ratio is (n-5+1=n-4), and (n-4=6) gives (n=10). In exams the ratio of consecutive permutations gives the last factor.
Step 3
Exam Tip
अनुपात (n-5+1=n-4) होता है और (n-4=6) से (n=10) है। परीक्षा में consecutive permutations का अनुपात अंतिम factor देता है।
\(^{11}C_8=^{11}C_3\) and \(^{11}C_3=\frac{^{11}P_3}{3!}\). In exams first use complement to get a smaller index.
Step 2
Why this answer is correct
The correct answer is A. \(^{11}C_8=\frac{^{11}P_3}{3!}\). \(^{11}C_8=^{11}C_3\) and \(^{11}C_3=\frac{^{11}P_3}{3!}\). In exams first use complement to get a smaller index.
Step 3
Exam Tip
\(^{11}C_8=^{11}C_3\) और \(^{11}C_3=\frac{^{11}P_3}{3!}\) है। परीक्षा में पहले complement से छोटा index लें।
Put (n=10) and (r=6) 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 A. \(^{9}C_6+^{9}C_5\). Put (n=10) and (r=6) 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=10) और (r=6) रखें। परीक्षा में दोनों पदों का upper index (1) कम होता है।
In Pascal's identity the second term is \(^{n-1}C_{r-1}\). In exams remember the included case of one special object.
Step 2
Why this answer is correct
The correct answer is C. \(^{n-1}C_{r-1}\). In Pascal's identity the second term is \(^{n-1}C_{r-1}\). In exams remember the included case of one special object.
Step 3
Exam Tip
पास्कल पहचान में दूसरा पद \(^{n-1}C_{r-1}\) होता है। परीक्षा में एक विशेष वस्तु के included case को याद रखें।
The coefficients are \(^{9}C_4\) and \(^{9}C_5\), which have complementary indices. In exams check symmetry in binomial coefficients.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (4+5=9) / Because (4+5=9). The coefficients are \(^{9}C_4\) and \(^{9}C_5\), which have complementary indices. In exams check symmetry in binomial coefficients.
Step 3
Exam Tip
Coefficient क्रमशः \(^{9}C_4\) और \(^{9}C_5\) हैं और ये पूरक indices हैं। परीक्षा में binomial coefficients में symmetry देखें।
The coefficients are \(^{8}C_3\) and \(^{8}C_5\), and (3+5=8). In exams coefficients of complementary powers are equal.
Step 2
Why this answer is correct
The correct answer is A. दोनों बराबर हैं / Both are equal. The coefficients are \(^{8}C_3\) and \(^{8}C_5\), and (3+5=8). In exams coefficients of complementary powers are equal.
Step 3
Exam Tip
Coefficients \(^{8}C_3\) और \(^{8}C_5\) हैं और (3+5=8) है। परीक्षा में पूरक powers के coefficients बराबर होते हैं।
A. क्योंकि even और odd indexed sums बराबर होते हैं/Because even and odd indexed sums are equal
Step 1
Concept
From ((1+1)^n) and ((1-1)^n), even and odd sums are equal. In exams remember the alternating sum identity.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि even और odd indexed sums बराबर होते हैं / Because even and odd indexed sums are equal. From ((1+1)^n) and ((1-1)^n), even and odd sums are equal. In exams remember the alternating sum identity.
Step 3
Exam Tip
((1+1)^n) और ((1-1)^n) से even और odd sums बराबर मिलते हैं। परीक्षा में alternating sum identity याद रखें।
There are (r=5) positions to fill, so there will be (5) factors. In exams connect the number of factors with selected positions.
Step 2
Why this answer is correct
The correct answer is C. (5). There are (r=5) positions to fill, so there will be (5) factors. In exams connect the number of factors with selected positions.
Step 3
Exam Tip
(r=5) स्थान भरने हैं इसलिए (5) गुणनखंड होंगे। परीक्षा में factors की संख्या को selected positions से जोड़ें।
After cancelling ((n-4)!), four factors from (n) to (n-3) remain. In exams the number of numerator factors is (r).
Step 2
Why this answer is correct
The correct answer is A. (n(n-1)(n-2)(n-3)). After cancelling ((n-4)!), four factors from (n) to (n-3) remain. In exams the number of numerator factors is (r).
Step 3
Exam Tip
((n-4)!) कटने के बाद चार factors (n) से (n-3) तक बचते हैं। परीक्षा में numerator factors की संख्या (r) होती है।
First there are \(^{9}C_4\) ways for the committee and (4) choices for leader. In exams multiply when a role follows selection.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}C_4\times4\). First there are \(^{9}C_4\) ways for the committee and (4) choices for leader. In exams multiply when a role follows selection.
Step 3
Exam Tip
पहले committee के \(^{9}C_4\) तरीके हैं और leader के (4) विकल्प हैं। परीक्षा में चयन के बाद role हो तो multiply करें।
One leader is distinct and the order of the other (3) members is irrelevant, so divide \(^{9}P_4\) by (3!). In exams separate roles and ordinary members.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}P_4\div3!\). One leader is distinct and the order of the other (3) members is irrelevant, so divide \(^{9}P_4\) by (3!). In exams separate roles and ordinary members.
Step 3
Exam Tip
एक leader अलग है और बाकी (3) सदस्यों का क्रम महत्वहीन है इसलिए \(^{9}P_4\) को (3!) से भाग देंगे। परीक्षा में roles और ordinary members अलग करें।
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 C. \(^{6}C_3\times{}^{5}C_2\). There is unordered selection in two categories, so combinations are multiplied. In exams count category-wise selections separately.
Step 3
Exam Tip
दो category में बिना क्रम selection हो रहा है इसलिए combinations multiply होंगे। परीक्षा में category-wise selection को अलग-अलग गिनें।
Exactly (2) girls and (3) boys must be selected. In exams take only that case for exact conditions.
Step 2
Why this answer is correct
The correct answer is A. \(^{5}C_2\times{}^{6}C_3\). Exactly (2) girls and (3) boys must be selected. In exams take only that case for exact conditions.
Step 3
Exam Tip
ठीक (2) girls और (3) boys चुनने हैं। परीक्षा में exact condition में केवल वही case लें।
A. \(^{6}C_3{}^{5}C_2+^{6}C_4{}^{5}C_1+^{6}C_5{}^{5}C_0\)
Step 1
Concept
The number of boys 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 A. \(^{6}C_3{}^{5}C_2+^{6}C_4{}^{5}C_1+^{6}C_5{}^{5}C_0\). The number of boys can be (3), (4), or (5). In exams add all valid cases in at least conditions.
Step 3
Exam Tip
Boys 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 (6) choices and (7) choices for each remaining place. In exams treat leading zero separately.
Step 2
Why this answer is correct
The correct answer is A. \(6\times7^3\). The first digit cannot be (0), so there are (6) choices and (7) choices for each remaining place. In exams treat leading zero separately.
Step 3
Exam Tip
पहला digit (0) नहीं हो सकता इसलिए (6) choices हैं और बाकी तीन स्थानों पर (7) choices हैं। परीक्षा में leading zero को अलग देखें।
There are (6) non-zero choices for the first place and then (3) places are filled from the remaining (6) digits. In exams handle the first place separately.
Step 2
Why this answer is correct
The correct answer is B. \(6\times{}^{6}P_3\). There are (6) non-zero choices for the first place and then (3) places are filled from the remaining (6) digits. In exams handle the first place separately.
Step 3
Exam Tip
पहले स्थान के लिए (6) non-zero choices हैं और फिर बाकी (3) स्थान (6) बची digits से भरते हैं। परीक्षा में पहले स्थान को अलग handle करें।
Different lower indices are complementary, so (r+(r-2)=n). In exams solve equal combinations using the complement rule.
Step 2
Why this answer is correct
The correct answer is A. (n=2r-2). Different lower indices are complementary, so (r+(r-2)=n). In exams solve equal combinations using the complement rule.
Step 3
Exam Tip
अलग lower indices पूरक होंगे इसलिए (r+(r-2)=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 (2). In exams form the equation directly from the ratio formula.
Step 2
Why this answer is correct
The correct answer is A. (n-r=2(r+1)). The ratio is \(\frac{n-r}{r+1}\) and it is set equal to (2). In exams form the equation directly from the ratio formula.
Step 3
Exam Tip
अनुपात \(\frac{n-r}{r+1}\) है और उसे (2) के बराबर रखा गया है। परीक्षा में ratio formula से सीधे equation बनाएं।
A quadrilateral needs an unordered selection of (4) 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 B. \(^{10}C_4\). A quadrilateral needs an unordered selection of (4) points. In exams do not count the order of points when forming a shape.
Step 3
Exam Tip
Quadrilateral के लिए (4) 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 B. \(^{10}P_2\). 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।
In an ordinary line segment the order of endpoints does not matter. In exams use \(^{n}C_2\) for unordered pairs.
Step 2
Why this answer is correct
The correct answer is C. \(^{10}C_2\). In an ordinary line segment the order of endpoints does not matter. In exams use \(^{n}C_2\) for unordered pairs.
Step 3
Exam Tip
सामान्य line segment में endpoints का order नहीं बदलता। परीक्षा में unordered pair के लिए \(^{n}C_2\) लगाएँ।
Each circular arrangement is counted (8) 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 \(=\frac{8!}{8}\). Each circular arrangement is counted (8) 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 (8) rotations से गिनी जाती है। परीक्षा में circle में rotations को extra count मानें।
First removing rotations gives (6!), 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. \(\frac{6!}{2}\). First removing rotations gives (6!), then mirror images being the same makes us divide by (2). In exams always check reflection in necklace problems.
Step 3
Exam Tip
पहले rotation हटाकर (6!) मिलता है और 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. (3!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 (3!2!). In exams multiply the factorials in the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9!}{3!2!}\). Internal orders of two identical groups are not different, so divide by (3!2!). In exams multiply the factorials in the denominator.
Step 3
Exam Tip
दो identical groups के internal orders अलग नहीं दिखते इसलिए (3!2!) से भाग देते हैं। परीक्षा में factorial denominator को multiply करें।
Because \(^{n}P_r=^{n}C_r r!\), and (r!=1) only for (r=0) or (r=1). In exams verify equality using factorials.
Step 2
Why this answer is correct
The correct answer is A. (0) या (1) / (0) or (1). Because \(^{n}P_r=^{n}C_r r!\), and (r!=1) only for (r=0) or (r=1). In exams verify equality using factorials.
Step 3
Exam Tip
क्योंकि \(^{n}P_r=^{n}C_r r!\) और (r!=1) केवल (r=0) या (r=1) पर होता है। परीक्षा में equality को factorial से जाँचें।
Different lower indices are complementary, so (x+2x=18) and (x=6). In exams solve equal (C) terms using the complement rule.
Step 2
Why this answer is correct
The correct answer is C. (6). Different lower indices are complementary, so (x+2x=18) and (x=6). In exams solve equal (C) terms using the complement rule.
Step 3
Exam Tip
अलग lower indices पूरक हैं इसलिए (x+2x=18) और (x=6) है। परीक्षा में equal (C) terms को complement rule से हल करें।
This is the sum of non-empty selections \(2^n-1\), and \(2^6-1=63\). In exams subtract (1) when the empty selection is removed.
Step 2
Why this answer is correct
The correct answer is B. (6). This is the sum of non-empty selections \(2^n-1\), and \(2^6-1=63\). 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 \(2^6-1=63\) है। परीक्षा में empty selection हटाने पर (1) घटाएँ।
A. न चुनी गई वस्तुओं के internal orders हटाने के लिए/To remove internal orders of unchosen objects
Step 1
Concept
In the (n!) count the orders inside the unchosen group are also counted extra. In exams understand both corrections (r!) and ((n-r)!).
Step 2
Why this answer is correct
The correct answer is A. न चुनी गई वस्तुओं के internal orders हटाने के लिए / To remove internal orders of unchosen objects. In the (n!) count the orders inside the unchosen group are also counted extra. In exams understand both corrections (r!) and ((n-r)!).
Step 3
Exam Tip
(n!) वाली गिनती में न चुने गए समूह के क्रम भी extra गिने जाते हैं। परीक्षा में (r!) और ((n-r)!) दोनों corrections समझें।
A. बची हुई वस्तुओं के क्रम को ignore करना/Ignoring the order of remaining objects
Step 1
Concept
Permutation needs the order of only the selected (r) objects and not the order of remaining objects. In exams remove the unwanted tail from (n!).
Step 2
Why this answer is correct
The correct answer is A. बची हुई वस्तुओं के क्रम को ignore करना / Ignoring the order of remaining objects. Permutation needs the order of only the selected (r) objects and not the order of remaining objects. In exams remove the unwanted tail from (n!).
Step 3
Exam Tip
Permutation में केवल चुनी (r) वस्तुओं का order चाहिए और बची वस्तुओं का order नहीं चाहिए। परीक्षा में (n!) से unwanted tail हटाएँ।
Order inside each group is irrelevant, so divide by (a!b!c!). In exams do not divide among labelled groups themselves.
Step 2
Why this answer is correct
The correct answer is A. (a!b!c!). Order inside each group is irrelevant, so divide by (a!b!c!). In exams do not divide among labelled groups themselves.
Step 3
Exam Tip
हर group के अंदर order महत्वहीन है इसलिए (a!b!c!) से भाग दिया जाता है। परीक्षा में labelled groups में groups को आपस में divide न करें।
The two equal groups are unlabelled, so interchanging the groups counts twice. In exams apply the extra (2!) correction for equal unlabelled groups.
Step 2
Why this answer is correct
The correct answer is A. (2!). The two equal groups are unlabelled, so interchanging the groups counts twice. In exams apply the extra (2!) correction for equal unlabelled groups.
Step 3
Exam Tip
दो equal groups unlabelled हैं इसलिए group interchange दो बार गिनता है। परीक्षा में equal unlabelled groups पर extra (2!) correction लगाएँ।
After choosing the first (4)-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. \(\frac{^{8}C_4}{2!}\). After choosing the first (4)-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 (4)-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. \(^{n}C_r\) selection गिनता है और \(^{n}P_r\) ordered selection गिनता है/\(^{n}C_r\) counts selection and \(^{n}P_r\) counts ordered selection
Step 1
Concept
Permutation adds the arrangement of chosen objects to combination. In exams remember \(P=C\times r!\) conceptually.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_r\) selection गिनता है और \(^{n}P_r\) ordered selection गिनता है / \(^{n}C_r\) counts selection and \(^{n}P_r\) counts ordered selection. Permutation adds the arrangement of chosen objects to combination. In exams remember \(P=C\times r!\) conceptually.
Step 3
Exam Tip
Permutation में combination के साथ चुनी वस्तुओं की व्यवस्था भी जुड़ती है। परीक्षा में \(P=C\times r!\) को concept से याद रखें।
