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goes back to 0, up to 2 again, and finally down to 0. So the absolute minimum
is −1, and it occurs in the first and third position. There are five u’s after
the third positon. So this Dyck path is in the block B 5 of our partition. Now
comes the crucial question. Why were there five u’s after that last absolute
minimum in position 3? Try with the same path and i =3. Figure out why
there are three u’s after the last absolute minimum in the resulting path. All
this discussion should explain why when i =5, the set of all Catalan paths
is mapped into the set B 5 . Keeping i =5for a while, try to see why this
correspondence between Catalan paths and B 5 is a bijection. Then, if you
need to, do the same thing with i =3. This should give you enough insight
to do the general case.
55. What would the lower limit of the sum have to be for this problem to be a
routine application of the binomial theorem?
56. What does the binomial theorem give you for (x − y) n ?
57. Consider (x + y) m (x + y) n .
Additional Hint: What does ( m+n )
k
count? What does ( m i )( m
k−1 ) count?
58. For example when n =3,wehave( 3 0 ) = (3 3 ) and (3 1 ) = (3 ).
2
The number
of subsets of even size is ( 3 0 ) + (3 )
2
and the number of subsets of odd size is
( 3 1 ) + (3 ),
3
and the two sums can be paired off into equal terms. When we
subtract the number of subsets of odd size from the number of subsets of
even size, the pairing also gives us ( 3 0 ) − (3 1 ) + (3 2 ) − (3 3 ) =0.
59. Take the derivative of something interesting.
61. To prove that each function from a set S of size n to a set of size less than
n is not one-to-one, we must prove that regardless of the function f that we
choose, there are always two elements, say x and y, such that f (x) = f (y).
62. The previous exercise could help you prove that if f is one-to-one, then it is
onto.
Additional Hint: The sum principle can help you show that if f is an onto
function, then f is one-to-one.
63. The statement of the generalized pigeonhole principle involves the number
of elements in a block, so a counting principle is likely to help you.
64. You may choose a specific number for n if you want to. Notice that the last
two digits of the powers of a prime other than two cannot represent an even
number.
65. While this sounds like a pigeonhole principle problem, the ordinary pigeonhole
principle doesn’t guarantee three of something.