Q.3 Find a particular solution to the recurrence.
Let
Let n=1
Let n=2
Let n=3
, substituting (1),
Let n=4
, substituting (1),
Now we make a difference table
0 0 7 2
0 7 9 2
7 16 11 2
23 27 13
50 40
90
It required the 3rd difference to get all
constants so we assume a 3rd degree polynomial
for this particular solution:
Thuse we have the system
That has solution , , ,
and we found earlier.
so we suspect that a particular solution is
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R = {(1,1), (2,1), (3,2), (4,3)}, find R2,R4
R2 is set of all ordered pair of ordered pairs in R: =
I'll use brackets to indicate an ordered pair of ordered pairs:
{[(1,1),(1,1)], [(1,1),(2,1)], [(1,1),(3,2)], [(1,1),(4,3)],
[(2,1),(1,1)], [(2,1),(2,1)], [(2,1),(3,2)], [(2,1),(4,3)],
[(3,2),(1,1)], [(3,2),(2,1)], [(3,1),(3,2)], [(3,1),(4,3)],
[(4,3),(1,1)], [(4,3),(2,1)], [(4,1),(3,2)], [(4,1),(4,3)]}
R4
No way Jose! That's the set of every ordered pair of those ordered
pairs of ordered pairs. It's ridiculous of your teacher to expect
you to list 256 ordered pairs of ordered pairs of ordered pairs of
ordered pairs. Any teacher that would ask a student to do that
should be fired! They're not qualified to teach! That's not
teaching. That's student abuse!
(b) How many permutations are there of the letters, taken all at a
time, of the word ALLAHABAD?
ALLAHABAD has 9 letters. If you could tell the A's apart and the L's apart,
then the answer would be 9!, but since you can't tell them apart,
you must:
1. divide by 4! for that's how many ways the A's can be permuted
in any one permutation, and they would all look just alike.
and in addition to that, you will also have to
2. divide by 2! for that's how many ways the L's can be permuted
in any one permutation, and they would both look just alike.
Answer: ways.
That's all I'm going to do now.
Edwin