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COT 3100
spring’2001
Practice problems for the test#1.
Solutions (partial)
Combinatorics and counting.
1) Let
A
= {
a, b
,
c
,
d
,
e
,
f
,
g
}.
a)
How many fiveelement subsets are there if
a
cannot be in a set?
Ans
. It is the number of combinations of size 5 out of 6 distinct objects,
i.e.
C
(6, 5)=6.
b)
How many subsets of
A
are there which contain at least three
elements?
Ans
. It is easier to find first the number of subsets of cardinality less then
three. We have 1 empty subset, 7 singletons and
C
(7, 2)=21 subsets
containing two elements. So, 1+7+21=29 is the number of subsets of
cardinality less then 3. Now we subtract it from the total number of
subsets to get the number of subsets, which contain three and more (or
at least three) elements:
2
7

(1+7+21) = 128

29=99.
Alternatively, you directly take the sum of subsets with cardinalities
from 3 to 7. The corresponding row of the Pascal’s triangle is 1, 7, 21,
35, 35, 34, 21, 7, 1. So, we have: 35+35+21+7+1=99.
c)
How many subsets of
A
are there which contain at most six
elements?
Ans
. We have only one subset of
A
with which contain more than 6
element, that is
A
itself, 
A
=7. So, 2
7

1 = 128

1=127 subsets contain
at most six elements (i. e. six and less).
2). A juggler plans to jungle 11 rings. In his bag he has 20 identical red
rings, 30 identical blue rings and 40 identical white rings. He must select at
least one of each color to be patriotic. An example of a patriotic choice
would be: 2 red, 3 white and 6 blue. How many different patriotic choices
are there for 11 rings?
Ans
. With the restriction to choose at least one ring of each color the number
of possible choices is reduced to the number of ways to chose the remaining
8 (= 11

3) rings from three different colors. (Given numbers of rings of each
color guarantees that we have “unlimited supply” for choosing 11 rings).
This is the number of ways to distribute 8 identical objects into 3
distinguished boxes. We can find this number as the number of permutations
1
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View Full Documentof 8 identical objects and 2 vertical bars (separators between boxes). So, we
have for the number of permutations of 8+2 objects, when 8 are of one kind
and 2 are of another kind: 10!/(8!
⋅
2!)=45.
3). Find the number of permutations of letters of the English alphabet (i.e.
strings of length 26 using each letter exactly once) meeting the following
conditions:
a)
The permutation does not contain the string
math
.
Ans.
We can find the number of permutations that do contain the string
math
. This number is 23!. So, 26!

23! is the number of permutations of 26
letters that do not contain the string
math
.
b)
The permutation contains both of the strings
math
and
bio
.
Ans
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 Spring '09

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