JEE Mathematics Permutations and Combinations MCQs Set 17

Answer: (b) 6

Explanation:
1. \( \frac{{}^nC_r}{{}^nC_{r-1}}=\frac{n-r+1}{r}=\frac{28}{56}=\frac12 \) gives \( n=\frac{3r-2}{2} \).
2. \( \frac{{}^nC_{r+1}}{{}^nC_r}=\frac{n-r}{r+1}=\frac{8}{28}=\frac27 \) gives \( 7n=9r+2 \).
3. Substituting: \( \frac{7(3r-2)}{2}=9r+2 \), so 21r − 14 = 18r + 4 and r = 6 (then n = 8).

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. r
10. a
11. t
12. i
13. o
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15. o
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17.
18. c
19. o
20. n
21. s
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23. c
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26. i
27. v
28. e
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30. b
31. i
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33. o
34. m
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53. o
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66. n
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70.
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81.
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85. c
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87.
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91. h
92.
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94.
95. =
96.
97. 8
98. :
99.
100. 5
101. 6
102. ,
103.
104. 2
105. 8
106. ,
107.
108. 8
109. .

Answer: (c) 6

Explanation:
1. Arrangements of 4 from n: \( {}^nP_4=n(n-1)(n-2)(n-3) \); of 2: \( {}^nP_2=n(n-1) \).
2. So \( (n-2)(n-3)=12 \), i.e. \( n^2-5n-6=0 \), giving n = 6.

Teacher's Note:
1. C
2. a
3. n
4. c
5. e
6. l
7.
8. t
9. h
10. e
11.
12. c
13. o
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18.
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20. a
21. c
22. t
23. o
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25.
26. n
27. (
28. n
29.
30. −
31.
32. 1
33. )
34.
35. a
36. n
37. d
38.
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40. o
41. l
42. v
43. e
44.
45. t
46. h
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48.
49. q
50. u
51. a
52. d
53. r
54. a
55. t
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57. c
58. .
59.
60. R
61. e
62. j
63. e
64. c
65. t
66.
67. t
68. h
69. e
70.
71. n
72. e
73. g
74. a
75. t
76. i
77. v
78. e
79.
80. r
81. o
82. o
83. t
84. .

Answer: (a) 13

Explanation:
1. \( {}^nC_5={}^nC_8 \) means \( 5+8=n \), so n = 13.

Teacher's Note:
1. I
2. f
3.
4. n
5. C
6. a
7.
8. =
9.
10. n
11. C
12. b
13.
14. w
15. i
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17. h
18.
19. a
20.
21. ≠
22.
23. b
24.
25. t
26. h
27. e
28. n
29.
30. n
31.
32. =
33.
34. a
35.
36. +
37.
38. b
39. .

Answer: (d) 11

Explanation:
1. We need p < q. For q = 2, 3, 4, 5, 6 the pairs are 1, 2, 3, 4, 5 in number, a total of 15.
2. Remove repeats of the same value (2/4 = 1/2, 2/6 = 1/3, 3/6 = 1/2, 4/6 = 2/3). The distinct values are 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6.
3. That gives 11 distinct values.

Teacher's Note:
1. C
2. o
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7. f
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11. t
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23.
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25. e
26. s
27. s
28.
29. t
30. h
31. a
32. n
33.
34. q
35. ,
36.
37. t
38. h
39. e
40. n
41.
42. r
43. e
44. m
45. o
46. v
47. e
48.
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50. h
51. o
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53. e
54.
55. t
56. h
57. a
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59.
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61. e
62. d
63. u
64. c
65. e
66.
67. t
68. o
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73.
74. s
75. a
76. m
77. e
78.
79. v
80. a
81. l
82. u
83. e
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85.
86. A
87.
88. w
89. r
90. i
91. t
92. t
93. e
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95.
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100.
101. a
102. v
103. o
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105. d
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107.
108. m
109. i
110. s
111. t
112. a
113. k
114. e
115. s
116. .

Answer: (c) 9(9!)

Explanation:
1. The first digit can be any of 9 non-zero digits.
2. The remaining 8 places can be filled in \( 9\times8\times\dots\times2=9! \) ways (the remaining 9 digits, including 0, in order).
3. Total = 9 × 9! = 9(9!).

Teacher's Note:
1. F
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3. x
4.
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17. d
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30. s
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33. c
34. e
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38.
39. c
40. a
41. n
42. n
43. o
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45.
46. b
47. e
48.
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50. .
51.
52. T
53. h
54. e
55. n
56.
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58. r
59. r
60. a
61. n
62. g
63. e
64.
65. t
66. h
67. e
68.
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75. r
76. o
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85. m
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87. i
88. n
89. i
90. n
91. g
92.
93. d
94. i
95. g
96. i
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98. s
99. .

Answer: (a) 24

Explanation:
1. The odd places are 1st, 3rd and 5th; the even digits 0, 2, 4 must occupy them. The first place cannot be 0.
2. Choose the first place: 2 ways (2 or 4); the other two odd places take the remaining 2 even digits in 2! = 2 ways, so 4 ways.
3. The three even places take 1, 3, 5 in 3! = 6 ways. Total = 4 × 6 = 24.

Teacher's Note:
1. H
2. a
3. n
4. d
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7.
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9. h
10. e
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12. r
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40. e
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50.
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52. h
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54. n
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58. r
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60. n
61. g
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79. i
80. g
81. i
82. t
83. s
84.
85. f
86. r
87. e
88. e
89. l
90. y
91. .

Answer: (b) \( {}^4P_4\times{}^3P_3 \)

Explanation:
1. Treat the three particular men as one block: 4 objects can be arranged in 4! = \( {}^4P_4 \) ways.
2. The three men can arrange among themselves in \( 3!={}^3P_3 \) ways.
3. Total = \( {}^4P_4\times{}^3P_3 \).

Teacher's Note:
1. G
2. l
3. u
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49. ,
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52. h
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57. o
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87. h
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89.
90. b
91. l
92. o
93. c
94. k
95. .

Answer: (d) none of these

Explanation:
1. Among the three positions occupied by A, B, C only one of the 3! orders is allowed, so the required number is \( \frac{7!}{3!}=840 \).
2. This is not any of the listed values.

Teacher's Note:
1. O
2. u
3. t
4.
5. o
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7.
8. a
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10. l
11.
12. a
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27.
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29. i
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31. e
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35. r
36. d
37. e
38. r
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42.
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79. /
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81. !
82.
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90. .

Answer: (d) none of these

Explanation:
1. If the middle digit is n (4 ≤ n ≤ 9), choose the other 4 digits from the n smaller digits 0 to n − 1 in \( {}^nP_4 \) ways, which includes cases with 0 first.
2. Cases with 0 in the first place: choose the other 3 from the remaining n − 1 digits: \( {}^{n-1}P_3 \) ways.
3. Total = \( \sum_{n=4}^9\left({}^nP_4-{}^{n-1}P_3\right)=6048-756=5292 \), which matches none of the options a to c.

Teacher's Note:
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34. l
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108. .

Answer: (a) 216

Explanation:
1. Divisibility by 3 needs a digit sum divisible by 3. The sum of all six digits is 15, so the omitted digit must be 0 or 3.
2. Omit 0: digits 1, 2, 3, 4, 5 give 5! = 120 numbers. Omit 3: digits 0, 1, 2, 4, 5 give 5! − 4! = 96 numbers.
3. Total = 120 + 96 = 216.

Teacher's Note:
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117. .

Answer: (c) 91

Explanation:
1. A has 10 elements: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
2. Ordered pairs (p, q) with p ≠ q give distinct fractions: 10 × 9 = 90.
3. All pairs with p = q give the single value 1, adding 1. Total = 91.

Teacher's Note:
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Answer: (c) 35

Explanation:
1. Place the six '+' signs first; this creates 7 gaps (including the ends).
2. Choose 4 gaps for the '−' signs: \( {}^7C_4=35 \).

Teacher's Note:
1. P
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Answer: (a) 36

Explanation:
1. Divisibility by 4 depends on the last two digits. From 1, 2, 3, 5, 6 the valid endings are 12, 16, 32, 36, 52, 56 — six endings.
2. The first two places are filled from the remaining 3 digits in \( {}^3P_2=6 \) ways.
3. Total = 6 × 6 = 36.

Teacher's Note:
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99. .

Answer: (b) \( k^k \)

Explanation:
1. Each of the k elements of A can be mapped to any of the k elements of A.
2. So the number of functions is \( k\times k\times\dots\times k=k^k \).

Teacher's Note:
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Answer: (a) 384

Explanation:
1. The last digit must be 0 or 2: 2 ways.
2. The first digit can be 1, 2 or 3: 3 ways.
3. Each of the three middle places has 4 choices: 4³ = 64.
4. Total = 2 × 3 × 64 = 384.

Teacher's Note:
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40.
41. f
42. i
43. r
44. s
45. t
46.
47. d
48. i
49. g
50. i
51. t
52.
53. f
54. o
55. r
56.
57. b
58. e
59. i
60. n
61. g
62.
63. n
64. o
65. n
66. -
67. z
68. e
69. r
70. o
71. .
72.
73. M
74. u
75. l
76. t
77. i
78. p
79. l
80. y
81.
82. t
83. h
84. e
85.
86. i
87. n
88. d
89. e
90. p
91. e
92. n
93. d
94. e
95. n
96. t
97.
98. c
99. h
100. o
101. i
102. c
103. e
104. s
105. .

Answer: (c) 36

Explanation:
1. Vowels: A, U, A (A repeated); consonants: P, R, B.
2. Alternating pattern can begin with a vowel or a consonant: 2 patterns.
3. Vowels can be arranged in \( \frac{3!}{2!}=3 \) ways and consonants in 3! = 6 ways.
4. Total = 2 × 3 × 6 = 36.

Teacher's Note:
1. T
2. w
3. o
4.
5. s
6. t
7. a
8. r
9. t
10. i
11. n
12. g
13.
14. p
15. a
16. t
17. t
18. e
19. r
20. n
21. s
22.
23. a
24. r
25. e
26.
27. p
28. o
29. s
30. s
31. i
32. b
33. l
34. e
35.
36. w
37. h
38. e
39. n
40.
41. t
42. h
43. e
44.
45. n
46. u
47. m
48. b
49. e
50. r
51. s
52.
53. o
54. f
55.
56. v
57. o
58. w
59. e
60. l
61. s
62.
63. a
64. n
65. d
66.
67. c
68. o
69. n
70. s
71. o
72. n
73. a
74. n
75. t
76. s
77.
78. a
79. r
80. e
81.
82. e
83. q
84. u
85. a
86. l
87. .
88.
89. A
90. c
91. c
92. o
93. u
94. n
95. t
96.
97. f
98. o
99. r
100.
101. t
102. h
103. e
104.
105. r
106. e
107. p
108. e
109. a
110. t
111. e
112. d
113.
114. A
115. .

Answer: (a) \( \frac{5(7!)}{2} \)

Explanation:
1. Consonants are C, C, L, L, T; the word has 9 letters with C, L, A repeated twice each.
2. Choose the first and last letters (consonants) and arrange the remaining 7 letters. Counting by cases (CC, LL, CL or LC, CT or TC, LT or TL) and summing gives 12600.
3. This equals \( \frac{5(7!)}{2}=\frac{5\times5040}{2}=12600 \) (checked by direct enumeration).

Teacher's Note:
1. S
2. p
3. l
4. i
5. t
6.
7. i
8. n
9. t
10. o
11.
12. c
13. a
14. s
15. e
16. s
17.
18. b
19. y
20.
21. w
22. h
23. i
24. c
25. h
26.
27. t
28. w
29. o
30.
31. c
32. o
33. n
34. s
35. o
36. n
37. a
38. n
39. t
40. s
41.
42. a
43. r
44. e
45.
46. p
47. l
48. a
49. c
50. e
51. d
52.
53. a
54. t
55.
56. t
57. h
58. e
59.
60. e
61. n
62. d
63. s
64. .
65.
66. E
67. a
68. c
69. h
70.
71. c
72. a
73. s
74. e
75.
76. l
77. e
78. a
79. v
80. e
81. s
82.
83. a
84.
85. d
86. i
87. f
88. f
89. e
90. r
91. e
92. n
93. t
94.
95. r
96. e
97. p
98. e
99. a
100. t
101. e
102. d
103. -
104. l
105. e
106. t
107. t
108. e
109. r
110.
111. p
112. a
113. t
114. t
115. e
116. r
117. n
118.
119. f
120. o
121. r
122.
123. t
124. h
125. e
126.
127. m
128. i
129. d
130. d
131. l
132. e
133.
134. s
135. e
136. v
137. e
138. n
139.
140. l
141. e
142. t
143. t
144. e
145. r
146. s
147. .

Answer: (b) 12

Explanation:
1. \( \frac{n(n-3)}{2}=54 \), so \( n^2-3n-108=0 \), i.e. (n − 12)(n + 9) = 0.
2. Hence n = 12.

Teacher's Note:
1. U
2. s
3. e
4.
5. t
6. h
7. e
8.
9. d
10. i
11. a
12. g
13. o
14. n
15. a
16. l
17.
18. f
19. o
20. r
21. m
22. u
23. l
24. a
25.
26. n
27. (
28. n
29.
30. −
31.
32. 3
33. )
34. /
35. 2
36. .
37.
38. T
39. h
40. e
41.
42. n
43. e
44. g
45. a
46. t
47. i
48. v
49. e
50.
51. r
52. o
53. o
54. t
55.
56. i
57. s
58.
59. r
60. e
61. j
62. e
63. c
64. t
65. e
66. d
67. .

Answer: (a) \( \sum_{k=1}^{n-1}k \)

Explanation:
1. Every pair of lines meets in exactly one point and no point is shared by three lines.
2. Number of points \( ={}^nC_2=\frac{n(n-1)}{2}=1+2+\dots+(n-1)=\sum_{k=1}^{n-1}k \).

Teacher's Note:
1. O
2. n
3. e
4.
5. i
6. n
7. t
8. e
9. r
10. s
11. e
12. c
13. t
14. i
15. o
16. n
17.
18. p
19. o
20. i
21. n
22. t
23.
24. p
25. e
26. r
27.
28. p
29. a
30. i
31. r
32.
33. o
34. f
35.
36. l
37. i
38. n
39. e
40. s
41. .
42.
43. T
44. h
45. e
46.
47. s
48. u
49. m
50.
51. 1
52.
53. +
54.
55. 2
56.
57. +
58.
59. …
60.
61. +
62.
63. (
64. n
65.
66. −
67.
68. 1
69. )
70.
71. i
72. s
73.
74. t
75. h
76. e
77.
78. s
79. a
80. m
81. e
82.
83. a
84. s
85.
86. n
87. C
88. 2
89. .

Answer: (c) 5

Explanation:
1. \( {}^{10}C_3-{}^nC_3=110 \), so \( {}^nC_3=120-110=10 \).
2. Since \( {}^5C_3=10 \), n = 5.

Teacher's Note:
1. C
2. o
3. l
4. l
5. i
6. n
7. e
8. a
9. r
10.
11. p
12. o
13. i
14. n
15. t
16. s
17.
18. c
19. a
20. n
21. n
22. o
23. t
24.
25. f
26. o
27. r
28. m
29.
30. a
31.
32. t
33. r
34. i
35. a
36. n
37. g
38. l
39. e
40. ,
41.
42. s
43. o
44.
45. s
46. u
47. b
48. t
49. r
50. a
51. c
52. t
53.
54. n
55. C
56. 3
57.
58. f
59. r
60. o
61. m
62.
63. t
64. h
65. e
66.
67. t
68. o
69. t
70. a
71. l
72. .