What is the counting interpretation of ({}^{n}C_r=\frac{n}{r}{}^{n-1}C_{r-1})?
Correct answer: A
Choose the marked member in (n) ways and remove overcount of (r) possible marks. In exams understand such identities by member marking.
Mathematics
Derivations of formulas and their connections
सूत्रों की व्युत्पत्तियाँ और उनके पारस्परिक संबंध
In Class 11 Mathematics, this topic explains how the formulas in Permutations and Combinations are derived and how they are connected. Students learn the meaning of factorial notation, develop the formulas for nPr and nCr from counting principles, and understand why nPr = r! nCr and nCr = nC(n−r). The topic also shows when to use arrangements or selections, helping students apply these relationships logically instead of relying on memorized formulas.
TOPIC PRACTICE
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Hard · Level 3
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- First choose one marked member and then choose remaining (r-1)
- First arrange (r) objects in a circle
- Repeat every object
- Count only rejected objects
Hard · Level 3
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- (\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{n-r+1}{r})
- (\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{r}{n-r+1})
- (\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=\frac{r!}{n!})
- (\frac{{}^{n}C_{r-1}}{{}^{n}C_r}=r(n-r))
Hard · Level 3
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- (\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{n-r}{r+1})
- ({}^{n}C_r={}^{r}C_n)
- ({}^{n}C_r={}^{n}P_r)
- ({}^{n}C_r=n^r)
Hard · Level 3
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- ({}^{n}C_r)
- ({}^{r}P_n)
- ({}^{n+r-1}C_{r-1})
- (r^n)
Hard · Level 3
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- ({}^{n-1}C_{r-1})
- ({}^{n+r-1}C_{r-1})
- ({}^{n}P_r)
- (r^n-r)
Hard · Level 3
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- Pascal identity
- Complement identity
- Product rule
- Circular identity
Hard · Level 3
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- (^{10}C_4)
- (^{10}P_4)
- (4^{10})
- ({}^{10}C_4+4!)
Hard · Level 3
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- First choose the group then order that group
- First treat all (7) people as identical
- Order is ignored so (3!) is removed
- Every person is repeated
Hard · Level 3
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- (\frac{8!}{3!3!2!})
- (\frac{8!}{2!})
- (^{8}C_3)
- (3!3!2!)
Hard · Level 3
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- (2!)
- (3!)
- (8!)
- (5!)
Hard · Level 3
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- (^{n}P_r)
- (^{n}C_r)
- (r^n)
- ({}^{n+r-1}C_{r-1})
Hard · Level 3
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- (\sum_{k=0}^{r}(-1)^k{}^{r}C_k(r-k)^n)
- (r^n+{}^{n}C_r)
- (^{n}P_r\cdot r!)
- ({}^{n+r-1}C_{r-1})
Hard · Level 3
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- Because each person has (r) independent room choices
- Because rooms are identical
- Because every room has exactly (1) person
- Because order of rooms is ignored
Hard · Level 3
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- Each child gets at least one coin
- A child may get zero coins
- Coins are distinct
- Children are identical
Hard · Level 3
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- ({}^{12}C_3)
- ({}^{14}C_2)
- (3^{12})
- (^{12}P_3)
Hard · Level 3
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- ({}^{14}C_3)
- ({}^{18}C_3)
- ({}^{11}C_3)
- (4^{15})
Hard · Level 3
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- (^{a}C_s\cdot{}^{n-a}C_{r-s}) where there are (a) special objects
- (^{n}P_r)
- (^{n}C_s\cdot r!)
- (a^s)
Hard · Level 3
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- (^{a}C_1{}^{n-a}C_{r-1}) only
- (^{n}C_r-{}^{n-a}C_r)
- (^{n}P_r-{}^{a}P_r)
- (a^r)
Hard · Level 3
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- (^{8}C_2+{}^{8}C_4)
- (^{10}C_4-{}^{8}C_4)
- (^{10}P_4)
- (^{8}C_3\cdot2!)
Hard · Level 3
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- (^{9}C_5-{}^{7}C_3)
- (^{9}C_5+{}^{7}C_3)
- (^{7}C_5) only
- (^{9}P_5)
Hard · Level 3
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- (8!-7!\cdot2!)
- (7!\cdot2!)
- (^{8}C_2\cdot6!)
- (8!+7!\cdot2!)
Hard · Level 3
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- Arrange girls first and put boys in gaps
- Arrange boys first and place (4) girls in (7) gaps
- Arrange all (10) people in a circle
- Treat girls as identical
Hard · Level 3
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- (^{8}C_5\cdot5!)
- (^{7}C_5\cdot5!)
- (^{12}C_5)
- (5^7)
Hard · Level 3
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- Among (n) objects, (p) of one type and (q) of another type are identical
- All (n) objects are distinct
- Order is ignored
- Repetition is allowed with independent choices
Hard · Level 3
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- Because (A) appears three times and (N) appears twice
- Because (B) appears three times
- Because total letters are (6)
- Because vowels are together
If (^{n}C_{r-1}:{}^{n}C_r=3:5), which expression is useful in the ratio derivation?
Correct answer: B
Canceling factorials in consecutive combinations gives (\frac{r}{n-r+1}). In exams do not calculate full values in ratio questions.
Which relation is most useful for identifying the largest term of ({}^{n}C_r)?
Correct answer: A
Near the largest term, the consecutive ratio changes around (1). In exams check the transition from increasing to decreasing.
What is the formula for distributing (n) identical balls into (r) distinct boxes when empty boxes are allowed?
Correct answer: C
In stars and bars, (n) stars and (r-1) bars are arranged. In exams use the bars method for identical distribution.
How does the formula change if (n) identical balls are distributed into (r) distinct boxes with every box non-empty?
Correct answer: A
Give (1) ball to each box first, then distribute the remaining (n-r) balls. In exams allot the minimum first for non-empty conditions.
The equality of ({}^{n+r-1}C_r) and ({}^{n+r-1}C_{n-1}) comes from which identity?
Correct answer: B
The two lower indices sum to (n+r-1). In exams accept complementary forms in stars and bars answers.
The count for seating (4) selected students in a row of (10) seats while leaving other seats empty is connected with which expression?
Correct answer: B
Choosing seats first and then arranging students gives (^{10}C_4\cdot4!=^{10}P_4). In exams count both positions and arrangements.
Why is the count for choosing (3) people from (7) into an ordered queue equal to (^{7}C_3\cdot3!)?
Correct answer: A
Order of selected people is meaningful in a queue. In exams derive ordered selection as combination times factorial.
What is the formula for dividing (8) distinct objects into labelled groups of (3), (3), and (2)?
Correct answer: A
For labelled groups, group names are fixed so only internal order is removed. In exams use factorial denominators for fixed group sizes.
When (8) distinct objects are divided into unlabelled groups of sizes (3), (3), and (2), by what extra factor do we divide?
Correct answer: A
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.
Which formula distributes (n) distinct objects into (r) distinct boxes when each object can go into one box?
Correct answer: C
Each distinct object has (r) independent choices. In exams use the power rule for distinct objects and distinct boxes.
Which inclusion-exclusion form counts onto distributions of (n) distinct objects into (r) distinct boxes?
Correct answer: A
Inclusion-exclusion is used to exclude empty boxes. In exams think of removing empty boxes when the word onto appears.
(n) people are to be placed into (r) distinct rooms and rooms may be empty. Why is the count (r^n)?
Correct answer: A
Independent choices of distinct people multiply. In exams connect distinct-person box problems with powers.
Which condition gives the formula ({}^{n+r-1}C_{n}) for distributing (n) identical coins among (r) children?
Correct answer: B
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.
What is the count of non-negative integer solutions of (x_1+x_2+x_3=12)?
Correct answer: B
Non-negative solutions by stars and bars are ({}^{12+3-1}C_{3-1}). In exams convert equation solutions into distribution problems.
If all (x_i\geq1) in (x_1+x_2+x_3+x_4=15), what is the count?
Correct answer: C
Give (1) to each variable first, then (11) remain. In exams subtract the number of variables for positive solutions.
What is the formula for choosing (r) objects from (n) objects with exactly (s) special objects?
Correct answer: A
Choose (s) objects from the special group and (r-s) from the non-special group. In exams split exactly conditions into product of choices.
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?
Correct answer: B
The complement of at least one special is no special. In exams total minus none is fast for at least conditions.
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?
Correct answer: A
Case (1): take both fixed books, case (2): take neither. In exams keep cases clear for paired restrictions.
Choose (5) people from (9) people, but two particular people cannot be selected together. What is the count?
Correct answer: A
Subtract selections containing both particular people from total selections. In exams solve not-together selection using complement.
Arrange (8) people in a row so that (A) and (B) are not together. Which formula gives the count?
Correct answer: A
Subtract block arrangements where (A) and (B) are together from total arrangements. In exams complement is easy for not-together permutations.
Arrange (6) boys and (4) girls in a row so that no two girls are together. What is the main step in the derivation?
Correct answer: B
Arranging boys creates (7) gaps where girls can be placed. In exams use the gap method for no two together.
A word is formed from (5) vowels and (7) consonants with no two vowels together. What is the vowel placement factor?
Correct answer: A
After arranging (7) consonants, (8) gaps are formed and (5) vowels are arranged. In exams choose gaps and permute vowels.
Under which condition does the repeated-object arrangement formula (\frac{n!}{p!q!}) apply?
Correct answer: A
Internal permutations of same-type objects do not create new arrangements. In exams derive repeated-letter counts using factorial division.
Why is the denominator for distinct arrangements of (BANANA) equal to (3!2!)?
Correct answer: A
Interchanging identical letters gives the same word. In exams make repeated counts into denominator factorials.
QUIZ COMPLETE