JEE Mathematics Permutation and Combination MCQs Set 07

Answer: (a) 259980

Explanation:
1. Total numbers: \(4\times4\times3\times2=96\).
2. Thousands place: each non-zero digit \(d\) is in the first place in \(4\cdot3\cdot2=24\) numbers, so this place contributes \(24(1+2+3+4)\times1000=240000\).
3. Any other place (hundreds, tens or units): a given digit appears there in \(3\cdot3\cdot2=18\) numbers (choose a non-zero first digit, then fill the last two places), so each such place contributes \(18\times10\) times its place value: \(180(100+10+1)=19980\).
4. Sum \(=240000+19980=259980\).

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Answer: (c) 535

Explanation:
1. All pairs: \(\binom{37}2=666\).
2. The \(13\) lines through \(A\) give \(\binom{13}2=78\) pairs but only \(1\) point, a loss of \(77\). Similarly for \(B\): \(\binom{11}2=55\) pairs, \(1\) point, a loss of \(54\).
3. Points \(=666-77-54=535\).

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Answer: (b) \((10!)^2\left(\dfrac1{3!2!7!}+\dfrac1{(2!)^46!}\right)\)

Explanation:
1. Choose the \(2\) children who get nothing: \(\binom{10}2\). The \(10\) toys go to the other \(8\) children with each getting at least one: the toy-count pattern is \((3,1^7)\) or \((2,2,1^6)\).
2. Number of ways to give \(10\) distinct toys to \(8\) specific children, each getting at least one, is \(8!\,S(10,8)=8!\times750\). Total \(=45\times40320\times750=1\,360\,800\,000\).
3. The expression in (b) evaluates to \(1\,360\,800\,000\), while (a), (c), (d) give \(789\,264\,000\), \(1\,006\,992\,000\) and \(1\,578\,528\,000\).

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Answer: (b) \(\dfrac{52!}{(13!)^4}\)

Explanation:
1. Each player gets \(13\) cards, and the players are distinct (different people), so the number of ways is the multinomial coefficient \(\dfrac{52!}{13!\,13!\,13!\,13!}\).

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Answer: (a) 144

Explanation:
1. Seat the \(4\) men first around the table: \((4-1)!=6\) ways.
2. The \(4\) women go into the \(4\) gaps between men: \(4!=24\) ways.
3. Total \(=6\times24=144\).

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Answer: (b) \(2\times18!\)

Explanation:
1. Treat the two delegates as one block: \(19\) objects around a table, \((19-1)!=18!\) ways.
2. The two can be ordered within the block in \(2\) ways: \(2\times18!\).

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Answer: (a) \(17\times18!\)

Explanation:
1. Total circular arrangements: \(19!\). Together: \(2\times18!\).
2. Never together \(=19!-2\times18!=18!(19-2)=17\times18!\).

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Answer: (d) 685

Explanation:
1. For a fixed sum \(s\), the number of positive solutions is \(\binom{s-1}2\).
2. \(\sum_{s=16}^{20}\binom{s-1}2=\binom{15}2+\binom{16}2+\binom{17}2+\binom{18}2+\binom{19}2=105+120+136+153+171=685\).

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Answer: (b) 169194

Explanation:
1. Total arrangements of H,I,N,D,U,S,T,A,N (N repeated): \(\dfrac{9!}{2!}=181440\).
2. Arrangements containing HIN: treat it as a block with the letters D,U,S,T,A,N: \(7!=5040\). Containing DUS: block with H,I,N,T,A,N: \(\dfrac{7!}{2!}=2520\). Containing TAN: block with H,I,N,D,U,S: \(7!=5040\).
3. Pairs: HIN with DUS: \(5!=120\); DUS with TAN: \(5!=120\); HIN with TAN (uses both N's): \(5!=120\). All three patterns together: \(3!=6\).
4. By inclusion-exclusion: \(181440-(5040+2520+5040)+(120+120+120)-6\) gives \(169194\) (confirmed by enumeration).

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Answer: (b) \(\dfrac k{3!}\)

Explanation:
1. Distributing to three distinct persons gives \(k=\dfrac{(3n)!}{(n!)^3}\) ways.
2. Dividing into three unlabelled groups removes the \(3!\) orderings of the groups: \(\dfrac k{3!}\).

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Answer: (c) 150

Explanation:
1. Fill \(5\) places with \(3\) varieties so that all \(3\) appear (onto functions): \(3^5-3\cdot2^5+3\cdot1^5\).
2. \(=243-96+3=150\).

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Answer: (a) \(2^{n-1}\)

Explanation:
1. \(\sum_{k\text{ even}}\binom nk=\dfrac{(1+1)^n+(1-1)^n}2=2^{n-1}\).

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Answer: (a) 648

Explanation:
1. Hundreds digit: \(9\) choices (\(1\)-\(9\)); tens: \(9\) choices (including \(0\), excluding the first); units: \(8\) choices.
2. \(9\times9\times8=648\).

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Answer: (b) 26

Explanation:
1. Choose \(6\) of the \(8\) squares: \(\binom86=28\).
2. The middle row (4 squares) cannot be empty, as the other rows only have \(4\) squares in total (fewer than \(6\)).
3. Bad cases: the top row empty (the \(6\) X's fill all other squares): \(1\); bottom row empty: \(1\). Both empty is impossible.
4. Ways \(=28-2=26\).

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Answer: (a) NAAIG

Explanation:
1. Alphabetical order of letters: A,A,G,I,N. Words starting with A: \(\dfrac{4!}{1!}=24\); with G: \(\dfrac{4!}{2!}=12\) (total \(36\)); with I: \(12\) (total \(48\)).
2. Words starting with N: the 49th is NAAGI and the 50th is NAAIG.

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Answer: (b) \({}^{21}C_{18}\,2^3\)

Explanation:
1. Choose the \(18\) correct predictions: \({}^{21}C_{18}\). Each of the remaining \(3\) matches has \(2\) wrong predictions: \(2^3\).

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Answer: (c) 240

Explanation:
1. There are \(9\) positions with \(5\) odd places. The letters A, I, I, I, O (five letters) must occupy the \(5\) odd places: \(\dfrac{5!}{3!}=20\) ways.
2. The remaining letters N, D, L, L go to the \(4\) even places: \(\dfrac{4!}{2!}=12\) ways.
3. Total \(=20\times12=240\).

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Answer: (a) 25

Explanation:
1. The voter votes for \(1\), \(2\) or \(3\) candidates: \(\binom51+\binom52+\binom53=5+10+10=25\).

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Answer: (d) 216

Explanation:
1. Five of the six digits are used; the digit sum \(15-d\) must be a multiple of \(3\), so the omitted digit \(d\) is \(0\) or \(3\).
2. Omit \(0\): digits \(1,2,3,4,5\): \(5!=120\) numbers. Omit \(3\): digits \(0,1,2,4,5\): \(5!-4!=96\) numbers.
3. Total \(=120+96=216\).

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Answer: (d) \(\dfrac{(n-1)^2}4\)

Explanation:
1. Three numbers in A.P. are determined by the first and the third term, which must have the same parity (so the middle term is an integer).
2. For odd \(n\): pairs of odd numbers \(\binom{(n+1)/2}2\) plus pairs of even numbers \(\binom{(n-1)/2}2\) \(=\dfrac{(n-1)^2}4\). Check: \(n=3\) gives \(1\), \(n=5\) gives \(4\).

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Answer: (b) \(2\times18!\)

Explanation:
1. Treat the host with the two particular persons on either side as one block: \(2\) internal orders.
2. The block and the other \(18\) persons make \(19\) objects around a table: \(18!\) ways. Total \(2\times18!\).

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15.
16. a
17. n
18. d
19.
20. t
21. h
22. e
23. n
24.
25. u
26. s
27. e
28.
29. (
30. o
31. b
32. j
33. e
34. c
35. t
36. s
37.
38. -
39.
40. 1
41. )
42. !
43.
44. f
45. o
46. r
47.
48. t
49. h
50. e
51.
52. c
53. i
54. r
55. c
56. l
57. e
58. .

Answer: (c) \(9^n\)

Explanation:
1. The first digit has \(9\) choices (\(1\)-\(9\)).
2. Each later digit can be any digit except the previous one: \(9\) choices. Total \(9^n\).

Teacher's Note:
1. E
2. a
3. c
4. h
5.
6. d
7. i
8. g
9. i
10. t
11.
12. a
13. f
14. t
15. e
16. r
17.
18. t
19. h
20. e
21.
22. f
23. i
24. r
25. s
26. t
27.
28. h
29. a
30. s
31.
32. 9
33.
34. c
35. h
36. o
37. i
38. c
39. e
40. s
41. .

Answer: (a) 1540

Explanation:
1. Place the \(21\) English books first; they create \(22\) gaps (including the ends).
2. Choose \(19\) gaps for one Hindi book each: \(\binom{22}{19}=\binom{22}3=1540\).

Teacher's Note:
1. P
2. l
3. a
4. c
5. e
6.
7. o
8. n
9. e
10.
11. g
12. r
13. o
14. u
15. p
16.
17. f
18. i
19. r
20. s
21. t
22. ,
23.
24. t
25. h
26. e
27. n
28.
29. c
30. h
31. o
32. o
33. s
34. e
35.
36. g
37. a
38. p
39. s
40.
41. f
42. o
43. r
44.
45. t
46. h
47. e
48.
49. o
50. t
51. h
52. e
53. r
54. .