Chapter 4: Permutations, Combinations and Binomial Theorem (Set-3)
In how many ways can 3 positions be filled from 9 distinct students, without repetition
A 720
B 84
C 504
D 126
Filling 3 distinct positions from 9 students is an ordered selection, so use permutations: 9P3 = 9×8×7 = 504. Order changes the outcome.
A team of 3 is chosen from 10 students without any role assignment
A 120
B 720
C 210
D 30
Choosing 3 students without roles is an unordered selection, so use combinations: 10C3 = 120. Different orders of the same 3 students are not counted separately.
How many 4-digit numbers can be formed using digits 0–9 without repetition, with first digit nonzero
A 5040
B 9000
C 3024
D 4536
First digit has 9 choices (1–9). Remaining three digits have 9, 8, 7 choices. Total = 9×9×8×7 = 4536.
From 1 to 100, how many numbers are divisible by 2 or 5
A 70
B 55
C 60
D 50
Count multiples of 2: 50, of 5: 20. Subtract multiples of 10 counted twice: 10. So total = 50+20−10 = 60.
How many length-5 strings can be made from {A,B,C} having at least one A
A 243
B 211
C 32
D 81
Total strings = 3^5 = 243. Strings with no A use only {B,C}: 2^5 = 32. So required = 243 − 32 = 211.
Number of derangements of 4 distinct objects is
A 8
B 6
C 9
D 24
A derangement means no object stays in its original place. For 4 objects, the standard count is !4 = 9, found using inclusion–exclusion or known values.
Number of nonnegative solutions of x1+x2+x3 = 5 is
A 21
B 15
C 10
D 35
Using stars and bars, solutions to x1+x2+x3=5 with xi≥0 are (5+3−1)C(3−1) = 7C2 = 21.
Number of positive integer solutions of x1+x2+x3 = 7 is
A 21
B 10
C 28
D 15
For xi≥1, set yi=xi−1. Then y1+y2+y3=4 with yi≥0. Count = (4+3−1)C2 = 6C2 = 15.
Number of shortest paths from (0,0) to (4,3) using only right and up moves
A 21
B 70
C 35
D 28
Total moves are 7 (4 right, 3 up). Choose where the 3 up moves occur: 7C3 = 35. This counts all shortest lattice paths.
Coefficient of x^3 in (2 + x)^5 is
A 40
B 20
C 10
D 80
The x^3 term comes from choosing x three times and 2 twice: coefficient = 5C3 × 2^2 = 10×4 = 40.
Coefficient of x^4 in (1 − x)^6 is
A −15
B 20
C 15
D 6
In (1−x)^6, coefficient of x^4 is 6C4(−1)^4 = 6C4 = 15. Even power keeps the sign positive.
Constant term in (x^2 + 1/x)^5 is
A 1
B 0
C 5
D 10
General term has power x^(2(5−r) − r) = x^(10−3r). For constant term exponent must be 0, but 10−3r=0 gives non-integer r, so none.
Term independent of x in (x^3 + 2/x)^4 has coefficient
A 32
B 16
C 24
D 64
General power is x^(3(4−r) − r)=x^(12−4r). Set 12−4r=0 → r=3. Coefficient = 4C3 × 2^3 = 4×8 = 32.
In (a+b)^8, the single middle term is the
A 4th term
B 6th term
C 8th term
D 5th term
For even n, the middle term position is (n/2 + 1). With n=8, it is (4+1)=5th term. This is a standard binomial theorem property.
The largest binomial coefficient value in (1+x)^7 is
A 21
B 28
C 35
D 42
Coefficients of (1+x)^7 are 1, 7, 21, 35, 35, 21, 7, 1. The greatest value is 35, occurring in the middle region.
Value of Σ r·6Cr from r=0 to 6 is
A 192
B 96
C 64
D 384
Identity: Σ r·nCr = n·2^(n−1). For n=6, value = 6·2^5 = 6·32 = 192. It follows from differentiating (1+x)^n.
Value of Σ r²·5Cr from r=0 to 5 is
A 160
B 80
C 240
D 120
Identity: Σ r²·nCr = n(n+1)2^(n−2). For n=5, value = 5·6·2^3 = 30·8 = 240.
Which expression equals nCr
A (n+r)C r
B (n/r)(n−1C r−1)
C (n−1C r+1)
D (n−r)C r
Using factorial form, nCr = n!/r!(n−r)!. It can be rewritten as (n/r)·(n−1C r−1). This identity helps simplify many combination expressions.
Value of 8C3 + 8C4 is
A 70
B 56
C 84
D 126
By Pascal identity, 8C3 + 8C4 = 9C4. Compute 9C4 = 126. This is the same “add two above” rule of Pascal’s triangle.
Number of ways to choose 3 students from 8 and assign Captain, Vice-Captain, Leader
A 56
B 120
C 336
D 168
Roles make order important, so use permutations: 8P3 = 8×7×6 = 336. Same three students in different roles count as different outcomes.
A 3-member committee from 10 must include a fixed person P. Total ways
A 36
B 84
C 45
D 120
Person P is fixed in the committee. Choose remaining 2 members from the other 9 students: 9C2 = 36. Order does not matter in committees.
Number of ways to select 4 balls from 3 colors when each color can repeat
A 12
B 18
C 15
D 9
This is combinations with repetition: choose 4 items from 3 types. Count = (3+4−1)C4 = 6C4 = 15, using stars and bars logic.
Six people sit in a row. If A and B must sit together, total arrangements
A 240
B 120
C 480
D 720
Treat (A,B) as one block, so total units = 5. Arrange them in 5! ways, and A,B can swap inside the block in 2 ways. Total = 5!×2 = 240.
Six people sit in a row. If A and B must not sit together, total arrangements
A 240
B 360
C 480
D 600
Total arrangements = 6! = 720. Arrangements with A and B together = 240. So not together = 720 − 240 = 480, a standard restriction method.
From 1 to 100, count numbers not divisible by 2 or 3
A 34
B 35
C 36
D 33
Divisible by 2: 50, by 3: 33, by 6: 16. Divisible by 2 or 3 = 50+33−16=67. Not divisible by both = 100−67=33.
How many 5-digit numbers can be formed using digits 1–5 with repetition allowed
A 625
B 3125
C 120
D 1024
Each of the 5 positions can be filled in 5 ways and repetition is allowed. Total = 5^5 = 3125 by the fundamental counting principle.
How many 5-digit numbers (0–9 allowed) have last digit even, repetition allowed
A 50000
B 40000
C 45000
D 40500
First digit cannot be 0 → 9 choices. Middle three digits each have 10 choices → 10^3. Last digit even → 5 choices. Total = 9×1000×5 = 45000.
Coefficient of x^2 in (1 + 2x)^4 is
A 24
B 16
C 12
D 32
Choose x^2 term: coefficient = 4C2·(2x)^2 = 6·4 = 24. Binomial coefficient selects how many times 2x is taken in the product.
Coefficient of x^5 in (x + 1)^7 is
A 35
B 7
C 21
D 28
In (1+x)^7, coefficient of x^k is 7Ck. Here k=5, so coefficient = 7C5 = 21. Symmetry also gives 7C2 = 21.
Coefficient of x^7 in (1 − x)^9 is
A 36
B −84
C 84
D −36
Coefficient of x^7 is 9C7(−1)^7. Since 9C7 = 36 and the power is odd, sign becomes negative. So coefficient = −36.
After simplifying (x + 1/x)^6, number of distinct terms is
A 6
B 7
C 13
D 4
Powers become x^(6−2r) for r=0 to 6, giving exponents 6,4,2,0,−2,−4,−6. All are distinct, so there are 7 terms.
In (a+b+c)^4, coefficient of a^2bc is
A 24
B 6
C 12
D 4
Use multinomial coefficient: 4!/(2!1!1!) = 12. It counts how many ways to arrange the choices giving a^2 b^1 c^1 in the expansion.
Number of distinct permutations of the word BANANA is
A 60
B 120
C 30
D 90
BANANA has 6 letters with A repeated 3 times and N repeated 2 times. Distinct arrangements = 6!/(3!2!) = 720/12 = 60.
Number of distinct permutations of the word STATISTICS is
A 100800
B 25200
C 50400
D 40320
STATISTICS has 10 letters with S repeated 3, T repeated 3, I repeated 2. Distinct permutations = 10!/(3!3!2!) = 50400.
Number of ways to choose 2 items from 6 types when repetition is allowed
A 15
B 12
C 36
D 21
This is combinations with repetition: selecting 2 from 6 types equals (6+2−1)C2 = 7C2 = 21. It matches the stars and bars method.
Value of 12C10 is
A 120
B 12
C 66
D 55
Use symmetry: 12C10 = 12C2. Compute 12C2 = 12×11/2 = 66. Symmetry reduces heavy calculation quickly and safely.
Value of 7P5 is
A 840
B 2520
C 720
D 504
7P5 = 7!/(7−5)! = 7!/2! = 5040/2 = 2520. It counts ordered selections of 5 from 7 without repetition.
Coefficient of x^4 in (1+x)^9 is
A 84
B 36
C 126
D 210
In (1+x)^9, coefficient of x^4 is 9C4. Compute 9C4 = 126. This is direct coefficient extraction from the binomial theorem.
Value of Σ 3Cr·4C(3−r) for r=0 to 3 is
A 35
B 21
C 28
D 49
By Vandermonde’s identity, Σ (mCr)(nC(k−r)) = (m+n)Ck. Here m=3, n=4, k=3, so result = 7C3 = 35.
Compute directly: 4C0=1, 4C1=4, 4C2=6. Sum = 1+4+6 = 11. This also matches partial sums of binomial coefficients.
In (1+x)^6, sum of coefficients of odd powers of x is
A 32
B 64
C 31
D 16
For n≥1, sum of odd-power coefficients equals sum of even-power coefficients, each being 2^(n−1). For n=6, each sum is 2^5 = 32.
For small x, best first-order approximation of (1+x)^5 is
A 1+x^5
B 1+5x
C 5+x
D 1+25x
Binomial approximation uses first two terms: (1+x)^n ≈ 1 + nx when |x| is small. Here n=5, so approximation is 1+5x.
Coefficient of x^2 in (1 − x)^−3 is
A 3
B 9
C 6
D 12
(1−x)^−3 expands as Σ C(r+2,2) x^r for r≥0. For r=2, coefficient is C(4,2)=6. This is a basic negative binomial series idea.
Coefficient of x^2 in (1 + x)^(1/2) is
A 1/8
B −1/4
C 1/4
D −1/8
Using binomial series, coefficient of x^2 is (1/2 choose 2) = (1/2)(−1/2)/2 = −1/8. This is the standard second-term coefficient in √(1+x).
Number of solutions to x1+x2+x3=4 with 0≤xi≤2 is
A 9
B 12
C 6
D 15
Without upper limits, solutions are 6C2 = 15. Subtract cases where some xi≥3: for each variable there are 3 solutions, total 9. No overlaps possible, so 15−9=6.
How many distinct 4-digit arrangements can be formed using digits 1,1,2,2
A 6
B 12
C 4
D 8
Total arrangements of 4 items are 4!, but identical digits reduce distinct outcomes. With two 1’s and two 2’s, count = 4!/(2!2!) = 6.
If 7 different books are arranged on a shelf, in how many ways can 3 specific books be kept together
A 1440
B 360
C 720
D 840
Treat the 3 specific books as one block. Then there are 5 units to arrange: 5! ways. Inside the block, 3 books can permute in 3! ways. Total = 5!×3! = 120×6 = 720? Actually 120×6 = 720.
In a row of 8 distinct people including A and B, how many arrangements place A and B with exactly one person between them
A 4320
B 7200
C 1440
D 8640
Possible position pairs for A and B are (1,3) to (6,8), giving 6 pairs. A and B can swap (×2). Remaining 6 people arrange in 6!6!. Total 6×2×720=86406×2×720=8640.
Coefficient of x^3 in (x + 2)^6 is
A 240
B 320
C 160
D 80
x^3 comes from choosing x three times and 2 three times. Coefficient = 6C3 × 2^3 = 20×8 = 160? Actually 20×8 = 160