Chapter 4: Permutations, Combinations and Binomial Theorem (Set-5)
In permutations of numbers 1 to 7, the order must satisfy 1 before 2 before 3
A 720
B 1260
C 840
D 1680
Total permutations are 7!. Among 1,2,3 all 3! relative orders are equally likely, and only one works. So required count is 7!/3!=8407!/3!=840.
Ten people stand in a row, and three specific people must be together but the group cannot be at either end
A 181440
B 151200
C 201600
D 241920
Treat the three as one block, giving 8 units. Total arrangements 8!×3!8!×3!. Block at an end: 2×7!×3!2×7!×3!. Subtract: (8!−2⋅7!)3!=181440(8!−2⋅7!)3!=181440.
Number of ways to choose 4 items from 10 arranged in a line so that no two chosen are consecutive
A 45
B 35
C 50
D 60
For choosing r non-adjacent items from n in a line, use C(n−r+1,r)C(n−r+1,r). Here C(10−4+1,4)=C(7,4)=35C(10−4+1,4)=C(7,4)=35.
From 1 to 1000, how many integers are divisible by 4 or 9
A 341
B 327
C 360
D 334
Multiples of 4: 250. Multiples of 9: 111. Multiples of lcm 36: 27. Using inclusion–exclusion: 250+111−27=334250+111−27=334.
Number of solutions of x1+x2+x3+x4+x5=12 where each xi is between 0 and 4 inclusive
A 320
B 280
C 300
D 350
Total nonnegative solutions: C(16,4)=1820C(16,4)=1820. Subtract cases where some xi≥5: 5⋅C(11,4)=16505⋅C(11,4)=1650. Add back two variables ≥5: C(5,2)C(6,4)=150C(5,2)C(6,4)=150. Result 320320.
For nonnegative integers x,y,z with x even, count solutions of x+y+z=10
A 30
B 40
C 36
D 45
Let x=2kx=2k with k=0k=0 to 5. Then y+z=10−2ky+z=10−2k has 10−2k+110−2k+1 solutions. Sum 11+9+7+5+3+1=3611+9+7+5+3+1=36.
Coefficient of x^6 in (1+x)^9(1−x)^5
A 45
B 35
C 40
D 55
Write (1+x)9(1−x)5=(1−x2)5(1+x)4(1+x)9(1−x)5=(1−x2)5(1+x)4. Only even powers come from (1−x2)5(1−x2)5. Combine matching degrees to get coefficient −5+60−10=45−5+60−10=45.
Coefficient of x^7 in (1+x)^12(1−x)^5
A 144
B 192
C 220
D 176
Convert to (1−x2)5(1+x)7(1−x2)5(1+x)7. To get x^7, pair x2kx2k with x7−2kx7−2k. Sum contributions for k=0 to 3: 1−105+350−70=1761−105+350−70=176.
Coefficient of x^10 in (1−x)^6(1+x)^14
A 572
B −462
C −572
D 462
Rewrite as (1−x2)6(1+x)8(1−x2)6(1+x)8. Only even powers from (1−x2)6(1−x2)6. Add coefficients: ∑6Ck(−1)k⋅8C(10−2k)∑6Ck(−1)k⋅8C(10−2k) for valid k gives −572.
Value of Σ r·8Cr from r=0 to 8
A 1024
B 512
C 768
D 2048
Identity: ∑r(nr)=n2n−1∑r(rn)=n2n−1. For n=8, value is 8⋅27=8⋅128=10248⋅27=8⋅128=1024. This comes from differentiating (1+x)n(1+x)n.
Value of Σ r(r−1)·10Cr from r=0 to 10
A 11520
B 23040
C 46080
D 5120
Identity: ∑r(r−1)(nr)=n(n−1)2n−2∑r(r−1)(rn)=n(n−1)2n−2. For n=10, this becomes 10⋅9⋅28=90⋅256=2304010⋅9⋅28=90⋅256=23040.
Coefficient of x^4 in (2+3x)^9
A 332640
B 295680
C 326592
D 362880
Choose x from 4 factors: coefficient =9C4⋅25⋅34=9C4⋅25⋅34. That is 126⋅32⋅81=326592126⋅32⋅81=326592. Binomial theorem gives this directly.
Constant term in (x − 1/x)^12
A 924
B 792
C 462
D 252
General term is (12r)(−1)rx12−2r(r12)(−1)rx12−2r. Constant term needs 12−2r=0⇒r=612−2r=0⇒r=6. Coefficient becomes (126)=924(612)=924.
Constant term in (2x^2 + 3/x)^9
A 362880
B 453600
C 504000
D 489888
General exponent is 2(9−r)−r=18−3r2(9−r)−r=18−3r. Set to 0 gives r=6. Coefficient =9C6⋅23⋅36=84⋅8⋅729=489888=9C6⋅23⋅36=84⋅8⋅729=489888.
Coefficient of x^4 in (1 + x + x^2)^5
A 40
B 45
C 50
D 55
It counts 5-tuples with entries 0,1,2 summing to 4. Without cap: C(8,4)=70C(8,4)=70. Subtract cases where one entry ≥3: 5⋅C(5,4)=255⋅C(5,4)=25. So 45.
Number of shortest lattice paths from (0,0) to (5,5) that never go above the line y=x
A 35
B 50
C 42
D 70
Such paths are counted by the Catalan number C5=16(105)C5=61(510). Since (105)=252(510)=252, result is 252/6=42252/6=42.
Number of shortest paths from (0,0) to (6,4) that never go above y=x
A 90
B 70
C 80
D 100
Use the ballot-path formula for m≥n: m−n+1m+1(m+nn)m+1m−n+1(nm+n). Here m=6, n=4, so 37(104)=37⋅210=9073(410)=73⋅210=90.
Number of permutations of 6 distinct objects having exactly 2 fixed points
A 90
B 120
C 135
D 150
Choose which 2 are fixed: 6C2=156C2=15. Remaining 4 must be deranged. Number of derangements of 4 is 9. Total 15×9=13515×9=135.
Ways to arrange 5 men and 5 women in a row so that no two women sit together
A 72000
B 90000
C 100800
D 86400
Arrange 5 men in 5!5! ways. This creates 6 gaps. Choose 5 gaps for women: 6C5=66C5=6. Arrange women in 5!5! ways. Total 5!⋅6⋅5!=864005!⋅6⋅5!=86400.
Number of 6-digit nondecreasing strings made using digits 0 to 9
A 5005
B 3003
C 8008
D 10010
A nondecreasing string corresponds to choosing 6 items from 10 digits with repetition allowed. Count is (10+6−1)C6=15C6=5005(10+6−1)C6=15C6=5005.
Ways to choose 5 numbers from 1 to 12 such that no two chosen numbers are consecutive
A 42
B 56
C 70
D 84
Non-adjacent selection count is C(n−r+1,r)C(n−r+1,r). Here C(12−5+1,5)=C(8,5)=56C(12−5+1,5)=C(8,5)=56. This ensures at least one gap between chosen numbers.
Value of Σ (−1)^r · (10Cr)^2 from r=0 to 10
A 252
B 0
C −252
D 924
Identity: ∑(−1)r(nr)2=0∑(−1)r(rn)2=0 for odd n, and =(−1)n/2(nn/2)=(−1)n/2(n/2n) for even n. For n=10, it is −(105)=−252−(510)=−252.
Number of 5-card selections from a 52-card deck that contain exactly 2 aces
A 103776
B 82368
C 116280
D 124740
Choose 2 aces from 4: 4C2=64C2=6. Choose remaining 3 cards from the 48 non-aces: 48C3=1729648C3=17296. Multiply: 6×17296=1037766×17296=103776.
In a circle of 9 distinct people, A,B,C must sit consecutively in the fixed order A-B-C
A 360
B 720
C 1440
D 5040
Treat A-B-C as one fixed block. Then there are 7 units around the circle. Circular arrangements are (7−1)!=6!=720(7−1)!=6!=720. No internal permutation is allowed.
Count of distinct arrangements of the word MATHEMATICS
A 2494800
B 9979200
C 3991680
D 4989600
MATHEMATICS has 11 letters with M, A, T each repeated twice. Distinct permutations =11!2!2!2!=399168008=4989600=2!2!2!11!=839916800=4989600.
Number of solutions of x1+x2+x3=15 where each xi is a multiple of 3 and nonnegative
A 15
B 18
C 21
D 28
Let xi=3yixi=3yi. Then y1+y2+y3=5y1+y2+y3=5 with nonnegative integers. Solutions are (5+3−1)C2=7C2=21(5+3−1)C2=7C2=21.
Number of 5-letter codes from A,B,C,D,E,F with exactly two vowels, where vowels are A and E only
A 2560
B 1280
C 1920
D 3200
Choose vowel positions: 5C2=105C2=10. Fill vowels: 22=422=4. Remaining 3 positions use 4 consonants with repetition: 43=6443=64. Total 10⋅4⋅64=256010⋅4⋅64=2560.
In permutations of 9 distinct people in a row, A must be left of B, B left of C, and C left of D
A 30240
B 15120
C 60480
D 90720
Out of the 4 people A,B,C,D, all 4! relative orders are equally likely in 9!. Only one order satisfies A
Value of Σ (−1)^r · 5Cr /(r+1) from r=0 to 5
A 1/5
B 1/7
C 1/6
D 1/8
This sum equals ∫01(1−x)5dx∫01(1−x)5dx. The integral is 1661. So the alternating combination fraction sum evaluates to 1/6.
Coefficient of x^5 in (1+x)^{10}(1+x^2)^3
A 642
B 612
C 672
D 702
Expand (1+x2)3=1+3×2+3×4+x6(1+x2)3=1+3×2+3×4+x6. For x^5: 10C5⋅1+3⋅10C3+3⋅10C1=252+360+30=64210C5⋅1+3⋅10C3+3⋅10C1=252+360+30=642.
Coefficient of x^8 in (1+x)^{10}(1−x)^4
A 21
B 25
C 30
D 27
Write as (1−x2)4(1+x)6(1−x2)4(1+x)6. Add contributions for even parts: −4⋅6C6+6⋅6C4−4⋅6C2+1⋅6C0=−4+90−60+1=27−4⋅6C6+6⋅6C4−4⋅6C2+1⋅6C0=−4+90−60+1=27.
Number of 4-element subsets from {1,2,…,20} having an even sum
A 2445
B 2240
C 2560
D 2625
Even sum occurs with 0,2,4 odd numbers. There are 10 odds and 10 evens. Count 10C4+10C2⋅10C2+10C4=210+2025+210=244510C4+10C2⋅10C2+10C4=210+2025+210=2445.
Number of permutations of ABCDEFG where A and E are not adjacent
A 2880
B 4320
C 3600
D 5040
Total permutations 7!=50407!=5040. Adjacent A and E treated as one block: 6!×2=14406!×2=1440. Not adjacent =5040−1440=3600=5040−1440=3600.
Coefficient of x^0 in (1+x)^8(1+1/x)^6
A 2002
B 3003
C 3432
D 4004
(1+1/x)6=x−6(1+x)6(1+1/x)6=x−6(1+x)6. Product becomes x−6(1+x)14x−6(1+x)14. Constant term needs x6x6 from (1+x)14(1+x)14, coefficient 14C6=300314C6=3003.
Coefficient of x^5 in (x^2 + x + 1)^4
A 12
B 14
C 18
D 16
Need 2a+b=52a+b=5 with a+b+c=4a+b+c=4, where a picks x^2, b picks x, c picks 1. Solutions: (a=1,b=3,c=0) gives 4 ways; (a=2,b=1,c=1) gives 12 ways. Total 16.
In a circle of 10 distinct people, A and B must not sit next to each other
A 282240
B 241920
C 302400
D 362880
Total circular arrangements: 9!=3628809!=362880. Adjacent A,B: treat as one block, giving 9 units in a circle → (9−1)!=8!=40320(9−1)!=8!=40320, times 2 orders = 80640. Subtract: 282240.
Number of shortest paths from (0,0) to (7,3) that pass through point (4,1)
A 40
B 45
C 50
D 60
Paths factor through the point. From (0,0) to (4,1): 5 steps with 1 up → 5C1=55C1=5. From (4,1) to (7,3): 5 steps with 2 ups → 5C2=105C2=10. Total 50.
Number of permutations of 9 distinct objects where 1 and 2 occupy positions of the same parity (both odd or both even)
A 141120
B 161280
C 151200
D 181440
Odd positions are 5, even positions are 4. Both in odd: 5P2⋅7!=20⋅5040=1008005P2⋅7!=20⋅5040=100800. Both in even: 4P2⋅7!=12⋅5040=604804P2⋅7!=12⋅5040=60480. Sum 161280.
Number of multisets of size 6 formed from 4 different types of objects
A 84
B 56
C 70
D 126
Choosing 6 items from 4 types with repetition allowed equals the number of nonnegative solutions of x1+x2+x3+x4=6×1+x2+x3+x4=6. Count is (4+6−1)C6=9C6=84(4+6−1)C6=9C6=84.
Number of 3-element subsets of {1,2,…,10} such that max − min = 4
A 12
B 15
C 18
D 21
Let min = a and max = a+4. Then a can be 1 to 6. The third number must be one of a+1, a+2, a+3 (3 choices). Total 6×3=186×3=18.
Coefficient of x^3 in (1−2x)^{−4}
A 80
B 120
C 200
D 160
For (1−t)−m(1−t)−m, coefficient of trtr is C(m+r−1,r)C(m+r−1,r). Here t=2xt=2x, m=4, r=3. Coefficient is C(6,3)⋅23=20⋅8=160C(6,3)⋅23=20⋅8=160.
Coefficient of x^3 in (1+2x)^{1/2}
A −1/2
B 1/2
C −3/8
D 3/8
Coefficient of x3x3 is (1/23)(23)(31/2)(23). (1/23)=(1/2)(−1/2)(−3/2)6=116(31/2)=6(1/2)(−1/2)(−3/2)=161. Multiply by 8 gives 1/21/2.
Coefficient of x^4 in (1+3x)^{1/3}
A −10/3
B 10/3
C −5/3
D 5/3
Coefficient is (1/34)34(41/3)34. Compute (1/34)=(1/3)(−2/3)(−5/3)(−8/3)24=−10243(41/3)=24(1/3)(−2/3)(−5/3)(−8/3)=−24310. Multiply by 8181 gives −10/3−10/3.
Constant term in (x^2 + 1/x)^9
A 70
B 126
C 84
D 210
General term: (9r)x2(9−r)x−r=(9r)x18−3r(r9)x2(9−r)x−r=(r9)x18−3r. Constant term needs 18−3r=0⇒r=618−3r=0⇒r=6. Coefficient is 9C6=849C6=84.
A committee of 5 is formed from 6 women and 8 men, and it must contain exactly 3 women
A 560
B 420
C 640
D 720
Choose 3 women from 6: 6C3=206C3=20. Choose 2 men from 8: 8C2=288C2=28. Multiply independent choices: 20×28=56020×28=560.
Number of diagonals in a 20-sided polygon
A 150
B 160
C 180
D 170
Total pairs of vertices: 20C2=19020C2=190. Subtract the 20 sides (not diagonals). So diagonals =190−20=170=190−20=170. Each diagonal connects two non-adjacent vertices.
Coefficient of x^8 in (1+x)^8(1−x)^8
A 0
B 70
C 28
D 105
Product is (1−x2)8(1−x2)8. The term x8x8 means (x2)4(x2)4, so take k=4. Coefficient =(84)(−1)4=70=(48)(−1)4=70.
Coefficient of x^5 in (1+x)^{12} equals coefficient of x^{r+2} in the same expansion. Find r
A 4
B 6
C 5
D 7
Need (12r)=(12r+2)(r12)=(r+212). Binomial coefficients match when indices are symmetric: r=12−(r+2)r=12−(r+2). Solving gives 2r=10⇒r=52r=10⇒r=5.
Coefficient of x^10 in (1+x)^{12}
A 66
B 220
C 792
D 924
Coefficient of x10x10 in (1+x)12(1+x)12 is (1210)(1012). By symmetry, (1210)=(122)=12⋅112=66(1012)=(212)=212⋅11=66. (So coefficient is 66.)
In (2x − 3)^7, the coefficient of x^6 is
A 1344
B −672
C −1344
D 672
For x6x6, choose (2x)(2x) from 6 factors and (−3)(−3) from 1 factor. Coefficient =7C1⋅26⋅(−3)=7⋅64⋅(−3)=−1344=7C1⋅26⋅(−3)=7⋅64⋅(−3)=−1344