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MATH 2320–Winter 2012. Problem set 8.
Instructions. Answer each question completely; justify your answers. This assignment
is due at 17:00 on Friday, March 30 in Assignment Box #45.
[1] Discrete Mathematics with Graph Theory (3rd Edition) Edgar G. Goodaire, Michael
M. Parmenter
1. Prove the following statements.
n
n
n+1
•
+
=
.
i
i+1
i+1
n!
n
n
n!
+
+
=
i!(n − i)! (n − i − 1)!(i + 1)!
i
i+1
=
(i + 1)n! + (n − i)n!
(i + 1)!(n − i)!
(n + 1)n!
=
=
(i + 1)!(n − i)!
n+1
i+1
n
n
•
=
.
i
n−i
n
n!
n
=
=
i
i!(n − i)!
n−i
• What is the interpretation of the above statements in relation with Pascal’s triangle?
The first statement represents the fact that each number in the Pascal’s triangle
is the sum of the two numbers directly above it. The second statement represents
the symmetry of the Pascal’s triangle along the vertical line passing through its
vertex.
2. Use the binomial theorem to verify the following statements.
n X
n 2i
• For every n ≥ 0,
2 = 5n .
i
i=0
n n X
n i n−i X n 2i
2 .
5 = (4 + 1) =
41
=
i
i
i=0
i=0
n
n
n
X
i n
• For every n ≥ 0,
(−1)
3n−i 4i = (−1)n .
i
i=0
n
(−1)n = ((−4) + 3) =
n X
n
i=0
i
(−4)i 3n−i =
n
X
n i n−i
(−1)n
43 .
i
i=0
3. Find the coefficient of x−6 in the binomial expansion of 16x2 −
16x2 −
1
2x
12
.
12−i
12
−1
1 12 X
=
(16x2 )i
2x
2x
i=0
2i 12
X
x
(−1)12−i 16i
=
12−i
2
x12−i
i=0
12 X
(−1)12−i 16i
=
x3i−12
12−i
2
i=0
Observe that 3i − 12 = −6 when i = 2; therefore the coefficient of x−6 is
(−1)10 162
−1
= (−1)10 2−2 =
210
4
4. A sequence is defined recursively by a0 = 1, a1 = −3, and an = 6an−1 − 9an−2 for
n ≥ 2.
• List the first five elements of the sequence.
1, −3, −27, −135, −567
• Prove by mathematical induction that an = 3n (1 − 2n) for all n ≥ 0.
We use strong induction.
Base cases: n = 0 and n = 1; it is clear that 1 = a0 = 30 and −3 = a1 = 31 (−1).
Inductive step: Let n ≥ 2 and suppose that ak = 3k (1 − 2k) for all k < n. We
shall verify that an = 3n (1 − 2n). Observe that
an = 6an−1 − 9an−2
= 6 · 3n−1 (2 − 2n) − 9 · 3n−2 (3 − 2n)
= 3n (2(2 − 2n) − (3 − 2n))
= 3n (1 − 2n)
By the strong induction principle, an = 3n (1 − 2n) for all n ≥ 0.
5. A cake is in the shape of a regular hexagon with each of its sides exactly 30cm long.
Seven flowers of icing adorn the top. Show that at least two flowers are no more than
30cm apart.
Divide the regular hexagon into six equilateral triangles whose sides are exactly 30cm
long. Observe that any two points in the same triangle are no more than 30cm apart.
By the Pigeon hole principle, if there are seven points in the interior of the hexagon,
then at least two of them are in the same triangle and, in particular, no more than
30cm apart.
6. In a party of 30 people, there are 104 different pairs of people who know each other.
• Show that some person must have at least seven acquintances.
We argue by contradiction. Let’s assume that each person has at most six acquintances. Then there are at most 12 · 30 · 6 = 90 different pairs of people who
know each other. But we know there are exactly 104 such pairs of people. Hence
104 ≤ 90, and we have reached a contradiction; therefore our assumption must
be false.
• Show that some person must have fewer than seven acquintances.
We argue by contradiction. Let’s assume that each person has at least seven
acquintances. Then there are at least 12 · 30 · 7 = 105 different pairs of people who
know each other. But we know there are exactly 104 such pairs of people. Hence
105 ≤ 104, and we have reached a contradiction; therefore our assumption must
be false.
7. In how many ways can two adjacent squares be selected from an 8 × 8 chess board.
A pair of adjacent squares is either in a row or a column of the chess board. There
are eight rows and eight columns. Each row or column has seven pairs of adjacent
squares. By the multiplication rule, there are
2 ∗ 8 ∗ 7 = 112.
different ways to select two adjacent squares.
8. How many three-digit numbers contain the digits 2 and 5 but none of the digits 0, 3,
7?
There are three disjoint cases to consider.
• The digits 2 and 5 occur only once. In this case there are three posibilities for
the position of 2, there are two posibilities for the position of 5, and then there
are five posibilities for the digit in the remainding position. By the multiplication
rule, this case has
3 ∗ 2 ∗ 5 = 30
three digit numbers.
• There are two ocurrences of the digit 2. In this case there are three posibilities
for the position of 5. This case has 3 three digit numbers.
• There are two ocurrences of the digit 5. As in the previous case, there are three
posibilities for the position of 2. This case has 3 three digit numbers.
By the addition rule, there are 36 three-digit numbers containing the digits 2 and 5
but none of the digits 0, 3, 7.
9. Use the principle of inclusion and exclusion to answer the following questions.
• How many integers between 1 and 1000 (inclusive) are not divisible by 2, 3, 5, 7?
This problem is from the textbook, section 6.1, 14.(a) [BB]. The complete solution
is on the back of the book. Answer is 228.
• How many integers between 1 and 1000 (exclusive) are not divisible by 2, 3, 5,
7?
Now 1 nad 1000 must be excluded. Since 1 is not divisible by 2,3,4 or 7 while
1000 is divisible by 5, the answer is 228-1=227.