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Math 510 Exam I
Time: 12:30pm–1:20pm, Monday, Feb. 26, 2007
Name:
Please read the problems carefully and do all you are asked to do. You need to show your work to indicate
how you do each problem. Your proofs and explanations should be literally clear.
1. (15pts) Prove that, among
any 50 points in a square of size 7cm × 7cm, there are two points whose
√
distance apart is at most 2cm.
Proof. Divide the 7cm × 7cm-square into 49 small 1cm
The largest
√
√ × 1cm-squares.
2
2
distance between any two points in a small square is 1 + 1 = 2. The 50 points
are placed into the 49 squares (a point lies on north or west wall or northwest corner
of a small square is regarded to in that square). By pigeonhole principle, there is at
least one small square containing at√least two of the 50 points. Thus the distance
between those two points is at most 2.
2. (20pts) Compute the following numbers:
(a). The number of the positive divisors of 611 × 715 × 117 .
Solution: 6 is not a prime number. 611 × 715 × 117 = 211 × 311 × 715 × 117 . By the fundmental theorem
of arithmatics, a divisor is of the form 2a · 3b · 7c · 11d with 0 ≤ a ≤ 11, 0 ≤ b ≤ 11, 0 ≤ c ≤ 15, 0 ≤ d ≤ 7.
The number of positive divisors is
(11 + 1) × (11 + 1) × (15 + 1) × (7 + 1) = 12 × 12 × 16 × 8 = 18432.
(b). The number of ways to place 8 indistinguishable non-attacking rooks on a 10 × 10 chess-board.
Solution: First choose 8 rows from the 10 rows to place the 8 rooks in. There are 10
ways of doing so.
8
For each of the choice of the 8 rows, there are 10 ways to place the rook in the first row (from top). 9 ways
to place the next chosen row, etc. There are P (10, 8) = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 ways to place the eight rooks
2
.
on those 8 chosen rows. Now the multi. principle implies there are total of 10
· P (10, 8) = (10·9·8·7·6·5·4·3)
2!
8
3. (20pts) A Kansas grade school has four grades with 10, 15, 17, 19 students respectively.
(a). The school wants to line up all students in one line so that students in each grade will stay together.
How many ways are there to do so.
Solution: First decide the order of the four grades, there are 4! ways to order the grades. Second, for
each order of the grades line up students in each of the four grades independently, there are respectively
10!, 15!, 17!, and 19! ways for the grades to do so. Since how to line up one grade does not depend on how
other grades are lined up. Hence the total is 4! · 10! · 15! · 17! · 19!.
(b). If four representatives are chosen with one from each grade, how many different outcomes are there?
Solution: The are 10 ways to choose one representative from grade 1. For each choice in grade 1, there
are 15 choices in grade 2. For each choice in grade 1 and 2, there are 17 choice in grade 3. For any choice
in grade 1, 2, 3, there are 19 choices in grade 4. Thus multiplication principle applies and the total is
10 · 15 · 17 · 19.
pg score
1
4. (10pts) In how many ways can 125 indistinguishable apples, 1 orange and 1 pear be all distributed to 5
children named A, B, C, D, and E, such that each child has at least one fruit and no one gets both orange
and pear?
Solution: Give one of 5 child the orange, there are 5 possible outcome. Then give one of the four remaining child the pear. There are 4 ways. Then give each of the remaining 3 child one apple. There is only
one way to give since all apples are
indistinguishable. Then there are 122 apples remain to be distributed
122+5−1
to 5 groups and there are
ways to do so. Hence the total number of ways to distribute the fruits
5−1
126·125·124·123
under the above given rule is 5 · 4 · 1 · 126
=
.
4·3·2·1
4
5. (15pts) Prove that in a group of n > 1 people, there are two with the same number of acquaintances.
(Assume that each person is not acquainted to him/herself.)
proof: Let a1 , a2 , . . . , an be the number of people acquainted with the first, second, ... , and the last
person respectively. Then 0 ≤ ai ≤ n − 1 for all n. If these n numbers are all distinct, then they have
to take each and all the values 0, 1, 2, . . . , n − 1 by Pigeonhole principle. Thus ai = 0 and aj = n − 1 for
some i and j. Since n > 1, then i 6= j. But this is impossible since the j-th person is acquainted with
all other n − 1 people in the group (everyone except himself), in particular the i-th person. But the i-th
person is acquainted with no one, in particular the j-th person. Since we assumed that acquaintance is a
mutual acquaintance. This is a contradiction and is impossible to both numbers 0 and n − 1 be taken as
the values of a1 , . . . , an .
6. (10pts) In how many ways can six women, eight men and a dog sit around a table such that no two
women sit next to each other?
solution: Treat the dog as a men. Since men (and the dog) are free to sit anywhere, get them seated in
= 8! ways to do so. Then have the six women to sit in
anyway around the table first. There are (8+1)!
9
the spaces between men (and/or dog). There are 9 such spaces and choose 6 of them to sit six different
women. There are P (9, 6) = 9 · 8 · 7 · 6 · 5 · 4 ways to do so. Thus the total number of ways to seat them
with no two women sitting next to each other is 8! · 9!/3!.
7. (10pts) Evaluate the following expression for any integer n > 1 by using integration or differentiation.
n
n
n
n
n
1−
+2
−3
+ · · · (−1) n
=
1
2
3
n
P
P
Solution: Note that (1 + x)n = nk=0 nk xk . Replace x by −x to get (1 − x)n = nk=0 (−1)k nk xk . Then
n−1
−n(1 − x)
n
X
n k−1
d
n
k
(1 − x) =
(−1) k
x .
=
dx
k
k=0
Evaluate both sides at x = 1 (with n − 1 > 0) to get
n
X
n
n
n
n
n
0=
(−1) k
=0−1·
+2
− · · · + (−1) n
.
k
1
2
n
k=0
Hence
k
n
n
n
n
n
1−
+2
−3
+ · · · (−1) n
= 1 + 0 = 1.
1
2
3
n
pg score
2