Download Math 574 Problem Set 1 summer 2008 solutions

Math 574 Problem Set #1
1. (a). How many license plates consist of two letters followed by 4 digits?
For example, DG 3221 or XX 4147
(b). What if the digits must be distinct? (So, AB 3413 would not be legal.)
Answer: (a). 26 2 ! 10 4
(b). 262 ! 10 ! 9 ! 8 ! 7
2. (a). There are 10 T-F questions on an exam. In how many ways can a student
answer the questions if she must answer all of them?
(b). What if she is allowed to omit as many as she likes?
10
Answer: (a). 2
10
(b). 3 (There are 3 choices for each question T, F, or omit)
3. How many Morse Code characters can be formed from a combination of one to at
most 4 dots and dashes?
Answer: 2 + 4 + 8 + 16 = 30
4. (a). How many numbers from 0 – 99,999 do not contain the digit 4?
(b). How many numbers from 0 – 99,999 contain the digit 3 at least once?
5
Answer: (a). 9
5
5
(b). 10 ! 9
5. How many ways are there to arrange the letters a, b, c, d, e, f in a row so that
(a). b is somewhere to the left of a?
(b). b is somewhere to the left of a and a is somewhere to the left of c?
(c). a is next to b (on the left or right).
(d). a, b and c are consecutive in some order. Example: d c b a f e or f d b c a e
Answer: (a). 360 (b). 120 (c). 240 (d). 144
6. How many subsets of {1, 2, 3, 4, 5, 6} contain 1 but not 2?
Examples: {1}, {1, 4, 5}, and {1, 4}
Answer: 16 (the number of subsets of {3, 4, 5, 6})
7. How many 2-element subsets are there of the set {1, 2, 3, 4, 5, 6}?
Some examples would be {1, 2}, {4, 5}, {3, 6}.
Answer: 15
8.
A string of length n using the letters in some set A as an alphabet is simply a sequence of n letters
(repetitions allowed unless stated otherwise) juxtaposed together. For example there are four
strings of length 2 that can be made from the letters a and b namely: a a
ab
ba
bb
9. (a). How many strings of length 6 are there using a’s, b’s, c’s, and d’s ?
Examples: a c c b d d OR a a a a a c OR b b a a c c a
(b). How many of these contain exactly one d?
(For example c c a b d a or b b b c d a or b d c c a b )
Answer:
6
(a). 4
5
(b). 6 ! 3
10. The number 72 has the 12 divisors 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72.
How many divisors does the integer 2 6 35 7 9113 have?
Hint: A divisor of an integer must be a product of the same primes as the number
2 4
6 2 4
2
6 5 9
3
itself. So a few divisors are: 3 7 11, 2 3 7 11 , and 2 3 7 11
Answer: 7 ! 6 ! 10 ! 4 = 1680
11. How many different arrangements are there of 1, 2, 3, 4, 4, 4, 4, 4, 4, 4?
Answer: 10 ! 9 ! 8 = 720
12. How many arrangements of the ten letters a, b, c, d, e, f, g, h, i, j
(a). Have a, b and c next to one another and in that order?
Example: d i e a b c g f h j and g j e i d f h a b c
Answer: 8!
(b). Have a, b and c next to one another in some order?
Examples: d i e b a c g f h j and g j e i e f h a b c and i e c b a g f h j
Answer: First choose some arrangement of a, b, and c and then arrange the 8
objects consisting of the abc-block and the other 7 letters for a total of 6 ! 8! .
13. How many permutations of the letters a,b,c,d,e,f,g,h,i,j,k,l are there in which
(a). Each of a, b, and c is somewhere to the left of both d and e?
(b). All of a, b and c appear together in some order and all three are to the left of
all of d, e, f and g which also appear together in some order?
(c). Each of a and b is somewhere to the left of both c and d and both of e and f
are somewhere to the left of both g and h?
! 12 $
! 12 $ ! 8 $
3!! 4!! 8!
Answers: (a). # & ' 3!' 2!' 7!
(b).
(c). # & ' # & ' 2 4 ' 4!
2
" 5%
" 4 % " 4%
14. A club contains 8 women and 9 men. How many ways are there to form a
committee that consists of 3 men and 4 women?
! 9$ ! 8 $
Solution: # & ' # &
" 3% " 4 %
15. How many strings of length 8 using the letters a, b, c, d, e
(a). contain exactly 3 d’s? Example: a d b b c d e d or d a b c c c d d
(b). contain at least one d?
(c), contain exactly 2 a’s and 3b’s? Example: a c d b b a c b
(d). contain exactly 2 a’s or exactly 3 b’s?
! 8$
! 8$ ! 6$
Answers: (a). # & ' 4 5
(b). 5 8 ! 4 8
(c). # & # & ' 33
" 3%
" 2 % " 3%
! 8$
! 8$
! 8$ ! 6$
(d). # & ' 4 6 + # & ' 4 5 ( # & # & ' 33
" 2%
" 3%
" 2 % " 3%
Suppose that we have two identical A’s and two identical B’s then there are six
different ways that these could be arranged namely:
AABB ABAB ABBA BABA BBAA BAAB
So even though there are 4 objects there are not 4! = 24 arrangements in this case.
You could have determined that there were just six such arrangements by thinking
that there are four positions to fill and there are six ways to choose some two of
them for the A’s (and then the B’s go in the other two spots).
16. (a). How many ways are there to arrange 5 identical A’s and 7 identical B’s in a row?
(b). How many ways are there to arrange 5 identical A’s and 7 identical B’s in a row
if no two A’s can be next to each other.
! 12 $
! 8$
Answers: (a). # &
(b). # &
" 5%
" 5%
17. (a). How many ways are there to arrange 4 identical A’s, 6 identical B’s and
5 identical C’s in a row?
(b). Answer (a) with the restriction that no two A’s can be next to each other.
! 15 $ ! 11$
! 11$ ! 12 $
Answers: (a). # & ' # &
(b). # & ' # &
" 4% " 6%
" 6% " 4%
18. If n points are equally distributed around a circle and all pairs of points are joined
by line segments. In terms of n, how many points of intersection occur inside the
circle? Hint: Look at a point of intersection and see where it comes from.
Solution: There is a 1-1 correspondence between the points of intersection and
! n$
the ways to choose four of the points. So the answer is # & .
" 4%
n
19. Find a simple formula for the sum
! n$
' k #" k &%
(in terms of n).
k=0
Hint: First note that this sum counts the number of ways to form a committee
with
one person selected as the chair of the committee. Now answer that same
question a different way. Here k represents the number of people on the
committee. So we might say that we are “conditioning on the number of
people on the committee.”
Solution: You could choose such a committee by first choosing the chairperson
in n ways, and then choosing some subset of the remaining members to serve as
the rest of the committee.
There are 2 n !1 subsets of an n – 1 element set and so there are n2 n !1 such
n
! n$
committees. Thus it must be that ' k # & = n2 n (1 .
k=0 " k%
20. How many 5-element subsets of {1, 2, 3, 4, 5, 6 …, 17, 18, 19, 20} do not contain
two consecutive integers? Examples: {1, 5, 7, 9, 14} or {3, 6, 8, 17, 19} but not
{2, 3, 6, 8, 12} or {1, 5, 6, 7, 8} or {2, 6, 12, 13, 20}.
Hint: Suppose that you indicate such a set by marking 5 of the 20 integers
with X’s.
21. Get an identity involving the binomial coefficients by counting the number of
k-element subsets of {1,2,3,…, n, a,b,c} in two different ways.
Solution: First the number of k-element subsets of this n + 3 element set is
! n + 3$
simply #
. On the other hand, the number of k-element subsets can be
" k &%
expressed as the sum of the number of k-element subsets that contain no letters +
the number of k-element subsets that contain exactly one letter + the number of kelement subsets that contain exactly two letters + the number of k-element subsets
that contain all three letters. This latter sum
! n$
! n $
! n $ ! n $
is # & + 3#
+
3
#" k ' 2&% + #" k ' 3&% .
" k%
" k ' 1&%
! n$
! n $
! n $ ! n $ ! n + 3$
Hence, we get the identity # & + 3#
.
+ 3#
+
=
&
" k%
" k ' 1%
" k ' 2&% #" k ' 3&% #" k &%