JEE Mathematics Permutation and Combination MCQs Set 08

Answer: (b) 4

Explanation:
1. \({}^{2n+1}P_{n-1}=\dfrac{(2n+1)!}{(n+2)!}\) and \({}^{2n-1}P_n=\dfrac{(2n-1)!}{(n-1)!}\).
2. Ratio \(=\dfrac{(2n+1)(2n)\,(n-1)!}{(n+2)!}=\dfrac{2n(2n+1)}{n(n+1)(n+2)}=\dfrac{2(2n+1)}{(n+1)(n+2)}=\dfrac35\).
3. \(10(2n+1)=3(n+1)(n+2)\Rightarrow3n^2-11n-4=0\Rightarrow(3n+1)(n-4)=0\Rightarrow n=4\).

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48. .

Answer: (b) 13860

Explanation:
1. Choose which letters go where: \(\dfrac{12!}{6!\,4!\,2!}=13860\).

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Answer: (a) 425

Explanation:
1. Choose any \(3\) of the \(15\) points: \(\binom{15}3=455\).
2. Subtract triples lying on one line (not triangles): \(3\binom53=30\). Triangles \(=425\).

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Answer: (c) 451

Explanation:
1. \(\binom{15}3=455\); the \(4\) collinear points give \(\binom43=4\) degenerate triples.
2. Triangles \(=455-4=451\).

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Answer: (d) 59

Explanation:
1. MEDITERRANEAN has M, D, I, T once; E three times; R, A, N twice each. After using one R and one E, the remaining letters are M, D, I, T, A, A, N, N, R, E, E, i.e. eight distinct letters with E, A, N still available twice.
2. Two middle letters: different letters \(8\times7=56\); identical letters (E, A or N): \(3\). Total \(=59\).

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Answer: (c) 9

Explanation:
1. \({}^{n+2}C_6=\dfrac{(n+2)!}{6!\,(n-4)!}\) and \({}^{n-2}P_2=(n-2)(n-3)\).
2. The ratio is \(\dfrac{(n+2)(n+1)n(n-1)}{720}=11\Rightarrow(n+2)(n+1)n(n-1)=7920=11\cdot10\cdot9\cdot8\).
3. So \(n=9\).

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Answer: (b) 108

Explanation:
1. Each digit appears in the units place in \(3!=6\) of the \(24\) numbers.
2. Sum \(=6(3+4+5+6)=108\).

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Answer: (d) \(36(7!)\)

Explanation:
1. The digit sum of \(0,\ldots,9\) is \(45\). Omitting two digits whose sum is \(9\) keeps the total divisible by \(9\): pairs \((0,9),(1,8),(2,7),(3,6),(4,5)\).
2. Omit \((0,9)\): digits \(1\)-\(8\): \(8!=8\cdot7!\) numbers. Each of the other four pairs leaves \(0\) among the digits: \(8!-7!=7\cdot7!\) numbers.
3. Total \(=8\cdot7!+4\cdot7\cdot7!=36\cdot7!\).

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Answer: (b) \([10,12)\)

Explanation:
1. Let there be \(n\) men. Games among men: \(2\binom n2=n(n-1)\). Games between men and women: \(2\cdot2n=4n\).
2. \(n(n-1)-4n=66\Rightarrow n^2-5n-66=0\Rightarrow n=11\).
3. \(11\) lies in \([10,12)\).

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Answer: (b) \({}^{21}C_7\)

Explanation:
1. Give each question \(2\) marks first, leaving \(30-16=14\) marks to be distributed among \(8\) questions with no restriction (whole marks).
2. Number of ways \(=\binom{14+7}7=\binom{21}7\).

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Answer: (d) 8

Explanation:
1. Total \(4!=24\). Arrangements with \(a,b\) adjacent: block \(\times\) \(c,d\): \(3!\cdot2=12\); with \(c,d\) adjacent: \(12\).
2. Both pairs adjacent: \(2!\) orders of the two blocks and \(2\cdot2\) orders inside: \(8\).
3. Neither adjacent \(=24-12-12+8=8\). (They are \(acbd\)-type arrangements: \(a,b\) and \(c,d\) alternate: \(2\cdot2\cdot2=8\).)

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Answer: (b) 162

Explanation:
1. The patterns are \(aab\) and \(abb\) with \(a\ne b\) (excluding \(aaa\); \(aba\) has no adjacent equal digits).
2. \(aab\): \(a\in1\)-\(9\), \(b\in0\)-\(9\), \(b\ne a\): \(9\times9=81\). \(abb\): \(a\in1\)-\(9\), \(b\ne a\): \(81\).
3. Total \(=162\).

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Answer: (a) 126

Explanation:
1. Descending: any \(5\) distinct digits from \(0\)-\(9\) arranged in decreasing order: \(m=\binom{10}5=252\).
2. Ascending: the first digit cannot be \(0\), so choose \(5\) of the digits \(1\)-\(9\): \(n=\binom95=126\).
3. \(m-n=126\).

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Answer: (c) 80

Explanation:
1. Choose which \(4\) pairs contribute a shoe: \(\binom54=5\). From each of these pairs choose left or right: \(2^4=16\).
2. Total \(=5\times16=80\).

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Answer: (c) 240

Explanation:
1. First four digits: arrangements of \(1,2,3,3\): \(\dfrac{4!}{2!}=12\). Fifth digit: \(2\) choices.
2. Sixth digit: any of \(10\) digits, and then the seventh digit is fixed (\(9-\) sixth, always a digit).
3. Total \(=12\times2\times10=240\).

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Answer: (b) 52

Explanation:
1. Count three-digit numbers containing both digits \(2\) and \(3\). Case: digits exactly \(\{2,3,x\}\) with \(x\) a third distinct digit (\(x\ne2,3\)): \(8\) choices of \(x\); arrangements of three distinct digits: \(3!=6\), minus those with leading zero if \(x=0\) (\(2\)): \(7\cdot6+4=46\).
2. Case: digits \(\{2,2,3\}\) (\(3\) arrangements) and \(\{2,3,3\}\) (\(3\) arrangements): \(6\).
3. Total \(=46+6=52\) (confirmed by listing all numbers from \(100\) to \(999\)).

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Answer: (c) 8

Explanation:
1. The first letter has \(3\) choices; each later letter has \(2\) choices (one letter is forbidden after the previous one).
2. Number \(=3\cdot2^{n-1}=384\Rightarrow2^{n-1}=128\Rightarrow n=8\).

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Answer: (c) 10

Explanation:
1. Sum of the pairwise intersections \(=\)(exactly two)\(+3\times\)(all three)\(=9+3=12\).
2. \(|A\cup B\cup C|=22+15+14-12+1=40\).
3. None \(=50-40=10\).

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Answer: (c) 144

Explanation:
1. Treat A, B as a block: \(5!\times2=240\) arrangements.
2. Subtract those with C and D adjacent: two blocks and E, F: \(4!\times2\times2=96\). Total \(=240-96=144\).

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Answer: (b) 30

Explanation:
1. A number is divisible by \(4\) if its last two digits form a multiple of \(4\): \(04,12,20,24,32,40\) (distinct digits from the set).
2. Zero among the last two places (\(04\), \(40\), \(20\)): the other three digits can be arranged freely, \(3!=6\) each: \(18\).
3. Last two digits \(12\), \(24\) or \(32\): the remaining digits include \(0\), which cannot lead: \(3!-2!=4\) each: \(12\). Total \(=18+12=30\).

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Answer: (c) 50

Explanation:
1. \(100=80+50+30-S+5\), where \(S\) is the sum of the pairwise intersections. So \(S=65\).
2. Exactly two \(=S-3\times5=65-15=50\).

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Answer: (c) 420

Explanation:
1. \(77=7\cdot11\). Let \(s=x_1+x_2+x_3\ge3\) and \(t=y_1+y_2\ge2\). Valid factor pairs: \((s,t)=(7,11)\) and \((11,7)\) (the pairs \((1,77)\) and \((77,1)\) fail the minimum sums).
2. \((7,11)\): \(\binom62\cdot\binom{10}1=15\cdot10=150\). \((11,7)\): \(\binom{10}2\cdot\binom61=45\cdot6=270\).
3. Total \(=150+270=420\).

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18. n
19. t
20.
21. p
22. o
23. s
24. i
25. t
26. i
27. v
28. e
29.
30. s
31. o
32. l
33. u
34. t
35. i
36. o
37. n
38. s
39.
40. o
41. f
42.
43. e
44. a
45. c
46. h
47.
48. s
49. u
50. m
51. .