Chapter 4: Permutations, Combinations and Binomial Theorem (Set-1)
Which symbol denotes factorial?
A n!
B n^2
C n+1
D n/2
Factorial of n is written as n! and means the product of all positive integers from 1 to n. It is used heavily in permutations and combinations formulas.
Value of 0! equals
A 0
B Undefined
C 1
D −1
By definition, 0! equals 1 to keep factorial-based formulas consistent, especially in combinations like nC0 = 1 and binomial expansions.
Value of 1! equals
A 0
B 2
C 10
D 1
1! means product of integers from 1 to 1, so it equals 1. It also fits the factorial pattern used in counting formulas.
What does nPr represent?
A Arrangements of r
B Selections of r
C Sum of r
D Product of r
nPr counts ordered arrangements of r objects chosen from n. Order matters, so it is used for tasks like seating and ranking.
What does nCr represent?
A Ordered selection
B Unordered selection
C Circular order
D Repeated order
nCr counts ways to choose r objects from n when order does not matter. It is used in selection problems and binomial coefficients.
Formula for nPr is
A n!/r!
B r!/(n−r)!
C n!/(n−r)!
D (n−r)!/n!
Permutations count ordered choices. After picking r objects from n, the remaining part is (n−r)!, so nPr = n!/(n−r)!.
Formula for nCr is
A n!/(r!(n−r)!)
B n!/(n−r)!
C r!/n!
D (n−r)!/r!
Combinations count unordered choices. We divide permutations by r! to remove ordering among selected r objects, giving nCr = n!/(r!(n−r)!).
Fundamental counting principle means
A Add choices
B Subtract cases
C Divide outcomes
D Multiply stages
If a task has multiple independent stages, total ways equal the product of ways at each stage. This is the base rule behind many counting problems.
If order matters, use
A nCr
B n! only
C nPr
D nC(n−r)
When different orders are counted as different outcomes, we use permutations nPr. In selections like teams, order doesn’t matter, so nCr is used.
If order does not matter, use
A nCr
B nPr
C n^r
D r^n
Combinations are used when only the chosen set matters, not the arrangement. Choosing committees or selecting items are standard combination situations.
Relation between nPr and nCr
A nPr = nCr/r!
B nPr = nCr × r!
C nCr = nPr × r!
D nCr = nPr − r!
For any chosen set of r objects, there are r! ways to arrange them. So ordered count nPr equals unordered count nCr multiplied by r!.
Value of nC0 equals
A 0
B n
C 1
D n!
There is exactly one way to choose nothing from n objects: select the empty set. This matches binomial coefficient behavior in expansions.
Value of nCn equals
A 1
B 0
C n
D n!
There is exactly one way to choose all n objects from n: select everything. Hence nCn = 1, consistent with binomial theorem endpoints.
Symmetry property of combinations is
A nCr = nC(r+1)
B nCr = (n−r)Cr
C nCr = rCn
D nCr = nC(n−r)
Choosing r objects to include is equivalent to choosing (n−r) objects to exclude. This symmetry is a key property used in simplifying expressions.
Value of 5P2 equals
A 10
B 30
C 20
D 5
5P2 = 5!/(5−2)! = 5!/3! = 5×4 = 20. It counts ordered pairs chosen from 5 objects.
Value of 5C2 equals
A 10
B 20
C 5
D 15
5C2 = 5!/(2!3!) = (5×4)/(2×1) = 10. It counts unordered pairs chosen from 5 objects.
Arrangements with repetition count uses
A nPr
B n^r
C nCr
D r^n
If each of r positions can be filled by any of n choices with repetition allowed, total outcomes are n×n×…×n (r times) = n^r.
Combinations with repetition of choosing r items from n types equals
A nPr
B nCr
C (n+r−1)Cr
D (n−r)Cr
Choosing r items from n types with repetition is same as counting non-negative solutions of x1+x2+⋯+xn=rx1+x2+⋯+xn=r. Stars and bars gives (n+r−1)Cr(n+r−1)Cr.
Permutations of identical objects use
A n!
B n^r
C nCr only
D n!/(p!q!)
If some objects repeat, swapping identical items doesn’t create new arrangements. Divide total n! by factorials of counts of identical objects.
Circular permutations of n distinct objects
A (n−1)!
B n!
C n^2
D nCr
In a circle, rotations are considered the same arrangement. Fix one object to break rotational symmetry, leaving (n−1)! circular permutations.
Circular arrangements with reflection same means
A (n−1)!
B n!
C (n−1)!/2
D n^r
If clockwise and anticlockwise arrangements are treated identical, each circular arrangement is counted twice. So the number becomes (n−1)!/2.
Inclusion–exclusion is used to
A Avoid overcounting
B Multiply cases
C Create factorial
D Find middle term
When sets overlap, simply adding counts double-counts common elements. Inclusion–exclusion corrects this by subtracting intersections and adding back higher overlaps.
Derangement means
A Same position allowed
B No fixed point
C Circular shift
D Repetition allowed
A derangement is a permutation where no object remains in its original position. It appears in problems like wrong letter in wrong envelope.
Counting by cases means
A Single formula only
B Ignore restrictions
C Always subtract
D Split into scenarios
When a problem has different possible situations, count each case separately and add the results. This helps handle restrictions cleanly without confusion.
Binomial theorem expands
A (a+b)^n
B (a−b)/n
C a^n+b^n
D (ab)^n
Binomial theorem gives the expansion of (a+b)^n as a sum of terms with binomial coefficients. It is used for algebraic expansions and coefficients.
General term in (a+b)^n is
A nCr a^r b^(n−r)
B nPr a^r b^r
C nCr a^(n−r) b^r
D a^n + b^n
The (r+1)th term is T(r+1)= nCr · a^(n−r) · b^r. Here r runs from 0 to n, covering all terms.
Number of terms in (a+b)^n is
A n+1
B n
C 2n
D 2^n
In binomial expansion, powers of b go from 0 to n, giving n+1 distinct terms. This is a standard property used in term counting.
Binomial coefficients are
A nPr values
B n! values
C prime numbers
D nCr values
The coefficients in (a+b)^n are combinations nC0, nC1, …, nCn. They appear in Pascal’s triangle and many identities.
Symmetry of coefficients means
A nCr = nC(r+1)
B nCr = nPr
C nCr = nC(n−r)
D nCr = n!
Binomial coefficients are symmetric because choosing r from n equals choosing n−r. This symmetry makes coefficients mirror around the middle term.
Sum of coefficients in (a+b)^n equals
A 2^n
B 0
C n!
D n^2
Put a=1 and b=1 in (a+b)^n, giving (1+1)^n = 2^n. This equals the sum of all binomial coefficients.
Alternating sum of coefficients equals
A 2^n
B 0
C n
D 1
Put a=1 and b=−1 in (a+b)^n giving (1−1)^n = 0 for n≥1. This equals alternating sum of binomial coefficients.
Coefficient of x^r in (1+x)^n is
A nCr
B nPr
C r!
D 2^n
In (1+x)^n, the term containing x^r is nCr x^r. So the coefficient of x^r is nCr, widely used in coefficient extraction.
Middle term exists uniquely when n is
A odd
B prime
C even
D zero
If n is even, there is one middle term: the (n/2 + 1)th term. If n is odd, there are two middle terms.
Two middle terms occur when n is
A even
B zero
C negative
D odd
When n is odd, the expansion has an even number of terms, so the “middle” lies between two terms, giving two middle terms.
Expansion of (a−b)^n uses
A alternating signs
B same signs
C only negative terms
D no coefficients
In (a−b)^n, terms have coefficients nCr but powers of b carry (−1)^r, so signs alternate depending on r being even or odd.
Pascal triangle relation is
A nCr = nC(r−1) + nC(r+1)
B nCr = (n−1)Cr
C nCr = (n−1)C(r−1) + (n−1)Cr
D nCr = nPr/r
Each binomial coefficient equals the sum of the two above it in Pascal’s triangle. This recurrence helps compute coefficients and prove identities.
nC1 equals
A n
B 1
C n!
D n−1
Choosing 1 item from n can be done in n ways, so nC1 = n. This also appears as the coefficient of x in (1+x)^n.
nC(n−1) equals
A 1
B n
C n−1
D n^2
By symmetry, nC(n−1) = nC1 = n. It counts ways to leave out exactly one item from n.
Term independent of x means power of x is
A 1
B n
C negative
D 0
A term independent of x has x^0, which equals 1. In expansions, set the exponent of x to zero to find such terms.
Coefficient comparison is used to
A equate same powers
B remove terms
C change base
D find factorial
If two polynomials are equal, coefficients of like powers must match. This method helps solve identities and unknown coefficients problems.
Vandermonde identity (intro) relates to
A product of sums
B circular counting
C sum of combinations
D derangements only
Vandermonde identity states a sum of products of combinations equals another combination, often written as Σ (mCr)(nC(k−r)) = (m+n)Ck.
Greatest term (basic) in expansion depends on
A term magnitude
B sign only
C r! only
D n−r only
The greatest term is the term with maximum value among all terms for given a and b. It is found by comparing successive term ratios.
Binomial coefficient maximum occurs near
A r=0
B r=n
C r≈n
D r≈n/2
nCr increases up to around r=n/2 and then decreases due to symmetry. Hence the maximum coefficient lies at the middle region.
Subset counting of an n-set equals
A n!
B 2^n
C n^2
D nCr only
Each element can be either included or excluded independently, giving 2 choices per element. So total subsets are 2^n.
Lattice paths (intro) on grid mainly count
A shortest routes
B rotations
C identical objects
D factorial sums
Basic lattice path counting often counts shortest paths using steps right and up. The number of such paths equals combinations based on step arrangement.
Stars and bars (intro) is used for
A circular seating
B binomial signs
C distribution counts
D derangements
Stars and bars counts ways to distribute identical objects into distinct groups under basic conditions. It converts distribution into choosing divider positions.
Selection constraints best handled by
A cases and counting
B random guessing
C only factorial
D only symmetry
When restrictions exist, splitting into valid cases and counting each carefully avoids missing or double-counting. This approach keeps the problem clear and accurate.
Permutations with restrictions often use
A only nCr
B only 2^n
C only n^r
D inclusion-exclusion
Restrictions like “not together” or “not in place” can be handled by counting total arrangements and subtracting invalid ones, often via inclusion–exclusion.
Binomial distribution link (intro) uses
A nPr terms
B circular terms
C nCr terms
D derangement terms
In binomial probability, the number of ways to get r successes in n trials is nCr. This counting factor multiplies probability terms.
Fractional index expansion (basic) uses
A Pascal triangle only
B binomial series
C nPr only
D derangements
For fractional or negative powers, (1+x)^n can be expanded as an infinite binomial series under suitable conditions. It generalizes the usual binomial theorem.