Chapter 4: Permutations, Combinations and Binomial Theorem (Set-2)
For nPr, r must satisfy
A r > n
B r ≤ n
C r < 0
D r = n+1
In nPr, you choose r items from n without repetition. You cannot select more items than available, so r must be less than or equal to n.
For nCr, r must satisfy
A r > n
B r is prime
C r ≤ n
D r is negative
In combinations, selecting r objects from n is possible only when r is between 0 and n. Otherwise the selection has no meaning in basic counting.
7C0 equals
A 0
B 7
C 49
D 1
Choosing zero objects from any set can be done in exactly one way: choose nothing. This is why nC0 is always 1.
7C7 equals
A 7
B 1
C 0
D 49
Choosing all available objects is only one selection: take everything. Therefore, nCn is always 1, matching binomial theorem endpoints.
6P6 equals
A 720
B 6
C 1
D 120
6P6 means arranging all 6 distinct objects, so it equals 6! = 720. It counts the total permutations of 6 different items.
6C1 equals
A 1
B 12
C 6
D 720
Selecting one item from 6 can be done in 6 ways. This is a direct application of nC1 = n.
6C5 equals
A 1
B 6
C 5
D 30
By symmetry, 6C5 = 6C1 because choosing 5 to include equals choosing 1 to exclude. Hence the value is 6.
8P2 equals
A 16
B 72
C 64
D 56
8P2 = 8×7 = 56? Wait, that is 56. Actually permutations for 2 places: 8 choices then 7 choices, so 8P2 = 56.
8C2 equals
A 56
B 16
C 28
D 14
8C2 = 8!/(2!6!) = (8×7)/2 = 28. It counts the number of unordered pairs chosen from 8 objects.
Difference between nPr and nCr is
A same always
B order matters
C repetition compulsory
D circle only
In permutations nPr, different orders are counted separately, while in combinations nCr, only selection matters. This single idea decides which formula is used.
If repetition allowed in r places
A n^r
B nPr
C nCr
D r^n
With repetition allowed, each of r positions can be filled independently in n ways. Multiplying n, r times gives n^r total sequences.
Arranging letters of “LEVEL” equals
A 120
B 30
C 60
D 20
LEVEL has 5 letters with L repeated twice and E repeated twice. Distinct permutations are 5!/(2!2!) = 120/4 = 30? Actually 120/4 = 30.
Distinct permutations of “BOOK” equals
A 24
B 6
C 12
D 4
BOOK has 4 letters where O repeats twice. Distinct arrangements are 4!/2! = 24/2 = 12, since swapping the O’s doesn’t create a new word.
Number of circular arrangements of 5 people
A 120
B 20
C 5
D 24
Circular permutations of n distinct objects are (n−1)!. For 5 people, arrangements are 4! = 24 when rotations are treated the same.
When reflections also same in circle
A n!
B (n−1)!/2
C (n−1)!
D n^r
If clockwise and anticlockwise are considered identical, each circular arrangement is counted twice. So the number becomes (n−1)!/2 for n>2.
Simple meaning of nCr
A ordered r-tuples
B multiply r items
C choose r items
D arrange all items
nCr counts how many ways to choose r objects from n without caring about order. It is the basic counting behind binomial coefficients.
nCr equals nC(n−r) because
A include vs exclude
B rotation same
C factorial cancels
D r always even
Selecting r to include automatically decides which n−r are excluded. Both choices uniquely correspond, so the counts are equal, giving the symmetry property.
First term of (a+b)^n expansion
A nCab^n
B a^n
C b^n
D n a^(n−1)b
In binomial expansion, the term with r=0 is nC0 a^n b^0 = a^n. It is always the first term.
Last term of (a+b)^n expansion
A a^n
B n a^(n−1)b
C 2^n
D b^n
The term with r=n is nCn a^0 b^n = b^n. It appears as the last term of the expansion.
Coefficients in (a+b)^n are from
A Fibonacci
B prime list
C Pascal triangle
D logarithms
The binomial coefficients nC0, nC1, …, nCn form Pascal’s triangle. Each coefficient equals the sum of the two directly above it.
Sum of coefficients of (a+b)^n is found by
A a=0,b=1
B a=1,b=1
C a=1,b=0
D a=−1,b=1
Setting a=1 and b=1 gives (1+1)^n = 2^n, which equals the sum of all coefficients in the expansion.
Alternating sum of coefficients uses
A a=1,b=1
B a=0,b=1
C a=2,b=1
D a=1,b=−1
Put a=1 and b=−1 to get (1−1)^n. For n≥1 it becomes 0, giving alternating sum of binomial coefficients as zero.
Coefficient of x in (1+x)^n is
A nC0
B nC1
C nCn
D nC2
In (1+x)^n, coefficient of x^1 is nC1. Since nC1 = n, the coefficient of x is n.
Coefficient of x^0 in (1+x)^n is
A 1
B 0
C n
D 2^n
The constant term corresponds to x^0, which is the r=0 term. Its coefficient is nC0 = 1 for any n.
Middle term position when n is even
A n/2
B n+1
C n/2 + 1
D 2n
Total terms are n+1. For even n, the single middle term is the (n/2 + 1)th term in the expansion.
Two middle terms positions when n is odd
A n and n+1
B (n+1)/2 and (n+3)/2
C 1 and n
D n/2 and n/2+1
When n is odd, n+1 is even, so there are two central terms. Their positions are (n+1)/2 and (n+3)/2 in the list of terms.
Sign pattern in (a−b)^n depends on
A n only
B a value
C b value
D r parity
In (a−b)^n, the term contains (−b)^r, so the sign is positive for even r and negative for odd r, creating alternating signs.
nCr recurrence 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!
This Pascal identity builds nCr from the previous row. It is used in proofs, in Pascal’s triangle construction, and to simplify sums.
Coefficient extraction means
A find n
B find specific coefficient
C solve inequality
D draw triangle
Coefficient extraction focuses on identifying the multiplier of a required term like x^k in an expansion. It is common in binomial theorem applications.
Term independent of x means exponent equals
A 0
B 1
C n
D −1
A term independent of x has x^0. In expansions, set the power of x to zero and solve for the term index to locate it.
Basic probability selection count uses
A nPr
B n! only
C n^r
D nCr
Many probability questions involve selecting outcomes without order, such as choosing r successes among n trials, which is counted by nCr.
Choosing 2 books from 5 equals
A 20
B 25
C 10
D 5
Choosing 2 from 5 without order uses combinations: 5C2 = 10. It represents selecting a pair where order does not create different outcomes.
Arranging 3 prizes among 3 students means
A combination
B permutation
C subset count
D repetition only
If prizes are distinct and each student gets one, assignment is ordered. Different assignments are different outcomes, so permutations or direct counting applies.
Distributing identical balls into 2 boxes (nonnegative)
A stars and bars
B 2^n
C circular rule
D derangement
Distribution of identical objects into distinct boxes is a classic stars and bars situation. It counts solutions of x1+x2 = n with x1,x2 ≥ 0.
Lattice shortest path with 3R and 2U equals
A 12
B 20
C 6
D 10
Total steps are 5 with 3 rights identical and 2 ups identical. Number of paths equals 5!/(3!2!) = 10 using combinations.
Binomial coefficient meaning in counting is
A arrange r items
B choose r items
C repeat r items
D rotate r items
nCr counts the number of r-element subsets from an n-element set. This is why it naturally appears in (a+b)^n expansion coefficients.
nC2 formula simplifies to
A n(n−1)
B n^2
C n(n−1)/2
D (n−1)!
nC2 = n!/(2!(n−2)!) = n(n−1)/2. It is widely used for counting pairs, handshakes, and edges in graphs.
Number of ways to choose 1 from n is
A 1
B n!
C 2^n
D n
Choosing one item from n has n possibilities. This matches nC1 = n and also reflects the direct counting of single selections.
Total subsets of a 5-element set
A 10
B 32
C 25
D 120
Each element can be chosen or not chosen, giving 2 choices per element. So subsets of a 5-element set are 2^5 = 32.
Basic identity: sum of nCr for r=0 to n
A n
B n!
C 2^n
D 0
The sum of all binomial coefficients in row n equals 2^n. This is the same as the sum of coefficients of (1+x)^n at x=1.
Coefficient of x^2 in (1+x)^5
A 10
B 5
C 20
D 2
Coefficient of x^2 in (1+x)^5 is 5C2 = 10. In general, coefficient of x^k is nCk.
Constant term of (2x + 1)^3
A 6
B 1
C 8
D 3
Constant term occurs when x power is 0, meaning choose the “1” from each factor. So constant term is 1^3 = 1.
Number of terms in (a+b)^7
A 7
B 14
C 128
D 8
The number of terms in a binomial expansion (a+b)^n is n+1. For n=7, total terms are 8.
Basic approximation using binomial applies when
A x is large
B n is zero
C x is small
D r is negative
Binomial approximation like (1+x)^n ≈ 1+nx works when |x| is small. Higher power terms become very small and are often ignored.
Negative binomial expansion is for
A n positive integer
B n negative integer
C only n=1
D only n=0
When n is negative, (1+x)^n expands into an infinite series for |x|<1. It is called the negative binomial series in basic introduction.
Fractional index binomial series requires
A |x| < 1
B x=0 only
C x=1 only
D n must be prime
For fractional powers, (1+x)^k is expanded as an infinite series that converges when |x|<1. This condition ensures the series terms decrease.
Coefficient comparison is valid when
A angles equal
B polynomials equal
C numbers prime
D circle same
If two polynomials are equal for all x, each corresponding power of x must have equal coefficients. This method solves identities and unknown parameters.
Counting arrangements “not together” often uses
A only nCr
B only 2^n
C inclusion-exclusion
D only n^r
“Not together” problems often count all arrangements and subtract those where items are forced together. This is a basic inclusion–exclusion style approach.
Vandermonde identity combines coefficients of
A (1−x)^n only
B a^n only
C factorial only
D (1+x)^m and (1+x)^n
Multiplying (1+x)^m and (1+x)^n gives (1+x)^(m+n). Matching coefficients of x^k leads to Vandermonde’s identity involving sums of products of combinations.
Generating functions (intro) mainly help in
A drawing graphs
B counting sequences
C solving geometry
D finding primes
Generating functions turn counting problems into algebraic series manipulations. Even at an introductory level, they help count combinations, partitions, and restricted selections.