Combinatorics
Exam Duration: 45 Mins Total Questions : 10
How many ways are there to arrange the nine letters in the word ALLAHABAD?
- (a)
7500
- (b)
7560
- (c)
4000
- (d)
2000
The number of diagonals which can be drawn by joining the angular points of a heptagon is
- (a)
14
- (b)
7
- (c)
10
- (d)
12
At certain college the housing office has decided to appoint, for each floor, one male and one female residental advisor. The pairs of advisors can be selected for a seven story building from 12 male and 15 female candidates are
- (a)
40000
- (b)
5096520
- (c)
568400
- (d)
70000
Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of sequences B and C are there such that (i) each is sorted in ascending order (ii) B has 5 and c has 3 elements and (iii) the result of merging B ans C given A?
- (a)
2
- (b)
30
- (c)
56
- (d)
256
A person deposiits Rs 300 in a saving account of SBI at an interest rate of 5% per annum compounded annually. The obatin amount in his account at the end of r yeras is
- (a)
300(1.05)r
- (b)
200
- (c)
100(1.02)r+1
- (d)
50\(\times \) (1.00)r
The generating function for thr sequence 2, 2, 2, 2, 2, 2 is
- (a)
\(\frac { { z }^{ 2 }+1 }{ (z-1) } \)
- (b)
\(\frac { 18(z-1) }{ { z }^{ 2 }+1 } \)
- (c)
\(\frac { 2({ z }^{ 5 }-1) }{ z-1 } \)
- (d)
None of these
The generating function for the sequence 0, 1, 2, 4, 8, ...... is
- (a)
\(\frac { 2z }{ 1+{ z }^{ 2 } } \)
- (b)
\(\frac {1+3z }{ 1+{ 8}{ z } } \)
- (c)
\(\frac { { Z }^{ 2 }+1 }{ { Z }^{ 2 }-1 } \)
- (d)
\(\frac {z }{ 1+{ 2}{ z } } \)
Solution of recurrence relation
ar - 5ar-1 + 8ar-2 - 4ar-3 = 0 is
- (a)
ar = A1(1)r + (A2 + A3r) (2)r
- (b)
ar = (A1 + A2r2 + A3r3) (3)r
- (c)
ar = (A1 + A2r + A3 r2) (-2)r
- (d)
None of the above
The solution of the recurrence relation
ar + 6ar-1 + 12ar-2 - 8ar-3 = 0 is
- (a)
ar = (A1r + A2r + 2A3 r3)(2)r
- (b)
ar = (A1 + A2r2 + A3r3)(3)r
- (c)
ar = (A1 + A2r + A3r2)(-2)r
- (d)
None of these
The solution of the recurrence relation ar = ar-1 + 2ar-2 with a0 = 2,a1 = 7 is
- (a)
ar = (3)r + (1)r
- (b)
2ar = \(\frac { (2{ ) }^{ r } }{ 3 } \)- (1)r
- (c)
ar = 3r+1 - (-1)r
- (d)
ar = 3(2)r - (-1)r
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