Why does arranging (7) distinct beads in a bracelet give (\frac{(7-1)!}{2})?
Correct answer: A
After removing circular duplicates, mirror images are also the same. In exams divide by (2) for reflection in bracelet problems.
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 4
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- Both rotation and reflection are considered the same
- Only rotation is different
- Every bead is identical
- Order is ignored
Hard · Level 4
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- Rotations are considered the same in circular seating
- Order is ignored at a round table
- People are identical in row seating
- Reflection is always same at a round table
Hard · Level 4
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- (2!(n-2)!)
- (2!(n-1)!)
- (n!)
- (^{n}C_2(n-2)!)
Hard · Level 4
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- (^{n}C_r)
- (^{n}P_r)
- (n^r)
- ({}^{n+r-1}C_r)
Hard · Level 4
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- Each position has (6) independent choices
- Choices are (6), then (5), then (4), then (3)
- Order is ignored
- Digits are identical
Hard · Level 4
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- Because if the first digit is (0), the number will not remain (4)-digit
- Because (0) is forbidden everywhere
- Because the last digit must always be (0)
- Because all digits are identical
Hard · Level 4
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- When repetition is not allowed and order is important
- When repetition is allowed and order is important
- When every position is independent
- When the same object can be used multiple times
Hard · Level 4
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- (^{10}P_4)
- (10^4)
- (^{10}C_4)
- ({}^{13}C_4)
Hard · Level 4
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- (10^4)
- (^{10}C_4)
- (^{10}P_4)
- ({}^{13}C_4)
Hard · Level 4
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- ({}^{13}C_9)
- (10^4)
- (^{10}P_4)
- (^{10}C_4)
Hard · Level 4
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- (8^5)
- ({}^{12}C_7)
- (^{8}C_5)
- (^{8}P_5)
Hard · Level 4
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- Because the consecutive ratio is first greater than (1) and later less than (1)
- Because end terms are always (n!)
- Because all terms are equal
- Because (^{n}C_r) never decreases
Hard · Level 4
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- ({}^{n}C_{\frac{n}{2}})
- ({}^{n}C_0)
- ({}^{n}C_1)
- ({}^{n}C_n+{}^{n}C_0)
Hard · Level 4
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- ({}^{n}C_0) and ({}^{n}C_n)
- ({}^{n}C_{\frac{n-1}{2}}) and ({}^{n}C_{\frac{n+1}{2}})
- ({}^{n}C_1) and ({}^{n}C_2)
- ({}^{n}C_{n-1}) and ({}^{n}C_n)
Hard · Level 4
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- (r+s=n)
- (rs=n)
- (r=s+1)
- (r-s=n)
Hard · Level 4
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- Because (^{n}P_r) usually changes by extra positive factors as (r) increases
- Because order is ignored in permutations
- Because (^{n}P_r={}^{n}C_r) always
- Because (r+s=n) is always needed
Hard · Level 4
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- Combination is obtained by dividing permutation count by (r!)
- Multiplying combination by (r!) gives a smaller count
- Repeated choices are independent
- Circular arrangements are counted
Hard · Level 4
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- Because (r!\geq1)
- Because (n!\leq r!)
- Because combinations count order
- Because (^{n}C_r=0) always
Hard · Level 4
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- ({}^{n+1}C_{r+1})
- ({}^{n+1}C_r)
- ({}^{n}C_{2r+1})
- ({}^{2n}C_{r+1})
Hard · Level 4
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- ({}^{12}C_4+{}^{12}C_5={}^{13}C_5)
- ({}^{12}C_4+{}^{12}C_5={}^{24}C_9)
- ({}^{12}C_4+{}^{12}C_5={}^{13}C_6)
- ({}^{12}C_4+{}^{12}C_5={}^{12}C_9)
Hard · Level 4
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- By adding ((1+1)^n) and ((1-1)^n)
- Only from ((1+0)^n)
- By multiplying ({}^{n}P_r) by (r!)
- By the fixed method of circular permutation
Hard · Level 4
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- (n-r)
- (n-r+1)
- (r)
- (\frac{1}{n-r+1})
Hard · Level 4
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- Counting by marking one selected member
- Arranging all objects in a circle
- Selecting every object twice
- Ignoring order and then multiplying
Hard · Level 4
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- (\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n-r}{r})
- (\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{r}{n})
- (\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=\frac{n}{r})
- (\frac{{}^{n}C_r}{{}^{n-1}C_{r-1}}=r!)
Hard · Level 4
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- (^{n+1}C_r)
- (^{n+1}C_{r-1})
- (^{2n}C_r)
- (^{n}C_{2r-1})
What causes the difference between ((n-1)!) for round-table seating and (n!) for row seating of (n) distinct people?
Correct answer: A
In a circle, fixing one person removes rotational overcount. In exams use the one-fixed method for round tables.
If (A) and (B) are adjacent at a round table, what is the block method count for (n) people?
Correct answer: A
Treating (A) and (B) as one block gives (n-1) circular objects. In exams remember ((n-2)!) in circular block counts.
If exactly (r) selected positions are filled by (n) distinct objects and the rest are empty, which formula naturally appears?
Correct answer: B
Positions are ordered and objects are not repeated, so falling choices arise. In exams think of permutations when positions are distinct.
(4)-digit numbers are formed from (6) digits and repetition is allowed. Why is the formula (6^4)?
Correct answer: A
Choices do not decrease when repetition is allowed. In exams use the power formula when repetition is allowed.
Why is the first digit restriction handled separately when forming (4)-digit numbers from digits including (0)?
Correct answer: A
A leading zero does not form a valid (4)-digit number. In exams check the first-place condition first in digit arrangements.
In which derivation would replacing (^{n}P_r) by (n^r) be wrong?
Correct answer: A
Without repetition, choices decrease, so (^{n}P_r) is needed. In exams do not apply powers without reading the repetition condition.
If (4) different prizes are distributed among (10) students and one student can receive multiple prizes, what is the count?
Correct answer: B
Each distinct prize has (10) independent student choices. In exams use the power rule for distinct prizes with repetition allowed.
If (4) different prizes are given to (10) students and no student can receive more than one prize, what is the count?
Correct answer: C
Prizes are different and recipients cannot repeat, so it is an ordered assignment. In exams treat distinct prizes without repetition as permutation.
(4) identical prizes are distributed among (10) students and one student can receive multiple prizes. Which count is connected with this?
Correct answer: A
Identical prize distribution is stars and bars, with (4) stars and (9) bars. In exams connect identical prizes with combinations with repetition.
(5) identical prizes are distributed among (8) students and each selected student gets at most (1) prize. What is the count?
Correct answer: C
Prizes are identical and only (5) students need to be chosen. In exams treat at most one with identical prizes as simple selection.
Why are middle terms largest in ({}^{n}C_0+{}^{n}C_1+\cdots+{}^{n}C_n)?
Correct answer: A
The ratio (\frac{n-r}{r+1}) shows the transition. In exams check the trend of binomial coefficients by ratios.
If (n) is even, which is the unique largest term of ({}^{n}C_r)?
Correct answer: A
For even (n), binomial coefficients peak at the middle. In exams identify the middle term using symmetry and ratio.
If (n) is odd, which are the two largest terms of ({}^{n}C_r)?
Correct answer: B
For odd (n), the two middle indices are complementary and equal. In exams remember two largest terms for the odd case.
If ({}^{n}C_r={}^{n}C_s) and (r\neq s), what relation generally follows?
Correct answer: A
In combination symmetry, unequal equal-values come from complementary indices. In exams check sum (n) for equal combinations.
Why is ({}^{n}P_r={}^{n}P_s) with (r\neq s) generally not possible?
Correct answer: A
Permutations do not have complement symmetry, and changing length changes the count. In exams do not apply combination symmetry to permutations.
Writing ({}^{n}C_r) as (\frac{n(n-1)\cdots(n-r+1)}{r!}) shows which connection?
Correct answer: A
The numerator is (^{n}P_r) and the denominator removes order. In exams use this form for quick numerical cancellation.
If (^{n}P_r=r!,{}^{n}C_r), why is (^{n}P_r) always greater than or equal to (^{n}C_r)?
Correct answer: A
Permutation counts all orders of each selected group. In exams equality is also possible for (r=0) or (r=1).
Using Pascal identity, ({}^{n}C_r+{}^{n}C_{r+1}) is written in which form?
Correct answer: A
Adjacent lower indices (r) and (r+1) give the (r+1) term of the next row. In exams identify Pascal pattern quickly.
What is the first correct step in simplifying ({}^{12}C_4+{}^{12}C_5+{}^{13}C_6)?
Correct answer: A
By Pascal identity, the same upper index with adjacent lower indices combines. In exams first combine the adjacent pair.
The identity ({}^{n}C_0+{}^{n}C_2+{}^{n}C_4+\cdots=2^{n-1}) comes from which derivation?
Correct answer: A
To separate even-index coefficients, two binomial substitutions are added. In exams use (x=1) and (x=-1) for even-odd sums.
If (^{n}P_r) is obtained from (^{n}P_{r-1}), which multiplier is added?
Correct answer: B
Filling one more position gives (n-r+1) choices. In exams identify the new last factor in the next permutation term.
Which idea explains the relation (^{n}C_r=\frac{n}{r}{}^{n-1}C_{r-1})?
Correct answer: A
Choosing the marked member first gives (n) choices and each group is counted (r) times. In exams treat the denominator (r) as overcount.
If (^{n}C_r) is compared with (^{n-1}C_{r-1}), which ratio is correct?
Correct answer: C
Canceling factorial forms gives (\frac{n}{r}). In exams use the ratio method for adjacent upper and lower indices.
Using Pascal identity, (^{n}C_r+{}^{n}C_{r-1}) is written in which form?
Correct answer: A
Adjacent terms of the same row give the middle term in the next row. In exams apply Pascal identity when (r) and (r-1) appear together.
QUIZ COMPLETE