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C. फैक्टोरियल रूपों का अनुपात लेना/Taking the ratio of factorial forms
Step 1
Concept
Dividing the two factorial formulas gives the factor \(\frac{n}{n-r}\). In exams verify related identities using factorials.
Step 2
Why this answer is correct
The correct answer is C. फैक्टोरियल रूपों का अनुपात लेना / Taking the ratio of factorial forms. Dividing the two factorial formulas gives the factor \(\frac{n}{n-r}\). In exams verify related identities using factorials.
Step 3
Exam Tip
दोनों फैक्टोरियल सूत्रों को भाग देने पर \(\frac{n}{n-r}\) गुणक आता है। परीक्षा में संबंधित पहचान को फैक्टोरियल से जाँचें।
\(^{n}P_r=^{n}C_r\times r!\), so the multiplier here is (4!). In exams count the arrangement of chosen objects separately.
Step 2
Why this answer is correct
The correct answer is D. (4!). \(^{n}P_r=^{n}C_r\times r!\), so the multiplier here is (4!). In exams count the arrangement of chosen objects separately.
Step 3
Exam Tip
\(^{n}P_r=^{n}C_r\times r!\) है इसलिए यहाँ गुणक (4!) है। परीक्षा में चुनी वस्तुओं की व्यवस्था अलग से गिनें।
Choosing (12) is like leaving (3), so \(^{15}C_{12}=^{15}C_3\). In exams reduce a large lower index using its complement.
Step 2
Why this answer is correct
The correct answer is A. \(^{15}C_3\). Choosing (12) is like leaving (3), so \(^{15}C_{12}=^{15}C_3\). In exams reduce a large lower index using its complement.
Step 3
Exam Tip
(12) चुनना (3) छोड़ने जैसा है इसलिए \(^{15}C_{12}=^{15}C_3\) है। परीक्षा में बड़े lower index को complement से छोटा करें।
In equal combinations, different lower indices are complementary, so (4+6=n). In exams check the sum of indices in equal combinations.
Step 2
Why this answer is correct
The correct answer is B. (10). In equal combinations, different lower indices are complementary, so (4+6=n). In exams check the sum of indices in equal combinations.
Step 3
Exam Tip
बराबर combinations में अलग lower indices पूरक होते हैं इसलिए (4+6=n) होगा। परीक्षा में equal combination में indices का योग देखें।
Put (n=9) and (r=5) 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. \(^{8}C_5+^{8}C_4\). Put (n=9) and (r=5) 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=9) और (r=5) रखें। परीक्षा में दोनों terms का upper index (1) घटता है।
Taking the ratio of factorial forms gives \(^{n}C_{r+1}=^{n}C_r\frac{n-r}{r+1}\). In exams use ratio method for adjacent terms.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{n-r}{r+1}\). Taking the ratio of factorial forms gives \(^{n}C_{r+1}=^{n}C_r\frac{n-r}{r+1}\). In exams use ratio method for adjacent terms.
Step 3
Exam Tip
फैक्टोरियल रूपों का अनुपात लेने पर \(^{n}C_{r+1}=^{n}C_r\frac{n-r}{r+1}\) मिलता है। परीक्षा में adjacent terms पर ratio method उपयोग करें।
The ratio is (n-4+1=n-3), and (n-3=7) 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-4+1=n-3), and (n-3=7) gives (n=10). In exams the ratio of consecutive permutations gives the last factor.
Step 3
Exam Tip
अनुपात (n-4+1=n-3) होता है और (n-3=7) से (n=10) है। परीक्षा में consecutive permutations का ratio last factor देता है।
In (^{n}C_3=\frac{n!}{3!(n-3)!}), cancelling ((n-3)!) leaves (n(n-1)(n-2)). In exams write the cancellation step clearly.
Step 2
Why this answer is correct
The correct answer is B. (n(n-1)(n-2)). In (^{n}C_3=\frac{n!}{3!(n-3)!}), cancelling ((n-3)!) leaves (n(n-1)(n-2)). In exams write the cancellation step clearly.
Step 3
Exam Tip
(^{n}C_3=\frac{n!}{3!(n-3)!}) में ((n-3)!) कटने पर (n(n-1)(n-2)) बचता है। परीक्षा में cancellation step साफ लिखें।
(3!=6), and it removes internal orders of three chosen objects. In exams remember small factorial values.
Step 2
Why this answer is correct
The correct answer is B. (3!) से / From (3!). (3!=6), and it removes internal orders of three chosen objects. In exams remember small factorial values.
Step 3
Exam Tip
(3!=6) होता है और यह चुनी गई तीन वस्तुओं के internal orders हटाता है। परीक्षा में छोटे factorial values याद रखें।
A triangle needs an unordered selection of (3) points. In exams do not count the order of points while forming a shape.
Step 2
Why this answer is correct
The correct answer is B. \(^{10}C_3\). A triangle needs an unordered selection of (3) points. In exams do not count the order of points while forming a shape.
Step 3
Exam Tip
Triangle बनाने के लिए (3) points का unordered selection चाहिए। परीक्षा में shape बनाने में points का order नहीं गिनें।
First choose (3) candidates and then assign the posts in (3!) ways. In exams add arrangement when posts are different.
Step 2
Why this answer is correct
The correct answer is B. \(^{7}C_3\times3!\). First choose (3) candidates and then assign the posts in (3!) ways. In exams add arrangement when posts are different.
Step 3
Exam Tip
पहले (3) candidates चुनें और फिर उन्हें (3!) ways से posts दें। परीक्षा में अलग posts हों तो arrangement जोड़ें।
A. दो groups के अंदर के क्रम हटाना/Removing orders inside two groups
Step 1
Concept
In (n!), internal orders of chosen and unchosen groups are counted separately. In exams remove internal order in group division.
Step 2
Why this answer is correct
The correct answer is A. दो groups के अंदर के क्रम हटाना / Removing orders inside two groups. In (n!), internal orders of chosen and unchosen groups are counted separately. In exams remove internal order in group division.
Step 3
Exam Tip
(n!) में चुने और न चुने groups के internal orders अलग गिने जाते हैं। परीक्षा में group division में internal order हटाएँ।
Once one group is chosen, the other group is fixed, so \(^{n}C_r\) is enough. In exams count only one selection when the complement is fixed.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_r\). Once one group is chosen, the other group is fixed, so \(^{n}C_r\) is enough. In exams count only one selection when the complement is fixed.
Step 3
Exam Tip
एक group चुनते ही दूसरा group तय हो जाता है इसलिए \(^{n}C_r\) पर्याप्त है। परीक्षा में complement fixed हो तो एक ही selection गिनें।
First there are \(^{8}C_3\) ways to form the team and (3) choices for captain. In exams multiply when an extra role follows selection.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_3\times3\). First there are \(^{8}C_3\) ways to form the team and (3) choices for captain. In exams multiply when an extra role follows selection.
Step 3
Exam Tip
पहले team के \(^{8}C_3\) तरीके हैं और captain के (3) विकल्प हैं। परीक्षा में selection के बाद extra role हो तो multiply करें।
One of the three is distinguished as captain, so removing the order of the remaining (2) members gives \(^{8}P_3\div2!\). In exams reduce overcounting by checking the roles.
Step 2
Why this answer is correct
The correct answer is C. \(^{8}P_3\div2!\). One of the three is distinguished as captain, so removing the order of the remaining (2) members gives \(^{8}P_3\div2!\). In exams reduce overcounting by checking the roles.
Step 3
Exam Tip
तीन में से एक captain distinguished है इसलिए बाकी (2) members का order हटाने पर \(^{8}P_3\div2!\) मिलता है। परीक्षा में roles की संख्या देखकर overcounting घटाएँ।
Internal orders of each repeated group are not different, so divide by (2!2!3!). In exams treat repeated letters as separate groups.
Step 2
Why this answer is correct
The correct answer is A. (2!2!3!). Internal orders of each repeated group are not different, so divide by (2!2!3!). In exams treat repeated letters as separate groups.
Step 3
Exam Tip
हर repeated group के internal orders अलग नहीं होते इसलिए (2!2!3!) से भाग देते हैं। परीक्षा में repeated letters को अलग-अलग group मानें।
A. Repeated letters के कारण division principle/Division principle due to repeated letters
Step 1
Concept
Interchanging identical letters does not make a new arrangement. In exams write frequencies of repeated letters and divide by factorials.
Step 2
Why this answer is correct
The correct answer is A. Repeated letters के कारण division principle / Division principle due to repeated letters. Interchanging identical letters does not make a new arrangement. In exams write frequencies of repeated letters and divide by factorials.
Step 3
Exam Tip
समान अक्षरों की अदला-बदली नया arrangement नहीं बनाती। परीक्षा में repeated letters की frequency लिखकर factorial से भाग दें।
A. (r)वें स्थान के विकल्प/Choices for the (r)th position
Step 1
Concept
After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 2
Why this answer is correct
The correct answer is A. (r)वें स्थान के विकल्प / Choices for the (r)th position. After filling the first (r-1) positions (n-r+1) choices remain for the (r)th position. In exams connect recurrence with the next-place idea.
Step 3
Exam Tip
पहले (r-1) स्थान भरने के बाद (r)वें स्थान पर (n-r+1) विकल्प बचते हैं। परीक्षा में recurrence को अगले स्थान के विचार से जोड़ें।
The circular count is (7!) and the linear count is (8!), so the ratio is \(\frac{7!}{8!}=\frac{1}{8}\). In exams rotations are removed in circular counting.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{8}\). The circular count is (7!) and the linear count is (8!), so the ratio is \(\frac{7!}{8!}=\frac{1}{8}\). In exams rotations are removed in circular counting.
Step 3
Exam Tip
Circular count (7!) और linear count (8!) है इसलिए ratio \(\frac{7!}{8!}=\frac{1}{8}\) है। परीक्षा में circular count में rotations हटते हैं।
Removing rotation first gives (5!), then reflection being same makes us divide by (2). In exams check both rotation and reflection in necklaces.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5!}{2}\). Removing rotation first gives (5!), then reflection being same makes us divide by (2). In exams check both rotation and reflection in necklaces.
Step 3
Exam Tip
पहले rotation हटाने से (5!) आता है फिर reflection same होने से (2) से भाग देते हैं। परीक्षा में necklace में rotation और reflection दोनों देखें।
In equal combinations the lower indices are equal or complementary. Here the indices are different so their sum is (n).
Step 2
Why this answer is correct
The correct answer is A. (r+(r+2)=n). In equal combinations the lower indices are equal or complementary. Here the indices are different so their sum is (n).
Step 3
Exam Tip
बराबर संचयों में निचले सूचक समान या पूरक होते हैं। यहाँ अलग सूचक हैं इसलिए उनका योग (n) होगा।
The selections of boys and girls are independent and order does not matter in a committee. In exams multiply combinations for category-wise selection.
Step 2
Why this answer is correct
The correct answer is C. \(^{5}C_2\times{}^{4}C_2\). The selections of boys and girls are independent and order does not matter in a committee. In exams multiply combinations for category-wise selection.
Step 3
Exam Tip
लड़कों और लड़कियों के selections स्वतंत्र हैं और committee में order नहीं है। परीक्षा में category-wise selection पर combinations multiply करें।
A. \(^{4}C_2{}^{5}C_2+^{4}C_3{}^{5}C_1+^{4}C_4{}^{5}C_0\)
Step 1
Concept
The number of girls can be (2), (3), or (4), and the cases are added. In exams write valid cases separately in at least questions.
Step 2
Why this answer is correct
The correct answer is A. \(^{4}C_2{}^{5}C_2+^{4}C_3{}^{5}C_1+^{4}C_4{}^{5}C_0\). The number of girls can be (2), (3), or (4), and the cases are added. In exams write valid cases separately in at least questions.
Step 3
Exam Tip
Girls की संख्या (2), (3) या (4) हो सकती है और cases जोड़े जाते हैं। परीक्षा में at least वाले प्रश्नों में valid cases अलग लिखें।
Even and odd indexed combination sums are equal and the total sum is \(2^n\). In exams remember ((1-1)^n) for the even-sum identity.
Step 2
Why this answer is correct
The correct answer is A. \(2^{n-1}\). Even and odd indexed combination sums are equal and the total sum is \(2^n\). In exams remember ((1-1)^n) for the even-sum identity.
Step 3
Exam Tip
Even और odd indexed combination sums बराबर होते हैं और कुल योग \(2^n\) है। परीक्षा में even-sum identity के लिए ((1-1)^n) याद रखें।
The sum of odd indexed combinations equals the even indexed sum and is \(2^{n-1}\). In exams prove it using the alternating identity.
Step 2
Why this answer is correct
The correct answer is B. \(2^{n-1}\). The sum of odd indexed combinations equals the even indexed sum and is \(2^{n-1}\). In exams prove it using the alternating identity.
Step 3
Exam Tip
Odd indexed combinations का योग even indexed combinations के बराबर \(2^{n-1}\) होता है। परीक्षा में alternating identity से इसे साबित करें।
Different lower indices must be complementary, so (2+4=n). In exams quickly check complementary indices in equal (C) terms.
Step 2
Why this answer is correct
The correct answer is C. (6). Different lower indices must be complementary, so (2+4=n). In exams quickly check complementary indices in equal (C) terms.
Step 3
Exam Tip
अलग lower indices पूरक होंगे इसलिए (2+4=n) है। परीक्षा में समान (C) terms में पूरक index तुरंत देखें।
Two objects have (2!) orders and (2!=2). In exams remember the relation between ordered and unordered counts in pairs.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (2!=2) / Because (2!=2). Two objects have (2!) orders and (2!=2). In exams remember the relation between ordered and unordered counts in pairs.
Step 3
Exam Tip
दो वस्तुओं के (2!) orders होते हैं और (2!=2) है। परीक्षा में pair में ordered और unordered count का संबंध याद रखें।
A. यह \(^{7}C_4\times4!\) है/It is \(^{7}C_4\times4!\)
Step 1
Concept
First choose (4) letters and then arrange them in (4!) orders. In exams order is important in words.
Step 2
Why this answer is correct
The correct answer is A. यह \(^{7}C_4\times4!\) है / It is \(^{7}C_4\times4!\). First choose (4) letters and then arrange them in (4!) orders. In exams order is important in words.
Step 3
Exam Tip
पहले (4) letters चुनते हैं फिर उन्हें (4!) orders में रखते हैं। परीक्षा में words में order महत्वपूर्ण होता है।
The first place cannot have (0), so it has (4) choices and the remaining places have (5) choices. In exams check the leading zero condition first.
Step 2
Why this answer is correct
The correct answer is A. \(4\times5\times5\). The first place cannot have (0), so it has (4) choices and the remaining places have (5) choices. In exams check the leading zero condition first.
Step 3
Exam Tip
पहले स्थान पर (0) नहीं आ सकता इसलिए (4) choices हैं और बाकी स्थानों पर (5) choices हैं। परीक्षा में leading zero condition पहले देखें।
There are (5) non-zero choices for the first digit and then the remaining (3) places are filled from (5) remaining digits. In exams treat the first place as a separate case.
Step 2
Why this answer is correct
The correct answer is A. \(5\times{}^{5}P_3\). There are (5) non-zero choices for the first digit and then the remaining (3) places are filled from (5) remaining digits. In exams treat the first place as a separate case.
Step 3
Exam Tip
पहले digit के लिए (5) non-zero choices हैं और फिर बाकी (3) स्थान (5) remaining digits से भरते हैं। परीक्षा में पहला स्थान अलग case की तरह लें।
Different lower indices are complementary, so (r+(r-1)=n). In exams this relation is useful for adjacent equal combinations.
Step 2
Why this answer is correct
The correct answer is A. (n=2r-1). Different lower indices are complementary, so (r+(r-1)=n). In exams this relation is useful for adjacent equal combinations.
Step 3
Exam Tip
अलग lower indices पूरक होंगे इसलिए (r+(r-1)=n) मिलता है। परीक्षा में adjacent equal combinations में यह relation उपयोगी है।
The ratio is \(\frac{n-r+1}{r}\), and setting it equal to (1) gives (n-r+1=r). In exams form an equation using the ratio identity.
Step 2
Why this answer is correct
The correct answer is A. (n-r+1=r). The ratio is \(\frac{n-r+1}{r}\), and setting it equal to (1) gives (n-r+1=r). In exams form an equation using the ratio identity.
Step 3
Exam Tip
अनुपात \(\frac{n-r+1}{r}\) है और इसे (1) रखने पर (n-r+1=r) मिलता है। परीक्षा में ratio identity से equation बनाइए।
In the recurrence, the extra factor is the choices for the (r)th place, which is (n-r+1). In exams treat the multiplier as next-place choices.
Step 2
Why this answer is correct
The correct answer is A. (n-r+1). In the recurrence, the extra factor is the choices for the (r)th place, which is (n-r+1). In exams treat the multiplier as next-place choices.
Step 3
Exam Tip
recurrence में extra factor (r)वें स्थान के choices यानी (n-r+1) होता है। परीक्षा में multiplier को next-place choices मानें।
A. Denominator में वही दो factorial रहते हैं/The same two factorials remain in the denominator
Step 1
Concept
In (\frac{n!}{r!(n-r)!}), changing the order of (r!) and ((n-r)!) does not change the value. In exams order does not matter in a product denominator.
Step 2
Why this answer is correct
The correct answer is A. Denominator में वही दो factorial रहते हैं / The same two factorials remain in the denominator. In (\frac{n!}{r!(n-r)!}), changing the order of (r!) and ((n-r)!) does not change the value. In exams order does not matter in a product denominator.
Step 3
Exam Tip
(\frac{n!}{r!(n-r)!}) में (r!) और ((n-r)!) का क्रम बदलने से मान नहीं बदलता। परीक्षा में multiplication denominator में order matter नहीं करता।
Once (4) selected objects are chosen, the (7) rejected objects are fixed. In exams connect selected-rejected split with combination.
Step 2
Why this answer is correct
The correct answer is A. \(^{11}C_4\). Once (4) selected objects are chosen, the (7) rejected objects are fixed. In exams connect selected-rejected split with combination.
Step 3
Exam Tip
(4) selected चुनते ही (7) rejected तय हो जाते हैं। परीक्षा में selected-rejected split को combination से जोड़ें।
In (n!), internal orders of the unchosen ((n-r)) objects are extra, so they are removed. In exams identify the remaining part in the permutation formula.
Step 2
Why this answer is correct
The correct answer is A. न चुनी गई वस्तुएँ / Unchosen objects. In (n!), internal orders of the unchosen ((n-r)) objects are extra, so they are removed. In exams identify the remaining part in the permutation formula.
Step 3
Exam Tip
(n!) में न चुनी गई ((n-r)) वस्तुओं के अंदरूनी क्रम extra हैं इसलिए उन्हें हटाया जाता है। परीक्षा में permutation formula में remaining part पहचानें।
One selected object has only (1) order. In exams permutation and combination both equal (n) when (r=1).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (1!=1) / Because (1!=1). One selected object has only (1) order. In exams permutation and combination both equal (n) when (r=1).
Step 3
Exam Tip
एक चुनी वस्तु का केवल (1) order होता है। परीक्षा में (r=1) पर permutation और combination दोनों (n) होते हैं।
A. क्योंकि खाली चयन और खाली व्यवस्था दोनों (1) हैं/Because empty selection and empty arrangement are both (1)
Step 1
Concept
For zero objects the count is taken as (1). In exams remember (0!=1) and the empty case.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि खाली चयन और खाली व्यवस्था दोनों (1) हैं / Because empty selection and empty arrangement are both (1). For zero objects the count is taken as (1). In exams remember (0!=1) and the empty case.
Step 3
Exam Tip
शून्य वस्तुओं के लिए count (1) माना जाता है। परीक्षा में (0!=1) और empty case याद रखें।
Combination is obtained by removing (r!) internal orders counted in permutation. In exams divide when going from ordered to unordered.
Step 2
Why this answer is correct
The correct answer is A. \(^{n}C_r=\frac{^{n}P_r}{r!}\). Combination is obtained by removing (r!) internal orders counted in permutation. In exams divide when going from ordered to unordered.
Step 3
Exam Tip
Combination permutation में गिने गए (r!) internal orders को हटाकर मिलता है। परीक्षा में ordered से unordered जाने पर divide करें।
B. न चुनी गई वस्तुओं की संख्या/Number of unchosen objects
Step 1
Concept
After choosing (r) objects (n-r) objects remain. In exams remember this meaning for complement counting.
Step 2
Why this answer is correct
The correct answer is B. न चुनी गई वस्तुओं की संख्या / Number of unchosen objects. After choosing (r) objects (n-r) objects remain. In exams remember this meaning for complement counting.
Step 3
Exam Tip
(r) वस्तुएँ चुनने के बाद (n-r) वस्तुएँ बचती हैं। परीक्षा में complement counting के लिए यह अर्थ याद रखें।
