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210
Solutions to Exercises
But
(~)
+
(n
~
1)
+ .
.. +
(i
~~)
+
(i)
(nl)
[(n2)
(n2)]
[(
~) (~)] (~1)
~ [(~ ~
1)++
(n
~
2)
:
..
~ (~~6l++
[(ni
2
)++
(n~13)
:
..
~+(:~:)
1
=
anl
+
an2,
by the induction hypothesis
=
an,
as before.
In either case, we obtain
an
=
anl
+
an2,
the recurrence relation which we know defines the
Fibonacci numbers. By the Principle of Mathematical Induction we conclude that this is true for all
n2':1.
24. (a) There is just one way to select 0 books from
n.
(b) We prove this by contradiction. So assume the result is not true. Then we can find
k
2':
~
+
1 such
that
[~]
=I
0; that is, a selection of the required type can be made. Label the books 1,2,
...
,n
from left to right and first assume
n
is even.
If
we partition the books into the
~
sets {I, 2}, {3, 4},
... ,
{n
 1,
n},
the PigeonHole Principle immediately tells us that two of the
k
books selected
must be in the same set, contradicting the fact that no two adjacent books are to be chosen.
Now assume
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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