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Class 11 Mathematics - Permutations and Combinations - Derivations of formulas and their connections Hard Quiz

Level 65 • 50/50 questions • 30 seconds per question.

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((1+x)^n) में \(x^r\) का coefficient \(^{n}C_r\) क्यों होता है?

Why is the coefficient of \(x^r\) in ((1+x)^n) equal to \(^{n}C_r\)?

Explanation opens after your attempt
Correct Answer

A. क्योंकि (r) brackets से (x) और बाकी से (1) चुनते हैंBecause (x) is chosen from (r) brackets and (1) from the rest

Explanation

Simple Explanation

\(x^r\) बनाने के लिए (n) brackets में से (r) brackets चुनते हैं। परीक्षा में coefficient को selection से जोड़ें। / To form \(x^r\), choose (r) brackets from (n) brackets. In exams connect coefficients with selection.

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\(\sum_{r=0}^{n} {}^{n}C_r=2^n\) किस combinatorial idea से निकला है?

The identity \(\sum_{r=0}^{n} {}^{n}C_r=2^n\) is derived from which combinatorial idea?

Explanation opens after your attempt
Correct Answer

B. हर वस्तु को चुनना या न चुननाSelecting or not selecting each object

Explanation

Simple Explanation

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। / 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.

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(\sum_{r=0}^{n} (-1)^r{}^{n}C_r=0) किस substitution से जुड़ा है?

The identity (\sum_{r=0}^{n} (-1)^r{}^{n}C_r=0) is connected with which substitution?

Explanation opens after your attempt
Correct Answer

C. ((1+x)^n) में (x=-1)(x=-1) in ((1+x)^n)

Explanation

Simple Explanation

(x=-1) रखने पर ((1-1)^n=0) मिलता है। परीक्षा में alternating binomial sum में (x=-1) सोचें। / Putting (x=-1) gives ((1-1)^n=0). In exams think of (x=-1) for alternating binomial sums.

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\({}^{m+n}C_r=\sum_{k=0}^{r}{}^{m}C_k{}^{n}C_{r-k}\) किस counting split से सिद्ध होता है?

The identity \({}^{m+n}C_r=\sum_{k=0}^{r}{}^{m}C_k{}^{n}C_{r-k}\) is proved by which counting split?

Explanation opens after your attempt
Correct Answer

B. दो groups से कुल (r) चुनने में (k) पहले group से लेनाTaking (k) from the first group while selecting total (r) from two groups

Explanation

Simple Explanation

कुल (r) selection में first group का count (k) बदलता है। परीक्षा में two-source selection को Vandermonde identity से जोड़ें। / In total (r) selections, the count from the first group varies as (k). In exams connect two-source selection with Vandermonde identity.

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\({}^{n}C_r=\frac{n}{r}{}^{n-1}C_{r-1}\) की counting व्याख्या क्या है?

What is the counting interpretation of \({}^{n}C_r=\frac{n}{r}{}^{n-1}C_{r-1}\)?

Explanation opens after your attempt
Correct Answer

A. पहले एक marked member चुनें और फिर बाकी (r-1) चुनेंFirst choose one marked member and then choose remaining (r-1)

Explanation

Simple Explanation

Marked member को (n) ways में चुनकर (r) possible marks के overcount को हटाते हैं। परीक्षा में member marking से ऐसी identities समझें। / Choose the marked member in (n) ways and remove overcount of (r) possible marks. In exams understand such identities by member marking.

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यदि \(^{n}C_{r-1}:{}^{n}C_r=3:5\), तो ratio derivation में कौन-सा expression उपयोगी है?

If \(^{n}C_{r-1}:{}^{n}C_r=3:5\), which expression is useful in the ratio derivation?

Explanation opens after your attempt
Correct Answer

B. \(\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{r}{n-r+1}\)

Explanation

Simple Explanation

Consecutive combinations में factorial cancel करने पर \(\frac{r}{n-r+1}\) मिलता है। परीक्षा में ratio questions में full values न निकालें। / Canceling factorials in consecutive combinations gives \(\frac{r}{n-r+1}\). In exams do not calculate full values in ratio questions.

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\({}^{n}C_r\) के largest term को पहचानने के लिए कौन-सा relation सबसे उपयोगी है?

Which relation is most useful for identifying the largest term of \({}^{n}C_r\)?

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Correct Answer

A. \(\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1}\)

Explanation

Simple Explanation

Largest term के पास consecutive ratio (1) के आसपास बदलता है। परीक्षा में increasing और decreasing transition देखें। / Near the largest term, the consecutive ratio changes around (1). In exams check the transition from increasing to decreasing.

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(n) identical balls को (r) distinct boxes में खाली box allowed होने पर बांटने का सूत्र क्या है?

What is the formula for distributing (n) identical balls into (r) distinct boxes when empty boxes are allowed?

Explanation opens after your attempt
Correct Answer

C. \({}^{n+r-1}C_{r-1}\)

Explanation

Simple Explanation

Stars and bars में (n) stars और (r-1) bars arrange होते हैं। परीक्षा में identical distribution में bars method लगाएं। / In stars and bars, (n) stars and (r-1) bars are arranged. In exams use the bars method for identical distribution.

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(n) identical balls को (r) distinct boxes में हर box non-empty हो तो formula कैसे बदलता है?

How does the formula change if (n) identical balls are distributed into (r) distinct boxes with every box non-empty?

Explanation opens after your attempt
Correct Answer

A. \({}^{n-1}C_{r-1}\)

Explanation

Simple Explanation

हर box को पहले (1) ball दें, फिर बाकी (n-r) balls distribute करें। परीक्षा में non-empty condition में पहले minimum allot करें। / Give (1) ball to each box first, then distribute the remaining (n-r) balls. In exams allot the minimum first for non-empty conditions.

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\({}^{n+r-1}C_r\) और \({}^{n+r-1}C_{n-1}\) का equality किस identity से आती है?

The equality of \({}^{n+r-1}C_r\) and \({}^{n+r-1}C_{n-1}\) comes from which identity?

Explanation opens after your attempt
Correct Answer

B. Complement identity

Explanation

Simple Explanation

दोनों lower indices का sum (n+r-1) है। परीक्षा में stars and bars answers में complementary forms accept करें। / The two lower indices sum to (n+r-1). In exams accept complementary forms in stars and bars answers.

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(10) seats की row में (4) selected students को बैठाने और बाकी seats empty छोड़ने की count किस expression से जुड़ती है?

The count for seating (4) selected students in a row of (10) seats while leaving other seats empty is connected with which expression?

Explanation opens after your attempt
Correct Answer

B. \(^{10}P_4\)

Explanation

Simple Explanation

पहले seats चुनना और फिर students arrange करना \(^{10}C_4\cdot4!=^{10}P_4\) है। परीक्षा में seat-position और arrangement दोनों गिनें। / Choosing seats first and then arranging students gives \(^{10}C_4\cdot4!=^{10}P_4\). In exams count both positions and arrangements.

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(7) people में से (3) को क्रम वाली queue में चुनने की count \(^{7}C_3\cdot3!\) क्यों है?

Why is the count for choosing (3) people from (7) into an ordered queue equal to \(^{7}C_3\cdot3!\)?

Explanation opens after your attempt
Correct Answer

A. पहले group चुनते हैं फिर उस group को order देते हैंFirst choose the group then order that group

Explanation

Simple Explanation

Queue में selected people की order meaningful होती है। परीक्षा में ordered selection को combination times factorial से derive करें। / Order of selected people is meaningful in a queue. In exams derive ordered selection as combination times factorial.

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(8) अलग objects को (3), (3), और (2) के labelled groups में बांटने का formula कौन-सा है?

What is the formula for dividing (8) distinct objects into labelled groups of (3), (3), and (2)?

Explanation opens after your attempt
Correct Answer

A. \(\frac{8!}{3!3!2!}\)

Explanation

Simple Explanation

Labelled groups में group names fixed होते हैं इसलिए सिर्फ internal order हटता है। परीक्षा में fixed group sizes के factorial denominator लगाएं। / For labelled groups, group names are fixed so only internal order is removed. In exams use factorial denominators for fixed group sizes.

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(8) अलग objects को sizes (3), (3), और (2) के unlabelled groups में बांटते समय अतिरिक्त division किससे होगा?

When (8) distinct objects are divided into unlabelled groups of sizes (3), (3), and (2), by what extra factor do we divide?

Explanation opens after your attempt
Correct Answer

A. (2!)

Explanation

Simple Explanation

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। / 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.

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किस formula से (n) distinct objects को (r) distinct boxes में distribute किया जाता है जब हर object किसी एक box में जा सकता है?

Which formula distributes (n) distinct objects into (r) distinct boxes when each object can go into one box?

Explanation opens after your attempt
Correct Answer

C. \(r^n\)

Explanation

Simple Explanation

हर distinct object के लिए (r) independent choices हैं। परीक्षा में distinct objects और distinct boxes में power rule लगाएं। / Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and distinct boxes.

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(n) distinct objects को (r) distinct boxes में onto distribution गिनने के लिए inclusion-exclusion का रूप कौन-सा है?

Which inclusion-exclusion form counts onto distributions of (n) distinct objects into (r) distinct boxes?

Explanation opens after your attempt
Correct Answer

A. (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n)

Explanation

Simple Explanation

Empty boxes को exclude करने के लिए inclusion-exclusion used होता है। परीक्षा में onto शब्द दिखे तो खाली boxes घटाने की सोचें। / Inclusion-exclusion is used to exclude empty boxes. In exams think of removing empty boxes when the word onto appears.

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(n) लोगों को (r) distinct rooms में रखना है और rooms empty हो सकते हैं। Count \(r^n\) क्यों है?

(n) people are to be placed into (r) distinct rooms and rooms may be empty. Why is the count \(r^n\)?

Explanation opens after your attempt
Correct Answer

A. क्योंकि हर person के लिए (r) room choices independent हैंBecause each person has (r) independent room choices

Explanation

Simple Explanation

Distinct persons के independent choices multiply होते हैं। परीक्षा में व्यक्ति distinct हों तो balls-and-boxes को power से जोड़ें। / Independent choices of distinct people multiply. In exams connect distinct-person box problems with powers.

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(n) identical coins को (r) बच्चों में बांटने में कौन-सी condition formula \({}^{n+r-1}C_{n}\) देती है?

Which condition gives the formula \({}^{n+r-1}C_{n}\) for distributing (n) identical coins among (r) children?

Explanation opens after your attempt
Correct Answer

B. किसी बच्चे को zero coin मिल सकता हैA child may get zero coins

Explanation

Simple Explanation

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। / 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.

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\(x_1+x_2+x_3=12\) के non-negative integer solutions की count कौन-सी है?

What is the count of non-negative integer solutions of \(x_1+x_2+x_3=12\)?

Explanation opens after your attempt
Correct Answer

B. \({}^{14}C_2\)

Explanation

Simple Explanation

Non-negative solutions stars and bars से \({}^{12+3-1}C_{3-1}\) होते हैं। परीक्षा में equation solutions को distribution problem बनाएं। / Non-negative solutions by stars and bars are \({}^{12+3-1}C_{3-1}\). In exams convert equation solutions into distribution problems.

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\(x_1+x_2+x_3+x_4=15\) में सभी \(x_i\geq1\) हों तो count कौन-सी है?

If all \(x_i\geq1\) in \(x_1+x_2+x_3+x_4=15\), what is the count?

Explanation opens after your attempt
Correct Answer

C. \({}^{11}C_3\)

Explanation

Simple Explanation

पहले हर variable को (1) दें, फिर (11) बचते हैं। परीक्षा में positive solutions में total से variables की संख्या घटाएं। / Give (1) to each variable first, then (11) remain. In exams subtract the number of variables for positive solutions.

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(n) objects में से (r) objects चुनने में exactly (s) special objects लेने का सूत्र क्या है?

What is the formula for choosing (r) objects from (n) objects with exactly (s) special objects?

Explanation opens after your attempt
Correct Answer

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

Explanation

Simple Explanation

Special group से (s) और non-special group से (r-s) objects चुनते हैं। परीक्षा में exactly condition को product of choices में तोड़ें। / Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.

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(n) objects में से (r) चुनते समय कम से कम (1) special object हो और special objects (a) हों, तो shortest expression कौन-सा है?

When choosing (r) objects from (n), if at least (1) special object is required and there are (a) special objects, what is the shortest expression?

Explanation opens after your attempt
Correct Answer

B. \(^{n}C_r-{}^{n-a}C_r\)

Explanation

Simple Explanation

At least one special का complement no special है। परीक्षा में at least के लिए total minus none तेज होता है। / The complement of at least one special is no special. In exams total minus none is fast for at least conditions.

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(10) books में से (4) books चुननी हैं लेकिन (2) fixed books साथ में या तो दोनों आएं या दोनों न आएं। Count कौन-सी है?

From (10) books, (4) books are to be selected, and (2) fixed books must either both appear or both not appear. What is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{8}C_2+{}^{8}C_4\)

Explanation

Simple Explanation

Case (1): दोनों fixed लें, case (2): दोनों fixed न लें। परीक्षा में paired restrictions में cases साफ रखें। / Case (1): take both fixed books, case (2): take neither. In exams keep cases clear for paired restrictions.

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(9) people में से (5) चुनने हैं और दो particular people साथ-साथ चयनित नहीं हो सकते। Count कौन-सी है?

Choose (5) people from (9) people, but two particular people cannot be selected together. What is the count?

Explanation opens after your attempt
Correct Answer

A. \(^{9}C_5-{}^{7}C_3\)

Explanation

Simple Explanation

Total selections से दोनों particular people वाले selections घटाएं। परीक्षा में not together in selection को complement से हल करें। / Subtract selections containing both particular people from total selections. In exams solve not-together selection using complement.

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(8) people को row में arrange करना है और (A) और (B) together न हों। Count का formula कौन-सा है?

Arrange (8) people in a row so that (A) and (B) are not together. Which formula gives the count?

Explanation opens after your attempt
Correct Answer

A. \(8!-7!\cdot2!\)

Explanation

Simple Explanation

Total arrangements से (A) और (B) together block arrangements घटते हैं। परीक्षा में not together permutation में complement आसान है। / Subtract block arrangements where (A) and (B) are together from total arrangements. In exams complement is easy for not-together permutations.

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(6) boys और (4) girls को row में arrange करना है ताकि no two girls together हों। Derivation का मुख्य step क्या है?

Arrange (6) boys and (4) girls in a row so that no two girls are together. What is the main step in the derivation?

Explanation opens after your attempt
Correct Answer

B. पहले boys arrange करें और फिर (7) gaps में (4) girls रखेंArrange boys first and place (4) girls in (7) gaps

Explanation

Simple Explanation

Boys arrange करने पर (7) gaps बनते हैं जिनमें girls बैठती हैं। परीक्षा में no two together के लिए gap method लगाएं। / Arranging boys creates (7) gaps where girls can be placed. In exams use the gap method for no two together.

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(5) vowels और (7) consonants से word बनाना है जिसमें no two vowels together हों। Vowel placement का factor क्या होगा?

A word is formed from (5) vowels and (7) consonants with no two vowels together. What is the vowel placement factor?

Explanation opens after your attempt
Correct Answer

A. \(^{8}C_5\cdot5!\)

Explanation

Simple Explanation

(7) consonants के arrangement के बाद (8) gaps बनते हैं और (5) vowels arrange होते हैं। परीक्षा में gaps चुनकर vowels permute करें। / After arranging (7) consonants, (8) gaps are formed and (5) vowels are arranged. In exams choose gaps and permute vowels.

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किस condition में repeated objects arrangement का सूत्र \(\frac{n!}{p!q!}\) लागू होगा?

Under which condition does the repeated-object arrangement formula \(\frac{n!}{p!q!}\) apply?

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Correct Answer

A. (n) objects में (p) एक प्रकार के और (q) दूसरे प्रकार के identical होंAmong (n) objects, (p) of one type and (q) of another type are identical

Explanation

Simple Explanation

Same-type objects की internal permutations नई arrangement नहीं देतीं। परीक्षा में repeated letters की count को factorial division से derive करें। / Internal permutations of same-type objects do not create new arrangements. In exams derive repeated-letter counts using factorial division.

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(BANANA) के distinct arrangements का denominator किस कारण (3!2!) है?

Why is the denominator for distinct arrangements of (BANANA) equal to (3!2!)?

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Correct Answer

A. क्योंकि (A) तीन बार और (N) दो बार आता हैBecause (A) appears three times and (N) appears twice

Explanation

Simple Explanation

समान letters की अदला-बदली same word देती है। परीक्षा में repeated counts को denominator factorials बनाएं। / Interchanging identical letters gives the same word. In exams make repeated counts into denominator factorials.

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(7) distinct beads को bracelet में arrange करने पर (\frac{(7-1)!}{2}) क्यों आता है?

Why does arranging (7) distinct beads in a bracelet give (\frac{(7-1)!}{2})?

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A. Rotation और reflection दोनों same माने जाते हैंBoth rotation and reflection are considered the same

Explanation

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Circular duplicates हटाने के बाद mirror images भी same हैं। परीक्षा में bracelet में necklace की तरह reflection by (2) divide करें। / After removing circular duplicates, mirror images are also the same. In exams divide by (2) for reflection in bracelet problems.

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(n) distinct people की round-table seating में ((n-1)!) और row seating में (n!) का अंतर किससे आता है?

What causes the difference between ((n-1)!) for round-table seating and (n!) for row seating of (n) distinct people?

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A. Circular seating में rotations same मानी जाती हैंRotations are considered the same in circular seating

Explanation

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Circle में एक व्यक्ति को fixed मानकर rotational overcount हटता है। परीक्षा में round table में one fixed method अपनाएं। / In a circle, fixing one person removes rotational overcount. In exams use the one-fixed method for round tables.

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यदि (A) और (B) round table पर adjacent हों, तो (n) people के लिए block method count क्या होगा?

If (A) and (B) are adjacent at a round table, what is the block method count for (n) people?

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A. (2!(n-2)!)

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(A) और (B) को one block मानने पर circular objects (n-1) होते हैं। परीक्षा में circular block count में ((n-2)!) याद रखें। / Treating (A) and (B) as one block gives (n-1) circular objects. In exams remember ((n-2)!) in circular block counts.

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(n) distinct objects की arrangements में exactly (r) selected positions filled हों और बाकी empty हों, तो कौन-सा formula naturally आता है?

If exactly (r) selected positions are filled by (n) distinct objects and the rest are empty, which formula naturally appears?

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B. \(^{n}P_r\)

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Positions ordered होते हैं और objects repeat नहीं होते, इसलिए falling choices मिलती हैं। परीक्षा में positions distinct हों तो permutation सोचें। / Positions are ordered and objects are not repeated, so falling choices arise. In exams think of permutations when positions are distinct.

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(6) digits से (4)-digit numbers बनते हैं और repetition allowed है। Formula \(6^4\) क्यों है?

(4)-digit numbers are formed from (6) digits and repetition is allowed. Why is the formula \(6^4\)?

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A. हर position पर (6) choices independent हैंEach position has (6) independent choices

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Repetition allowed होने से choices कम नहीं होतीं। परीक्षा में allowed repetition में power formula use करें। / Choices do not decrease when repetition is allowed. In exams use the power formula when repetition is allowed.

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(0) included digits से (4)-digit numbers बनाते समय first digit restriction क्यों अलग handle होती है?

Why is the first digit restriction handled separately when forming (4)-digit numbers from digits including (0)?

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A. क्योंकि first digit (0) होने पर number (4)-digit नहीं रहेगाBecause if the first digit is (0), the number will not remain (4)-digit

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Leading zero valid (4)-digit number नहीं बनाता। परीक्षा में digit arrangement में first place condition पहले देखें। / A leading zero does not form a valid (4)-digit number. In exams check the first-place condition first in digit arrangements.

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किस derivation में \(^{n}P_r\) को \(n^r\) से बदलना गलत होगा?

In which derivation would replacing \(^{n}P_r\) by \(n^r\) be wrong?

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A. जब repetition not allowed हो और order important होWhen repetition is not allowed and order is important

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Without repetition में choices घटती हैं, इसलिए \(^{n}P_r\) चाहिए। परीक्षा में repetition condition पढ़े बिना power न लगाएं। / Without repetition, choices decrease, so \(^{n}P_r\) is needed. In exams do not apply powers without reading the repetition condition.

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यदि (4) different prizes (10) students में बांटने हैं और एक student multiple prizes पा सकता है, तो count कौन-सा है?

If (4) different prizes are distributed among (10) students and one student can receive multiple prizes, what is the count?

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B. \(10^4\)

Explanation

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हर different prize के लिए (10) independent student choices हैं। परीक्षा में distinct prizes with repetition allowed को power rule से करें। / Each distinct prize has (10) independent student choices. In exams use the power rule for distinct prizes with repetition allowed.

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यदि (4) different prizes (10) students को देने हैं और कोई student एक से अधिक prize नहीं पा सकता, तो count क्या है?

If (4) different prizes are given to (10) students and no student can receive more than one prize, what is the count?

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C. \(^{10}P_4\)

Explanation

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Prizes different हैं और recipients repeat नहीं हो सकते इसलिए ordered assignment है। परीक्षा में distinct prizes without repetition को permutation समझें। / Prizes are different and recipients cannot repeat, so it is an ordered assignment. In exams treat distinct prizes without repetition as permutation.

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(4) identical prizes (10) students में बांटने हैं और एक student multiple prizes पा सकता है। Count किससे जुड़ी है?

(4) identical prizes are distributed among (10) students and one student can receive multiple prizes. Which count is connected with this?

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A. \({}^{13}C_9\)

Explanation

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Identical prizes distribution stars and bars है, (4) stars और (9) bars। परीक्षा में identical prizes को combinations with repetition से जोड़ें। / Identical prize distribution is stars and bars, with (4) stars and (9) bars. In exams connect identical prizes with combinations with repetition.

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(5) identical prizes (8) students में बांटने हैं और हर selected student को at most (1) prize मिले। Count क्या है?

(5) identical prizes are distributed among (8) students and each selected student gets at most (1) prize. What is the count?

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C. \(^{8}C_5\)

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Prizes identical हैं और सिर्फ (5) students choose करने हैं। परीक्षा में at most one with identical prizes को simple selection मानें। / Prizes are identical and only (5) students need to be chosen. In exams treat at most one with identical prizes as simple selection.

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\({}^{n}C_0+{}^{n}C_1+\cdots+{}^{n}C_n\) में middle terms largest क्यों होते हैं?

Why are middle terms largest in \({}^{n}C_0+{}^{n}C_1+\cdots+{}^{n}C_n\)?

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A. क्योंकि consecutive ratio पहले (1) से बड़ा और बाद में (1) से छोटा होता हैBecause the consecutive ratio is first greater than (1) and later less than (1)

Explanation

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Ratio \(\frac{n-r}{r+1}\) transition दिखाता है। परीक्षा में binomial coefficient trend ratio से check करें। / The ratio \(\frac{n-r}{r+1}\) shows the transition. In exams check the trend of binomial coefficients by ratios.

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यदि (n) even है तो \({}^{n}C_r\) का unique largest term कौन-सा होता है?

If (n) is even, which is the unique largest term of \({}^{n}C_r\)?

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A. \({}^{n}C_{\frac{n}{2}}\)

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Even (n) में binomial coefficients middle पर peak करते हैं। परीक्षा में symmetry और ratio दोनों से middle term पहचानें। / For even (n), binomial coefficients peak at the middle. In exams identify the middle term using symmetry and ratio.

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यदि (n) odd है तो \({}^{n}C_r\) के two largest terms कौन-से होते हैं?

If (n) is odd, which are the two largest terms of \({}^{n}C_r\)?

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B. \({}^{n}C_{\frac{n-1}{2}}\) और \({}^{n}C_{\frac{n+1}{2}}\)\({}^{n}C_{\frac{n-1}{2}}\) and \({}^{n}C_{\frac{n+1}{2}}\)

Explanation

Simple Explanation

Odd (n) में two middle indices complementary और equal होते हैं। परीक्षा में odd case में दो largest terms याद रखें। / For odd (n), the two middle indices are complementary and equal. In exams remember two largest terms for the odd case.

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\({}^{n}C_r={}^{n}C_s\) और \(r\neq s\) हो तो generally कौन-सा relation मिलता है?

If \({}^{n}C_r={}^{n}C_s\) and \(r\neq s\), what relation generally follows?

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A. (r+s=n)

Explanation

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Combination symmetry में unequal equal-values complementary indices से आते हैं। परीक्षा में equal combinations में sum (n) check करें। / In combination symmetry, unequal equal-values come from complementary indices. In exams check sum (n) for equal combinations.

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\({}^{n}P_r={}^{n}P_s\) और \(r\neq s\) generally क्यों संभव नहीं होता?

Why is \({}^{n}P_r={}^{n}P_s\) with \(r\neq s\) generally not possible?

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A. क्योंकि \(^{n}P_r\) usually (r) बढ़ने पर extra positive factors से बदलता हैBecause \(^{n}P_r\) usually changes by extra positive factors as (r) increases

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Permutation में complement symmetry नहीं होती और length बदलने से count बदलता है। परीक्षा में combination symmetry को permutation पर न लगाएं। / Permutations do not have complement symmetry, and changing length changes the count. In exams do not apply combination symmetry to permutations.

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\({}^{n}C_r\) को (\frac{n(n-1)\cdots(n-r+1)}{r!}) लिखना किस connection को दिखाता है?

Writing \({}^{n}C_r\) as (\frac{n(n-1)\cdots(n-r+1)}{r!}) shows which connection?

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A. Permutation count को (r!) से divide करके combination मिलता हैCombination is obtained by dividing permutation count by (r!)

Explanation

Simple Explanation

Numerator \(^{n}P_r\) है और denominator order हटाता है। परीक्षा में इस form से numerical cancellation जल्दी करें। / The numerator is \(^{n}P_r\) and the denominator removes order. In exams use this form for quick numerical cancellation.

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यदि \(^{n}P_r=r!,{}^{n}C_r\), तो \(^{n}P_r\) हमेशा \(^{n}C_r\) से बड़ा या बराबर क्यों होता है?

If \(^{n}P_r=r!,{}^{n}C_r\), why is \(^{n}P_r\) always greater than or equal to \(^{n}C_r\)?

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A. क्योंकि \(r!\geq1\) होता हैBecause \(r!\geq1\)

Explanation

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Permutation हर selected group के all orders गिनता है। परीक्षा में (r=0) या (r=1) पर equality भी संभव है। / Permutation counts all orders of each selected group. In exams equality is also possible for (r=0) or (r=1).

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\({}^{n}C_r+{}^{n}C_{r+1}\) को Pascal identity से किस रूप में लिखा जाएगा?

Using Pascal identity, \({}^{n}C_r+{}^{n}C_{r+1}\) is written in which form?

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A. \({}^{n+1}C_{r+1}\)

Explanation

Simple Explanation

Adjacent lower indices (r) और (r+1) मिलकर next row का (r+1) term देते हैं। परीक्षा में Pascal pattern तुरंत पहचानें। / Adjacent lower indices (r) and (r+1) give the (r+1) term of the next row. In exams identify Pascal pattern quickly.

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\({}^{12}C_4+{}^{12}C_5+{}^{13}C_6\) को simplify करने में कौन-सा पहला सही step है?

What is the first correct step in simplifying \({}^{12}C_4+{}^{12}C_5+{}^{13}C_6\)?

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A. \({}^{12}C_4+{}^{12}C_5={}^{13}C_5\)

Explanation

Simple Explanation

Pascal identity से same upper index और adjacent lower indices combine होते हैं। परीक्षा में पहले adjacent pair को combine करें। / By Pascal identity, the same upper index with adjacent lower indices combines. In exams first combine the adjacent pair.

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\({}^{n}C_0+{}^{n}C_2+{}^{n}C_4+\cdots=2^{n-1}\) किस derivation से आता है?

The identity \({}^{n}C_0+{}^{n}C_2+{}^{n}C_4+\cdots=2^{n-1}\) comes from which derivation?

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A. ((1+1)^n) और ((1-1)^n) को जोड़ने सेBy adding ((1+1)^n) and ((1-1)^n)

Explanation

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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। / To separate even-index coefficients, two binomial substitutions are added. In exams use (x=1) and (x=-1) for even-odd sums.

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