Chapter 4: Permutations, Combinations and Binomial Theorem (Set-4)
A school chooses President, Vice-President, and two different captains from 10 students, all roles distinct
A B. 210
B C. 720
C A. 5040
D D. 1260
Four distinct posts mean order matters. Count permutations: 10P4=10×9×8×7=504010P4=10×9×8×7=5040. Different students in different roles give different outcomes.
How many ways can 2 students be selected from 15 for a simple pair activity
A A. 105
B B. 120
C C. 90
D D. 60
A pair selection is unordered, so use combinations: 15C2=15×142=10515C2=215×14=105. Only the chosen two matter, not their order.
A librarian displays 5 different books chosen from 9 distinct books in a row, order matters
A B. 126
B C. 3024
C D. 945
D A. 15120
Choose and arrange 5 out of 9 distinct books: 9P5=9×8×7×6×5=151209P5=9×8×7×6×5=15120. Every different order is counted.
Number of distinct arrangements of letters in the word BALLOON
A B. 2520
B A. 1260
C C. 630
D D. 720
BALLOON has 7 letters with L repeated twice and O repeated twice. Distinct permutations are 7!2! 2!=50404=12602!2!7!=45040=1260.
How many distinct arrangements can be made from digits 1,1,1,2,2,3
A B. 120
B C. 30
C A. 60
D D. 90
There are 6 digits with 1 repeated 3 times and 2 repeated 2 times. Distinct permutations are 6!3! 2!=72012=603!2!6!=12720=60.
Seven people sit in a row and A and B must be on the two ends in any order
A A. 240
B B. 120
C C. 720
D D. 60
Place A and B at the ends in 2 ways. Arrange the remaining 5 people in the middle in 5!=1205!=120 ways. Total 2×120=2402×120=240.
Eight people sit in a row and person A is not allowed at either end
A B. 40320
B C. 20160
C D. 10080
D A. 30240
Total arrangements are 8!=403208!=40320. If A is at an end, there are 2 ends and 7!7! arrangements each, so subtract 2×7!=100802×7!=10080. Result 40320−10080=3024040320−10080=30240.
How many circular arrangements are possible for 7 distinct people around a round table
A B. 5040
B C. 1440
C A. 720
D D. 120
Circular permutations count rotations as same. For 7 distinct people, arrangements are (7−1)!=6!=720(7−1)!=6!=720. Fix one person to remove rotational duplicates.
Eight people sit around a round table and A must be exactly opposite B
A B. 5040
B A. 720
C C. 1440
D D. 360
Fix A to remove rotation. B has exactly one opposite seat. The remaining 6 people can be arranged in 6!=7206!=720 ways in the remaining seats.
Six people sit around a round table and A and B must sit together
A A. 48
B B. 24
C C. 72
D D. 120
Treat A and B as one block, giving 5 units around a circle. Circular arrangements are (5−1)!=24(5−1)!=24. Inside the block, AB or BA gives 2 ways. Total 24×2=4824×2=48.
How many 3-digit numbers (no repetition) formed from 0–9 are divisible by 5
A B. 144
B C. 150
C A. 136
D D. 120
Last digit must be 0 or 5. If last is 0: first has 9 choices, middle 8 → 72. If last is 5: first has 8 choices, middle 8 → 64. Total 72+64=13672+64=136.
Number of 6-letter codes from A,B,C,D,E with repetition allowed and exactly two A’s
A B. 2560
B C. 1920
C D. 1536
D A. 3840
Choose positions for two A’s: 6C2=156C2=15. Remaining 4 positions can use B,C,D,E (4 choices each): 44=25644=256. Total 15×256=384015×256=3840.
How many length-4 strings can be formed using digits 0,1,2,3 without repetition
A A. 24
B B. 16
C C. 12
D D. 36
Four positions filled with 4 distinct digits without repetition gives 4P4=4!=244P4=4!=24. Every different order creates a different string.
From 8 people standing in a line, how many ways to choose 3 so that no two chosen are adjacent
A B. 28
B C. 16
C A. 20
D D. 18
For choosing r non-adjacent from n in a line, count is C(n−r+1,r)C(n−r+1,r). Here C(8−3+1,3)=C(6,3)=20C(8−3+1,3)=C(6,3)=20.
Count integers from 1 to 300 that are divisible by 7 or 11
A B. 69
B A. 66
C C. 63
D D. 70
Multiples of 7: ⌊300/7⌋=42. Multiples of 11: ⌊300/11⌋=27. Common multiples of 77: ⌊300/77⌋=3. Total 42+27−3=6642+27−3=66.
Count integers from 1 to 200 that are not divisible by 4 or 6
A B. 120
B C. 140
C D. 150
D A. 133
Divisible by 4: 50. Divisible by 6: 33. Divisible by lcm 12: 16. Divisible by 4 or 6: 50+33−16=6750+33−16=67. Not divisible: 200−67=133200−67=133.
Number of nonnegative integer solutions of x + y + z = 10
A B. 55
B C. 45
C A. 66
D D. 78
Using stars and bars for nonnegative solutions: (10+3−1)C(3−1)=12C2=66(10+3−1)C(3−1)=12C2=66. It counts ways to distribute 10 identical units among 3 variables.
Number of positive integer solutions of a + b + c + d = 12
A A. 165
B B. 220
C C. 120
D D. 84
For positive solutions, set a′=a−1a′=a−1 etc. Then a′+b′+c′+d′=8a′+b′+c′+d′=8 with nonnegative integers. Count (8+4−1)C3=11C3=165(8+4−1)C3=11C3=165.
Number of solutions of x1+x2+x3+x4 = 12 where each xi ≥ 2
A B. 84
B A. 35
C C. 56
D D. 28
Put yi=xi−2yi=xi−2. Then y1+y2+y3+y4=4y1+y2+y3+y4=4 with yi≥0yi≥0. Count (4+4−1)C3=7C3=35(4+4−1)C3=7C3=35.
Number of solutions of x + y + z = 8 with 0 ≤ x,y,z ≤ 4
A B. 18
B C. 21
C A. 15
D D. 12
Total nonnegative solutions are 10C2=4510C2=45. Subtract solutions with a variable ≥5. For one variable ≥5, reduce by 5: solutions 5C2=105C2=10. Three variables → 30. No overlaps possible, so 45−30=1545−30=15.
Number of derangements of 5 distinct objects
A A. 44
B B. 45
C C. 120
D D. 24
A derangement means no item stays in its original position. The standard value is !5=44!5=44, obtained using inclusion–exclusion or known derangement sequence values.
From 5 red and 3 blue balls, number of ways to choose 2 balls with at least one blue
A B. 10
B C. 28
C D. 15
D A. 18
Total ways to choose 2 from 8 is 8C2=288C2=28. Ways with no blue means both red: 5C2=105C2=10. So at least one blue: 28−10=1828−10=18.
A committee of 4 is chosen from 5 men and 4 women, with at least 2 women
A B. 75
B C. 85
C A. 81
D D. 90
Count women 2,3,4. 4C2⋅5C2=604C2⋅5C2=60, 4C3⋅5C1=204C3⋅5C1=20, 4C4⋅5C0=14C4⋅5C0=1. Total 60+20+1=8160+20+1=81.
Number of ways to get exactly 4 successes in 9 independent trials
A B. 84
B A. 126
C C. 210
D D. 72
Choose which 4 trials are successes. That count is 9C4=1269C4=126. This is the basic binomial counting factor used in binomial probability.
Coefficient of x^5 in (1 + x)^12
A A. 792
B B. 495
C C. 220
D D. 924
In (1+x)n(1+x)n, coefficient of xkxk is nCknCk. Here it is 12C5=79212C5=792. This comes directly from the binomial theorem.
Coefficient of x^9 in (1 + x)^12
A B. 495
B C. 792
C A. 220
D D. 165
Coefficient of x9x9 is 12C912C9. By symmetry, 12C9=12C3=22012C9=12C3=220. Symmetry reduces calculation and avoids mistakes.
Coefficient of x^4 in (2x − 3)^7
A B. 15120
B C. −7560
C D. 7560
D A. −15120
In (2x−3)7(2x−3)7, x-power is 7−r7−r. For x4x4, take r=3r=3. Coefficient =7C3⋅24⋅(−3)3=35⋅16⋅(−27)=−15120=7C3⋅24⋅(−3)3=35⋅16⋅(−27)=−15120.
Coefficient of x^6 in (x^2 + 1/x)^6
A B. 20
B A. 15
C C. 6
D D. 10
General power is x2(6−r)−r=x12−3rx2(6−r)−r=x12−3r. Set 12−3r=612−3r=6 gives r=2r=2. Coefficient is 6C2=156C2=15.
Coefficient of x^9 in (x^2 + 1/x)^6
A A. 6
B B. 15
C C. 20
D D. 30
Power is x12−3rx12−3r. For x9x9, solve 12−3r=912−3r=9 giving r=1r=1. Coefficient equals 6C1=66C1=6 because other factors are 1.
Constant term in (2x − 3/x)^8
A B. 45360
B C. 70560
C A. 90720
D D. 80640
Power becomes x8−2rx8−2r. For constant term, 8−2r=08−2r=0 so r=4r=4. Coefficient =8C4⋅24⋅(−3)4=70⋅16⋅81=90720=8C4⋅24⋅(−3)4=70⋅16⋅81=90720.
Term independent of x in (x^3 + 1/x)^8 has coefficient
A B. 56
B C. 70
C D. 35
D A. 28
Power is x3(8−r)−r=x24−4rx3(8−r)−r=x24−4r. Set exponent 0 gives r=6r=6. The coefficient is 8C6=288C6=28, since no extra numeric multiplier exists.
Number of distinct terms after simplifying (x + 1/x)^9
A A. 10
B B. 9
C C. 8
D D. 11
Exponents are x9−2rx9−2r for r=0r=0 to 9, giving 10 different powers: 9,7,5,…,−9. Each is distinct, so there are 10 terms.
In (a+b)^9, sum of coefficients of terms with odd powers of b
A B. 512
B C. 128
C A. 256
D D. 1024
For any n≥1n≥1, the sum of coefficients of odd-power terms equals the sum of even-power terms, each 2n−12n−1. For n=9n=9, it is 28=25628=256.
Sum of coefficients of even powers of x in (1+x)^10
A B. 1024
B A. 512
C C. 256
D D. 768
Even-power coefficient sum equals (1+1)10+(1−1)102=210+02=29=5122(1+1)10+(1−1)10=2210+0=29=512. This uses standard binomial parity property.
Value of Σ r·10Cr from r=0 to 10
A B. 2560
B C. 10240
C D. 4096
D A. 5120
Identity: ∑r⋅nCr=n⋅2n−1∑r⋅nCr=n⋅2n−1. For n=10n=10, value is 10⋅29=10⋅512=512010⋅29=10⋅512=5120.
Value of Σ (8Cr)^2 from r=0 to 8
A B. 6435
B C. 17160
C A. 12870
D D. 11440
Identity: ∑r=0n(nCr)2=(2nCn)∑r=0n(nCr)2=(2nCn). For n=8n=8, sum =16C8=12870=16C8=12870. It follows from coefficient matching in (1+x)n(1+x)n(1+x)n(1+x)n.
Coefficient of x^4 in (1+x)^5(1+x)^7
A A. 495
B B. 792
C C. 220
D D. 330
Multiply to get (1+x)12(1+x)12. Coefficient of x4x4 is 12C4=49512C4=495. This is a direct application of adding exponents in same-base products.
Coefficient of x^5 in (1−x)^4(1+x)^6
A B. −12
B A. 12
C C. 24
D D. 6
Write as (1−x)4(1+x)4(1+x)2=(1−x2)4(1+x)2(1−x)4(1+x)4(1+x)2=(1−x2)4(1+x)2. Only even powers from (1−x2)4(1−x2)4. Need x4x4 term (coefficient 6) and x1x1 from (1+x)2(1+x)2 (coefficient 2). Product 6×2=126×2=12.
Coefficient of a^3 b^2 c in (a+b+c)^6
A B. 90
B C. 120
C D. 30
D A. 60
Use multinomial coefficient: 6!3! 2! 1!=7206⋅2=603!2!1!6!=6⋅2720=60. It counts distinct arrangements of the selected powers.
Coefficient of x^3y^2 in (x+2y)^5
A B. 20
B C. 80
C A. 40
D D. 10
Choose 2 factors of 2y2y and 3 factors of x. Coefficient =5C2⋅22=10⋅4=40=5C2⋅22=10⋅4=40. Remaining power automatically becomes x3y2x3y2.
Number of different terms in expansion of (a+b+c)^4
A A. 15
B B. 12
C C. 10
D D. 20
Different terms correspond to nonnegative solutions of i+j+k=4i+j+k=4. Count is (4+3−1)C2=6C2=15(4+3−1)C2=6C2=15. Each solution gives a unique monomial.
Ways to distribute 6 identical candies among 4 children, each gets at least one
A B. 20
B A. 10
C C. 15
D D. 35
Let each child get 1 first, remaining 6−4=26−4=2 candies are distributed freely. Nonnegative solutions to x1+x2+x3+x4=2×1+x2+x3+x4=2 are (2+4−1)C3=5C3=10(2+4−1)C3=5C3=10.
Ways to distribute 6 identical candies among 4 children with no minimum condition
A B. 56
B C. 70
C A. 84
D D. 90
Nonnegative solutions of x1+x2+x3+x4=6×1+x2+x3+x4=6 are (6+4−1)C3=9C3=84(6+4−1)C3=9C3=84. This is the standard stars and bars result.
Nine people sit in a row and three specific people must sit together as a group
A A. 30240
B B. 15120
C C. 60480
D D. 10080
Treat the three specific people as one block. Then there are 7 units to arrange in a row: 7!7!. Inside the block, the 3 people can permute in 3!3!. Total 7!×3!=5040×6=302407!×3!=5040×6=30240.
Number of ways to choose 4 hearts from a standard deck of cards
A B. 286
B C. 1820
C D. 495
D A. 715
Hearts are 13 cards. Choosing 4 hearts is an unordered selection: 13C4=71513C4=715. This is a direct application of combinations in probability counting.
Number of diagonals in a 15-sided polygon
A B. 105
B A. 90
C C. 75
D D. 120
Any diagonal is formed by choosing 2 vertices, excluding sides. Total pairs 15C2=10515C2=105. Subtract 15 sides gives 105−15=90105−15=90 diagonals.
From 7 distinct items including special A and B, number of ways to choose 3 items with at most one of A or B
A B. 35
B C. 25
C A. 30
D D. 20
Total selections 7C3=357C3=35. Subtract selections containing both A and B. If both chosen, choose 1 more from remaining 5: 5C1=55C1=5. So 35−5=3035−5=30.
Number of distinct arrangements of letters in the word SUCCESS
A B. 840
B C. 210
C D. 1260
D A. 420
SUCCESS has 7 letters with S repeated 3 times and C repeated 2 times. Distinct permutations are 7!3! 2!=504012=4203!2!7!=125040=420.
Coefficient of x^10 in (1+x)^15
A A. 3003
B B. 1365
C C. 5005
D D. 2002
Coefficient of x10x10 is 15C1015C10. By symmetry 15C10=15C515C10=15C5. Compute 15C5=300315C5=3003. This is standard coefficient extraction from binomial expansion.
If the sum of coefficients of (1+x)^n is 1024, then n equals
A B. 9
B A. 10
C C. 8
D D. 11
Sum of coefficients of (1+x)n(1+x)n equals (1+1)n=2n(1+1)n=2n. Given 2n=1024=2102n=1024=210, so n=10n=10.