A. क्योंकि (r) brackets से (x) और बाकी से (1) चुनते हैं/Because (x) is chosen from (r) brackets and (1) from the rest
Step 1
Concept
To form \(x^r\), choose (r) brackets from (n) brackets. In exams connect coefficients with selection.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (r) brackets से (x) और बाकी से (1) चुनते हैं / Because (x) is chosen from (r) brackets and (1) from the rest. To form \(x^r\), choose (r) brackets from (n) brackets. In exams connect coefficients with selection.
Step 3
Exam Tip
\(x^r\) बनाने के लिए (n) brackets में से (r) brackets चुनते हैं। परीक्षा में coefficient को selection से जोड़ें।
B. हर वस्तु को चुनना या न चुनना/Selecting or not selecting each object
Step 1
Concept
The left side adds all selection sizes and the right side gives two choices for each object. In exams remember this identity through subset counting.
Step 2
Why this answer is correct
The correct answer is B. हर वस्तु को चुनना या न चुनना / Selecting or not selecting each object. The left side adds all selection sizes and the right side gives two choices for each object. In exams remember this identity through subset counting.
Step 3
Exam Tip
Left side all possible selection sizes को जोड़ता है और right side हर object के two choices देता है। परीक्षा में subset counting से यह identity याद रखें।
Putting (x=-1) gives ((1-1)^n=0). In exams think of (x=-1) for alternating binomial sums.
Step 2
Why this answer is correct
The correct answer is C. ((1+x)^n) में (x=-1) / (x=-1) in ((1+x)^n). Putting (x=-1) gives ((1-1)^n=0). In exams think of (x=-1) for alternating binomial sums.
Step 3
Exam Tip
(x=-1) रखने पर ((1-1)^n=0) मिलता है। परीक्षा में alternating binomial sum में (x=-1) सोचें।
B. दो groups से कुल (r) चुनने में (k) पहले group से लेना/Taking (k) from the first group while selecting total (r) from two groups
Step 1
Concept
In total (r) selections, the count from the first group varies as (k). In exams connect two-source selection with Vandermonde identity.
Step 2
Why this answer is correct
The correct answer is B. दो groups से कुल (r) चुनने में (k) पहले group से लेना / Taking (k) from the first group while selecting total (r) from two groups. In total (r) selections, the count from the first group varies as (k). In exams connect two-source selection with Vandermonde identity.
Step 3
Exam Tip
कुल (r) selection में first group का count (k) बदलता है। परीक्षा में two-source selection को Vandermonde identity से जोड़ें।
A. पहले एक marked member चुनें और फिर बाकी (r-1) चुनें/First choose one marked member and then choose remaining (r-1)
Step 1
Concept
Choose the marked member in (n) ways and remove overcount of (r) possible marks. In exams understand such identities by member marking.
Step 2
Why this answer is correct
The correct answer is A. पहले एक marked member चुनें और फिर बाकी (r-1) चुनें / First choose one marked member and then choose remaining (r-1). Choose the marked member in (n) ways and remove overcount of (r) possible marks. In exams understand such identities by member marking.
Step 3
Exam Tip
Marked member को (n) ways में चुनकर (r) possible marks के overcount को हटाते हैं। परीक्षा में member marking से ऐसी identities समझें।
B. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{r}{n-r+1}\)
Step 1
Concept
Canceling factorials in consecutive combinations gives \(\frac{r}{n-r+1}\). In exams do not calculate full values in ratio questions.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{r}{n-r+1}\). Canceling factorials in consecutive combinations gives \(\frac{r}{n-r+1}\). In exams do not calculate full values in ratio questions.
Step 3
Exam Tip
Consecutive combinations में factorial cancel करने पर \(\frac{r}{n-r+1}\) मिलता है। परीक्षा में ratio questions में full values न निकालें।
A. \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}\)
Step 1
Concept
Near the largest term, the consecutive ratio changes around (1). In exams check the transition from increasing to decreasing.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}\). Near the largest term, the consecutive ratio changes around (1). In exams check the transition from increasing to decreasing.
Step 3
Exam Tip
Largest term के पास consecutive ratio (1) के आसपास बदलता है। परीक्षा में increasing और decreasing transition देखें।
In stars and bars, (n) stars and (r-1) bars are arranged. In exams use the bars method for identical distribution.
Step 2
Why this answer is correct
The correct answer is C. \({}^{n+r-1}C_{r-1}\). In stars and bars, (n) stars and (r-1) bars are arranged. In exams use the bars method for identical distribution.
Step 3
Exam Tip
Stars and bars में (n) stars और (r-1) bars arrange होते हैं। परीक्षा में identical distribution में bars method लगाएं।
Give (1) ball to each box first, then distribute the remaining (n-r) balls. In exams allot the minimum first for non-empty conditions.
Step 2
Why this answer is correct
The correct answer is A. \({}^{n-1}C_{r-1}\). Give (1) ball to each box first, then distribute the remaining (n-r) balls. In exams allot the minimum first for non-empty conditions.
Step 3
Exam Tip
हर box को पहले (1) ball दें, फिर बाकी (n-r) balls distribute करें। परीक्षा में non-empty condition में पहले minimum allot करें।
Choosing seats first and then arranging students gives \(^{10}C_4\cdot4!=^{10}P_4\). In exams count both positions and arrangements.
Step 2
Why this answer is correct
The correct answer is B. \(^{10}P_4\). Choosing seats first and then arranging students gives \(^{10}C_4\cdot4!=^{10}P_4\). In exams count both positions and arrangements.
Step 3
Exam Tip
पहले seats चुनना और फिर students arrange करना \(^{10}C_4\cdot4!=^{10}P_4\) है। परीक्षा में seat-position और arrangement दोनों गिनें।
A. पहले group चुनते हैं फिर उस group को order देते हैं/First choose the group then order that group
Step 1
Concept
Order of selected people is meaningful in a queue. In exams derive ordered selection as combination times factorial.
Step 2
Why this answer is correct
The correct answer is A. पहले group चुनते हैं फिर उस group को order देते हैं / First choose the group then order that group. Order of selected people is meaningful in a queue. In exams derive ordered selection as combination times factorial.
Step 3
Exam Tip
Queue में selected people की order meaningful होती है। परीक्षा में ordered selection को combination times factorial से derive करें।
For labelled groups, group names are fixed so only internal order is removed. In exams use factorial denominators for fixed group sizes.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{8!}{3!3!2!}\). For labelled groups, group names are fixed so only internal order is removed. In exams use factorial denominators for fixed group sizes.
Step 3
Exam Tip
Labelled groups में group names fixed होते हैं इसलिए सिर्फ internal order हटता है। परीक्षा में fixed group sizes के factorial denominator लगाएं।
Two groups have equal size (3), so interchanging those groups gives the same distribution. In exams divide extra by factorial of equal-sized unlabelled groups.
Step 2
Why this answer is correct
The correct answer is A. (2!). Two groups have equal size (3), so interchanging those groups gives the same distribution. In exams divide extra by factorial of equal-sized unlabelled groups.
Step 3
Exam Tip
दो groups का size (3) समान है इसलिए उन groups की अदला-बदली same distribution देती है। परीक्षा में equal-sized unlabelled groups के factorial से extra divide करें।
Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and distinct boxes.
Step 2
Why this answer is correct
The correct answer is C. \(r^n\). Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and distinct boxes.
Step 3
Exam Tip
हर distinct object के लिए (r) independent choices हैं। परीक्षा में distinct objects और distinct boxes में power rule लगाएं।
Inclusion-exclusion is used to exclude empty boxes. In exams think of removing empty boxes when the word onto appears.
Step 2
Why this answer is correct
The correct answer is A. (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n). Inclusion-exclusion is used to exclude empty boxes. In exams think of removing empty boxes when the word onto appears.
Step 3
Exam Tip
Empty boxes को exclude करने के लिए inclusion-exclusion used होता है। परीक्षा में onto शब्द दिखे तो खाली boxes घटाने की सोचें।
A. क्योंकि हर person के लिए (r) room choices independent हैं/Because each person has (r) independent room choices
Step 1
Concept
Independent choices of distinct people multiply. In exams connect distinct-person box problems with powers.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि हर person के लिए (r) room choices independent हैं / Because each person has (r) independent room choices. Independent choices of distinct people multiply. In exams connect distinct-person box problems with powers.
Step 3
Exam Tip
Distinct persons के independent choices multiply होते हैं। परीक्षा में व्यक्ति distinct हों तो balls-and-boxes को power से जोड़ें।
B. किसी बच्चे को zero coin मिल सकता है/A child may get zero coins
Step 1
Concept
When zero is allowed, place (n) stars and (r-1) bars. In exams treat \({}^{n+r-1}C_n\) and \({}^{n+r-1}C_{r-1}\) as the same.
Step 2
Why this answer is correct
The correct answer is B. किसी बच्चे को zero coin मिल सकता है / A child may get zero coins. When zero is allowed, place (n) stars and (r-1) bars. In exams treat \({}^{n+r-1}C_n\) and \({}^{n+r-1}C_{r-1}\) as the same.
Step 3
Exam Tip
Zero allowed हो तो (n) stars और (r-1) bars रखे जाते हैं। परीक्षा में \({}^{n+r-1}C_n\) और \({}^{n+r-1}C_{r-1}\) same मानें।
Non-negative solutions by stars and bars are \({}^{12+3-1}C_{3-1}\). In exams convert equation solutions into distribution problems.
Step 2
Why this answer is correct
The correct answer is B. \({}^{14}C_2\). Non-negative solutions by stars and bars are \({}^{12+3-1}C_{3-1}\). In exams convert equation solutions into distribution problems.
Step 3
Exam Tip
Non-negative solutions stars and bars से \({}^{12+3-1}C_{3-1}\) होते हैं। परीक्षा में equation solutions को distribution problem बनाएं।
Give (1) to each variable first, then (11) remain. In exams subtract the number of variables for positive solutions.
Step 2
Why this answer is correct
The correct answer is C. \({}^{11}C_3\). Give (1) to each variable first, then (11) remain. In exams subtract the number of variables for positive solutions.
Step 3
Exam Tip
पहले हर variable को (1) दें, फिर (11) बचते हैं। परीक्षा में positive solutions में total से variables की संख्या घटाएं।
A. \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) जहां special objects (a) हैं/\(^{a}C_s\cdot{}^{n-a}C_{r-s}\) where there are (a) special objects
Step 1
Concept
Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.
Step 2
Why this answer is correct
The correct answer is A. \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) जहां special objects (a) हैं / \(^{a}C_s\cdot{}^{n-a}C_{r-s}\) where there are (a) special objects. Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.
Step 3
Exam Tip
Special group से (s) और non-special group से (r-s) objects चुनते हैं। परीक्षा में exactly condition को product of choices में तोड़ें।
The complement of at least one special is no special. In exams total minus none is fast for at least conditions.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}C_r-{}^{n-a}C_r\). The complement of at least one special is no special. In exams total minus none is fast for at least conditions.
Step 3
Exam Tip
At least one special का complement no special है। परीक्षा में at least के लिए total minus none तेज होता है।
Case (1): take both fixed books, case (2): take neither. In exams keep cases clear for paired restrictions.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_2+{}^{8}C_4\). Case (1): take both fixed books, case (2): take neither. In exams keep cases clear for paired restrictions.
Step 3
Exam Tip
Case (1): दोनों fixed लें, case (2): दोनों fixed न लें। परीक्षा में paired restrictions में cases साफ रखें।
Subtract selections containing both particular people from total selections. In exams solve not-together selection using complement.
Step 2
Why this answer is correct
The correct answer is A. \(^{9}C_5-{}^{7}C_3\). Subtract selections containing both particular people from total selections. In exams solve not-together selection using complement.
Step 3
Exam Tip
Total selections से दोनों particular people वाले selections घटाएं। परीक्षा में not together in selection को complement से हल करें।
Subtract block arrangements where (A) and (B) are together from total arrangements. In exams complement is easy for not-together permutations.
Step 2
Why this answer is correct
The correct answer is A. \(8!-7!\cdot2!\). Subtract block arrangements where (A) and (B) are together from total arrangements. In exams complement is easy for not-together permutations.
Step 3
Exam Tip
Total arrangements से (A) और (B) together block arrangements घटते हैं। परीक्षा में not together permutation में complement आसान है।
B. पहले boys arrange करें और फिर (7) gaps में (4) girls रखें/Arrange boys first and place (4) girls in (7) gaps
Step 1
Concept
Arranging boys creates (7) gaps where girls can be placed. In exams use the gap method for no two together.
Step 2
Why this answer is correct
The correct answer is B. पहले boys arrange करें और फिर (7) gaps में (4) girls रखें / Arrange boys first and place (4) girls in (7) gaps. Arranging boys creates (7) gaps where girls can be placed. In exams use the gap method for no two together.
Step 3
Exam Tip
Boys arrange करने पर (7) gaps बनते हैं जिनमें girls बैठती हैं। परीक्षा में no two together के लिए gap method लगाएं।
After arranging (7) consonants, (8) gaps are formed and (5) vowels are arranged. In exams choose gaps and permute vowels.
Step 2
Why this answer is correct
The correct answer is A. \(^{8}C_5\cdot5!\). After arranging (7) consonants, (8) gaps are formed and (5) vowels are arranged. In exams choose gaps and permute vowels.
Step 3
Exam Tip
(7) consonants के arrangement के बाद (8) gaps बनते हैं और (5) vowels arrange होते हैं। परीक्षा में gaps चुनकर vowels permute करें।
A. (n) objects में (p) एक प्रकार के और (q) दूसरे प्रकार के identical हों/Among (n) objects, (p) of one type and (q) of another type are identical
Step 1
Concept
Internal permutations of same-type objects do not create new arrangements. In exams derive repeated-letter counts using factorial division.
Step 2
Why this answer is correct
The correct answer is A. (n) objects में (p) एक प्रकार के और (q) दूसरे प्रकार के identical हों / Among (n) objects, (p) of one type and (q) of another type are identical. Internal permutations of same-type objects do not create new arrangements. In exams derive repeated-letter counts using factorial division.
Step 3
Exam Tip
Same-type objects की internal permutations नई arrangement नहीं देतीं। परीक्षा में repeated letters की count को factorial division से derive करें।
A. क्योंकि (A) तीन बार और (N) दो बार आता है/Because (A) appears three times and (N) appears twice
Step 1
Concept
Interchanging identical letters gives the same word. In exams make repeated counts into denominator factorials.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (A) तीन बार और (N) दो बार आता है / Because (A) appears three times and (N) appears twice. Interchanging identical letters gives the same word. In exams make repeated counts into denominator factorials.
Step 3
Exam Tip
समान letters की अदला-बदली same word देती है। परीक्षा में repeated counts को denominator factorials बनाएं।
A. Rotation और reflection दोनों same माने जाते हैं/Both rotation and reflection are considered the same
Step 1
Concept
After removing circular duplicates, mirror images are also the same. In exams divide by (2) for reflection in bracelet problems.
Step 2
Why this answer is correct
The correct answer is A. Rotation और reflection दोनों same माने जाते हैं / Both rotation and reflection are considered the same. After removing circular duplicates, mirror images are also the same. In exams divide by (2) for reflection in bracelet problems.
Step 3
Exam Tip
Circular duplicates हटाने के बाद mirror images भी same हैं। परीक्षा में bracelet में necklace की तरह reflection by (2) divide करें।
A. Circular seating में rotations same मानी जाती हैं/Rotations are considered the same in circular seating
Step 1
Concept
In a circle, fixing one person removes rotational overcount. In exams use the one-fixed method for round tables.
Step 2
Why this answer is correct
The correct answer is A. Circular seating में rotations same मानी जाती हैं / Rotations are considered the same in circular seating. In a circle, fixing one person removes rotational overcount. In exams use the one-fixed method for round tables.
Step 3
Exam Tip
Circle में एक व्यक्ति को fixed मानकर rotational overcount हटता है। परीक्षा में round table में one fixed method अपनाएं।
Treating (A) and (B) as one block gives (n-1) circular objects. In exams remember ((n-2)!) in circular block counts.
Step 2
Why this answer is correct
The correct answer is A. (2!(n-2)!). Treating (A) and (B) as one block gives (n-1) circular objects. In exams remember ((n-2)!) in circular block counts.
Step 3
Exam Tip
(A) और (B) को one block मानने पर circular objects (n-1) होते हैं। परीक्षा में circular block count में ((n-2)!) याद रखें।
Positions are ordered and objects are not repeated, so falling choices arise. In exams think of permutations when positions are distinct.
Step 2
Why this answer is correct
The correct answer is B. \(^{n}P_r\). Positions are ordered and objects are not repeated, so falling choices arise. In exams think of permutations when positions are distinct.
Step 3
Exam Tip
Positions ordered होते हैं और objects repeat नहीं होते, इसलिए falling choices मिलती हैं। परीक्षा में positions distinct हों तो permutation सोचें।
A. हर position पर (6) choices independent हैं/Each position has (6) independent choices
Step 1
Concept
Choices do not decrease when repetition is allowed. In exams use the power formula when repetition is allowed.
Step 2
Why this answer is correct
The correct answer is A. हर position पर (6) choices independent हैं / Each position has (6) independent choices. Choices do not decrease when repetition is allowed. In exams use the power formula when repetition is allowed.
Step 3
Exam Tip
Repetition allowed होने से choices कम नहीं होतीं। परीक्षा में allowed repetition में power formula use करें।
A. क्योंकि first digit (0) होने पर number (4)-digit नहीं रहेगा/Because if the first digit is (0), the number will not remain (4)-digit
Step 1
Concept
A leading zero does not form a valid (4)-digit number. In exams check the first-place condition first in digit arrangements.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि first digit (0) होने पर number (4)-digit नहीं रहेगा / Because if the first digit is (0), the number will not remain (4)-digit. A leading zero does not form a valid (4)-digit number. In exams check the first-place condition first in digit arrangements.
Step 3
Exam Tip
Leading zero valid (4)-digit number नहीं बनाता। परीक्षा में digit arrangement में first place condition पहले देखें।
A. जब repetition not allowed हो और order important हो/When repetition is not allowed and order is important
Step 1
Concept
Without repetition, choices decrease, so \(^{n}P_r\) is needed. In exams do not apply powers without reading the repetition condition.
Step 2
Why this answer is correct
The correct answer is A. जब repetition not allowed हो और order important हो / When repetition is not allowed and order is important. Without repetition, choices decrease, so \(^{n}P_r\) is needed. In exams do not apply powers without reading the repetition condition.
Step 3
Exam Tip
Without repetition में choices घटती हैं, इसलिए \(^{n}P_r\) चाहिए। परीक्षा में repetition condition पढ़े बिना power न लगाएं।
Each distinct prize has (10) independent student choices. In exams use the power rule for distinct prizes with repetition allowed.
Step 2
Why this answer is correct
The correct answer is B. \(10^4\). Each distinct prize has (10) independent student choices. In exams use the power rule for distinct prizes with repetition allowed.
Step 3
Exam Tip
हर different prize के लिए (10) independent student choices हैं। परीक्षा में distinct prizes with repetition allowed को power rule से करें।
Prizes are different and recipients cannot repeat, so it is an ordered assignment. In exams treat distinct prizes without repetition as permutation.
Step 2
Why this answer is correct
The correct answer is C. \(^{10}P_4\). Prizes are different and recipients cannot repeat, so it is an ordered assignment. In exams treat distinct prizes without repetition as permutation.
Step 3
Exam Tip
Prizes different हैं और recipients repeat नहीं हो सकते इसलिए ordered assignment है। परीक्षा में distinct prizes without repetition को permutation समझें।
Identical prize distribution is stars and bars, with (4) stars and (9) bars. In exams connect identical prizes with combinations with repetition.
Step 2
Why this answer is correct
The correct answer is A. \({}^{13}C_9\). Identical prize distribution is stars and bars, with (4) stars and (9) bars. In exams connect identical prizes with combinations with repetition.
Step 3
Exam Tip
Identical prizes distribution stars and bars है, (4) stars और (9) bars। परीक्षा में identical prizes को combinations with repetition से जोड़ें।
Prizes are identical and only (5) students need to be chosen. In exams treat at most one with identical prizes as simple selection.
Step 2
Why this answer is correct
The correct answer is C. \(^{8}C_5\). Prizes are identical and only (5) students need to be chosen. In exams treat at most one with identical prizes as simple selection.
Step 3
Exam Tip
Prizes identical हैं और सिर्फ (5) students choose करने हैं। परीक्षा में at most one with identical prizes को simple selection मानें।
A. क्योंकि consecutive ratio पहले (1) से बड़ा और बाद में (1) से छोटा होता है/Because the consecutive ratio is first greater than (1) and later less than (1)
Step 1
Concept
The ratio \(\frac{n-r}{r+1}\) shows the transition. In exams check the trend of binomial coefficients by ratios.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि consecutive ratio पहले (1) से बड़ा और बाद में (1) से छोटा होता है / Because the consecutive ratio is first greater than (1) and later less than (1). The ratio \(\frac{n-r}{r+1}\) shows the transition. In exams check the trend of binomial coefficients by ratios.
Step 3
Exam Tip
Ratio \(\frac{n-r}{r+1}\) transition दिखाता है। परीक्षा में binomial coefficient trend ratio से check करें।
For even (n), binomial coefficients peak at the middle. In exams identify the middle term using symmetry and ratio.
Step 2
Why this answer is correct
The correct answer is A. \({}^{n}C_{\frac{n}{2}}\). For even (n), binomial coefficients peak at the middle. In exams identify the middle term using symmetry and ratio.
Step 3
Exam Tip
Even (n) में binomial coefficients middle पर peak करते हैं। परीक्षा में symmetry और ratio दोनों से middle term पहचानें।
B. \({}^{n}C_{\frac{n-1}{2}}\) और \({}^{n}C_{\frac{n+1}{2}}\)/\({}^{n}C_{\frac{n-1}{2}}\) and \({}^{n}C_{\frac{n+1}{2}}\)
Step 1
Concept
For odd (n), the two middle indices are complementary and equal. In exams remember two largest terms for the odd case.
Step 2
Why this answer is correct
The correct answer is B. \({}^{n}C_{\frac{n-1}{2}}\) और \({}^{n}C_{\frac{n+1}{2}}\) / \({}^{n}C_{\frac{n-1}{2}}\) and \({}^{n}C_{\frac{n+1}{2}}\). For odd (n), the two middle indices are complementary and equal. In exams remember two largest terms for the odd case.
Step 3
Exam Tip
Odd (n) में two middle indices complementary और equal होते हैं। परीक्षा में odd case में दो largest terms याद रखें।
In combination symmetry, unequal equal-values come from complementary indices. In exams check sum (n) for equal combinations.
Step 2
Why this answer is correct
The correct answer is A. (r+s=n). In combination symmetry, unequal equal-values come from complementary indices. In exams check sum (n) for equal combinations.
Step 3
Exam Tip
Combination symmetry में unequal equal-values complementary indices से आते हैं। परीक्षा में equal combinations में sum (n) check करें।
A. क्योंकि \(^{n}P_r\) usually (r) बढ़ने पर extra positive factors से बदलता है/Because \(^{n}P_r\) usually changes by extra positive factors as (r) increases
Step 1
Concept
Permutations do not have complement symmetry, and changing length changes the count. In exams do not apply combination symmetry to permutations.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(^{n}P_r\) usually (r) बढ़ने पर extra positive factors से बदलता है / Because \(^{n}P_r\) usually changes by extra positive factors as (r) increases. Permutations do not have complement symmetry, and changing length changes the count. In exams do not apply combination symmetry to permutations.
Step 3
Exam Tip
Permutation में complement symmetry नहीं होती और length बदलने से count बदलता है। परीक्षा में combination symmetry को permutation पर न लगाएं।
A. Permutation count को (r!) से divide करके combination मिलता है/Combination is obtained by dividing permutation count by (r!)
Step 1
Concept
The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.
Step 2
Why this answer is correct
The correct answer is A. Permutation count को (r!) से divide करके combination मिलता है / Combination is obtained by dividing permutation count by (r!). The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.
Step 3
Exam Tip
Numerator \(^{n}P_r\) है और denominator order हटाता है। परीक्षा में इस form से numerical cancellation जल्दी करें।
A. क्योंकि \(r!\geq1\) होता है/Because \(r!\geq1\)
Step 1
Concept
Permutation counts all orders of each selected group. In exams equality is also possible for (r=0) or (r=1).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(r!\geq1\) होता है / Because \(r!\geq1\). Permutation counts all orders of each selected group. In exams equality is also possible for (r=0) or (r=1).
Step 3
Exam Tip
Permutation हर selected group के all orders गिनता है। परीक्षा में (r=0) या (r=1) पर equality भी संभव है।
Adjacent lower indices (r) and (r+1) give the (r+1) term of the next row. In exams identify Pascal pattern quickly.
Step 2
Why this answer is correct
The correct answer is A. \({}^{n+1}C_{r+1}\). Adjacent lower indices (r) and (r+1) give the (r+1) term of the next row. In exams identify Pascal pattern quickly.
Step 3
Exam Tip
Adjacent lower indices (r) और (r+1) मिलकर next row का (r+1) term देते हैं। परीक्षा में Pascal pattern तुरंत पहचानें।
By Pascal identity, the same upper index with adjacent lower indices combines. In exams first combine the adjacent pair.
Step 2
Why this answer is correct
The correct answer is A. \({}^{12}C_4+{}^{12}C_5={}^{13}C_5\). By Pascal identity, the same upper index with adjacent lower indices combines. In exams first combine the adjacent pair.
Step 3
Exam Tip
Pascal identity से same upper index और adjacent lower indices combine होते हैं। परीक्षा में पहले adjacent pair को combine करें।
A. ((1+1)^n) और ((1-1)^n) को जोड़ने से/By adding ((1+1)^n) and ((1-1)^n)
Step 1
Concept
To separate even-index coefficients, two binomial substitutions are added. In exams use (x=1) and (x=-1) for even-odd sums.
Step 2
Why this answer is correct
The correct answer is A. ((1+1)^n) और ((1-1)^n) को जोड़ने से / By adding ((1+1)^n) and ((1-1)^n). To separate even-index coefficients, two binomial substitutions are added. In exams use (x=1) and (x=-1) for even-odd sums.
Step 3
Exam Tip
Even-index coefficients को अलग करने के लिए two binomial substitutions जोड़े जाते हैं। परीक्षा में even-odd sums के लिए (x=1) और (x=-1) use करें।
