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Math Investigations
Counting and Probability
Name
Overcounting
When counting distinct arrangements with n - indistinguishable items, you divide by n!
For example, suppose you were asked to find the number of distinct arrangements of the word TATTER.
There are 6 total letters, which would yield _6!_ total arrangements, but there are 3 T's, which means we
over counted. Therefore, we divide by 3!, giving us the answer of _
6!
 120 _.
3!
Counting with symmetries
When confronted with a situation which involves symmetries, we first ignore the symmetry, then divide
to account for the symmetries.
For example, suppose you were asked to find the number of different ways 6 people can be seated at a
round table? The number of ways that one can rearrange 6 people is 6!. However, there are 6 possible
rotations of the same arrangement. Therefore we must divide by 6, yielding 5!.
Exercise: In how many distinct ways can 4 keys be placed on a keychain? Two arrangements are not
considered different if the keys are in the same order?
Solution: Since there are 4 keys, the number of rearrangements would be 4!. However, since there are 4
rotational symmetries, the answer would be
4!
 3!  6 .
4
Challenge Problems:
1) How many diagonals of a regular octagon are not parallel to one of the sides?
Solution:
1
2
We can see that the dashed lines are parallel to one of the edges, and the
3
8
solid lines aren’t. By the properties of a regular n-gon, if n is even, then
if diagonals have an even number of vertices between them, then the
4
7
6
5
diagonal will either be the base of an isosceles trapezoid or a
parallelogram adjoined with an isosceles trapezoid.
To see the number of total diagonals of a given n –gon, we only consider the diagonals emanating from
one half of the vertices. This is to avoid overcounting. Given a vertex, there are n – 3 vertices which are
the endpoints of a diagonal because we don’t count the original vertex or the two adjacent vertices. In
general, there are
n
(n  3) diagonals for a given n –gon. For each given vertex, number of parallel
2
diagonals starting at that vertex are in one-to-one correspondence to the number of even numbers from 1
Math Investigations
to n – 3, inclusive, which is
Counting and Probability
Name
n4
nn4
. Therefore, there are 
 diagonals which are parallel to one of
2 2 
2
the sides. Hence, the number of non-parallel diagonals is
n( n  3) n(n  4) 2(n)( n  3)  ( n)( n  4) n
n


 (2n  6  n  4)  ( n  2)
2
4
4
4
4
In this particular case, there are 8 sides, so the answer would be 12.
2) A regular tetrahedron is a triangular pyramid whose faces are all equilateral triangles. How many
distinguishable ways can we paint the four face of a regular tetrahedron with red, blue, green, and orange
paint such that no two faces are the same color?
Solution: There are 4! ways to arrange 4 colors, but we have to consider the symmetries of a tetrahedron.
Consider the line through the center of the tetrahedron and the vertex, 1.
This is an axis of rotation, yielding rotations (234) and (243). For those
of you unfamiliar with this notation, (234) means the vertex 2 goes to 3,
the vertex 3 goes to 4, and the vertex 4 goes to 2, or 2  3  4  2 .
Similarly, there are rotations (124) and (142), and (134) and (143). If
we compose the rotation (124) with (123), we achieve the rotation of
180 degrees around the axes through the midpoint of 2 and 3 and the
midpoint of 1 and 4, yielding the rotation (14)(23). A similar
construction gives us other rotations (12)(34) and (13)(24). For those of
you familiar with group theory, this is the group called A4 , which is the
group of even permutations in S 4 . So there are
4!
2
12
distinguishable ways to paint a tetrahedron with 4 colors.
3) In how many distinguishable ways can 8 people sit a square table, 2 people to a side?
Solution: Once again, there are 8! ways to rearrange 8 people, but there are 4 symmetries of a square. So
x
y
there are
d
a
e
b
g
h
8!
distinguishable rearrangements.
4
Math Investigations
Counting and Probability
Name
4) Twenty married couples are at a party. Every man shakes hand with everyone except himself and his
spouse. Half of the women refuse to shake hands with any other women. The other 10 women all shake
hands with each other, but not with themselves. How many handshakes are there at the party?
Solution: Let d ( x) = the number of handshakes done by person x. If mi = the ith man,
wi *  the ith "nice" woman, and wi  the ith "mean" woman, then the number of total handshakes,
H, is given by
20
10
10
i 1
i 1
i 1
2 H   d (mi )   d ( wi *)   d ( wi )
since for each pair of people shaking hands, there is only one handshake. By the information given,
d (mi )  38, since there are 38 people in the party besides themselves and their spouse.
d ( wi *)  28,since there are 19 men in the party, other than their hubsand, and 9 other nice women.
d ( wi )  19, since there are 19 men in the party, other than their husband.
Therefore,
20
10
10
2 H   d (mi )   d ( wi *)   d ( wi )
i 1
i 1
i 1
20
10
10
i 1
i 1
i 1
  38   28  19
 (20)(38)  (10)(28)  (10)(19)
 (10)(76  28  19)
 (10)(123)
H  (5)(123)  615
5) An integer is called snakelike if its decimal representation a1a2 a3  ak satisfies ai  ai 1 if i is odd,
and ai  ai 1 if i is even. How many snakelike integers between 1000 and 9999 have 4 distinct digits.
Solution: Let us suppose a four digit number is snakelike. Then it is of the form a b c d, where
a < b, b > c, c < d. Then there are five possibilities, a < c < b < d, or a < c < d < b, or c < a < b < d, or
c < a < d < b, or a < c < d < b. This means that the number of four digit snakelike integers is 5 times the
number of sets of four increasing integers from 1 through 9, if we don’t include 0. To see this more
clearly, note that if we take x < y < z < w, then we can create snakelike integers, x w y z, or x z y w, or y z
x w, or y w x z, or z w x y. We would like to exclude zero at first so as not to count the three digit numbers
9
4
snakelike numbers. Therefore, there are 5   snakelike integers of four distinct digits without zero.
Now, we will consider the four digit snakelike integers that include zero. If x < y < z, then the
9
 3
possibilities are x z 0 y, or y z 0 x, or x y 0 z. This means we need to add 3   . So the answer is
9 9
5    3    630  252  882
 4  3