Download Math 210 Spring 2010. Homework 12. Comments and

Math 210 Spring 2010. Homework 12. Comments and Solutions.
3. Twenty points are taken on the circumference of a circle. Any two of them are endpoints of a chord.
How many chords are there?
Solution. There is a bijection between the set of 2-subsets of the set of point on the circle and the set
of chords those points define: every two points define a chord, and, conversely, any chord corresponds
to a pair point on the circle. That is, the number of chords is 20
2 = 190.
6. Prove that for n ∈ N,
Pn
i=0
n
i
4i = 5n .
Solution. By the binomial theorem
5n = (1 + 4)n =
n X
n
i=0
i
1n−i 4i =
n X
n
i=0
i
4i .
8. (a) Expand 0 = (1 − 1)n .
(b) Prove that for every n ≥ 1
n
n
n
n
n
n
+
+
+ ··· =
+
+
+ · · · = 2n−1
0
2
4
1
3
5
(c) Show that there exist as many subsets of an n-element set which contain an odd number of
elements as there are those which contain an even number of elements.
Solution (a). By the binomial theorem
n n X
X
n n−i
n
0 = (1 − 1)n =
1 (−1)i =
(−1)i
i
i
i=0
i=0
n
n
n
n
n
n
n−1
n n
=
−
+
−
+
+ · · · + (−1)
+ (−1)
.
0
1
2
3
4
n−1
n
Solution (b). Define A and B as follows
n
n
n
A=
+
+
+ ...;
0
2
4
n
n
n
B=
+
+
+ ....
1
3
5
As we know, A+B = 2n , and so it suffices to show that A = B. From this it would follow immediately
that A + B = 2A = 2B = 2n , and so A = B = 2n−1 . The equality A = B follows immediately from
(a).
Below we also present several other ways of proving that A = B. When n is odd, the number
n
of addends in both sums above is equal to bn/2c + 1, and trivially ni = n−i
. Alternatively, one
could notice that if n is odd, there is a bijective correspondence between even cardinality and odd
cardinality subsets of an n-set: if C ⊆ [n] is even (odd), then the compliment C c is odd (even). In
case when n is even these arguments are of no help. A beautiful but more elaborate argument along
the same lines can be read in Problem 8.1 on
http://www.cs.brown.edu/courses/cs022/hw/hw08-sol.pdf
1
2
On slides 21–23 of
http://www.cs.princeton.edu/courses/archive/fall02/cs341/lec3− 4.pdf
a bijection is constructed between the set of all odd cardinality subsets of an n-element sets and
the power set of an (n − 1)-element set, which proves B = 2n−1 without relating it to A.
For those who like algebraic manipulations, here is one more argument.
In this case we will use Pascal’s identity:
n
n−1
n−1
=
+
.
k
k−1
k
Suppose n = 2k, k ∈ N, is even. Then
2k
2k
2k
B=
+
+ ··· +
= Pascal’s identity
1
3
2k − 1
h2k − 1 2k − 1i h2k − 1 2k − 1i h2k − 1 2k − 1i
h2k − 1 2k − 1
+
+
+
+
+
··· +
+
0
1
2
3
4
5
2k − 2
2k − 1
h
i h
i
2k − 1
2k − 1
2k − 1
2k − 1
2k − 1
=
+
+
+
+
+ · · · = Pascals’s identity
0
1
2
3
4
2k
2k
2k
=
+
+
+ · · · = A.
0
2
4
Solution (c). It suffices to note that A and B as defined in part (b) count the number of even and
odd subsets of an n-element set, respectively. As we know from part (b), A = B.
22(b). How many solutions (x1 , x2 , . . . , x6 ) (ordered 6-tuples) does the equation
x1 + x2 + . . . x6 = 50
have if all xi ’s are positive integers?
Solution. For every 1 ≤ i ≤ 6 define yi = xi − 1, and consider the following problem:
How many solutions (y1 , y2 , . . . , y6 ) (ordered 6-tuples) does the equation
y1 + y2 + . . . y6 = 44
have if all yi ’s are nonnegative integers?
Clearly there is a bijective correspondence between the sets of solutions of the two problems above.
But the last problem is precisely problem 22 (a). As we saw in class, the number of solutions of the
last problem is 49
5 .
The problem also is equivalent of counting the number of distributions of 50 identical coins in 5
distinct wallets with no wallet being empty.
30. Fifteen points are taken on the circumference of a circle. Through any two of them a chord is drawn.
If no three of the chords intersect at a point inside the circle, how many points of intersections of
these chords are there (including points themselves) ?
Solution. Let A be the set of the fifteen points on the circumference of a circle, and let B be the set
of the points at which the chords defined by points from A intersect inside the circle. Clearly (easy
to understand, but not easy to think about using it!), there is a bijective correspondence between
3
B and the set of all 4-subsets of A: every point of intersection of two chords inside the circle is
defined uniquely by some four points lying on the circumference (the endpoints of the two chords),
and conversely, every four points on the circumference define uniquely a point of intersection of two
chords which is inside the circle. In the uniqueness part of both statements we use that no three
chords pass through the same point inside the circle. The number of 4-element subsets of A is 15
4 ,
and so, the total number of intersection is 15
4 + 15.