ECE 316 – Probability Theory and Random Processes
Assignment 1 Solutions
Xinxin Fan
Problems
1. (a) How many different 7-place license plates are possible if the first 2 places are for letters
and other 5 for numbers?
Solution. By the generalized version of the basic principle the answer is
26 · 26 · 10 · 10 · 10 · 10 · 10 = 67, 600, 000.
(b) Repeat part (a) under the assumption that no letter or number can be repeated in a single
licence plate.
Solution. By the generalized basic principle of counting the answer is
26 · 25 · 10 · 9 · 8 · 7 · 6 = 19, 656, 000.
2. How many outcome sequences are possible when a die is rolled four times, where we say, for
instance, that the outcome is 3, 4, 3, 1 if the first roll landed on 3, the second on 4, the third
on 3, and the fourth on 1?
Solution. Note that there are 6 possible outcomes when a die is rolled. Therefore, By the
generalized basic principle of counting the answer is
6 · 6 · 6 · 6 = 1296.
3. Twenty workers are to be assigned to 20 different jobs, one to each job. How many different
assignments are possible?
Solution. An assignment is a sequence i1 , . . . , i20 where ij is the job to which person j is
assigned. Since only one person can be assigned to a job, it follows that the sequence is a
permutation of the numbers 1, . . . , 20 and so there are 20! different possible assignments.
4. John, Jim, Jay, and Jack have formed a band consisting of 4 instruments. If each of the boys
can play all 4 instruments, how many different arrangements are possible? What if John and
Jim can play all 4 instruments, but Jay and Jack can each play only piano and drums?
Solution. If each of the boys can play all 4 instruments, there are 4! = 24 possible arrangements. By first assigning the piano and drum to Jay and Jack, John and Jim can only play
the remaining 2 instruments. Therefore, by generalized basic principle there are 2 · 1 · 2 · 1 = 4
possibilities.
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7. (a) In how may ways can 3 boys and 3 girls sit in a row?
Solution. It is a permutation of 6 people. Therefore, there are 6! = 720 possible ways.
(b) In how may ways can 3 boys and 3 girls sit in a row if the boys and the girls are each to
sit together?
Solution. We can first ask 3 boys to sit and there are 3! possible ways. Then 3 girls have
the following two choices for their positions.
0 1 0 1 0 1 or 1 0 1 0 1 0,
where “1” denotes boys and “0” represents girls. Furthermore, 3 girls can permutate their
positions. Therefore, there are totally 3! · 2 · 3! = 72 possible ways.
(c) In how may ways if only the boys must sit together?
Solution. We can first ask girls to sit and there are 3! possible ways. Since the 3 boys must
sit together, they have the following four choices for their positions.
1 1 1 0 0 0 or 0 1 1 1 0 0 or 0 0 1 1 1 0 or 0 0 0 1 1 1,
where “1” denotes boys and “0” represents girls. Furthermore, 3 boys can permutate their
positions. Therefore, there are totally 3! · 4 · 3! = 144 possible ways.
(d) In how may ways if no two people of the same sex are allowed to sit together?
Solution. Assume that there are 6 positions in a row. The first position can be occupied
by any of 6 people. Then the second position can be occupied by any of 3 people who has
the different sex with the person in the first position; and so on. Hence it follows from the
generalized basic principle of counting that there are 6 · 3 · 2 · 2 · 1 · 1 = 72 possible ways.
10. (a) In how may ways can 8 people be seated in a row if there are no restrictions on the seating
arrangement?
Solution. It is a permutation of 8 people. Therefore, there are 8! = 40, 320 possible ways.
(b) In how may ways can 8 people be seated in a row if persons A and B must sit next to each
other?
Solution. We can first ask 7 people including one of persons A and B to be seated in a row
and there are 7! possible ways. Then the remaining person (A or B) only has two choices for
his position since he must next to the particular person (B or A). Therefore, there are totally
7! · 2 = 10, 080 possible ways.
(c) In how may ways can 8 people be seated in a row if there are 4 men and 4 women and no
2 men and 2 women can sit next to each other?
Solution. This question can be solved following the same reasoning as the question 7(d).
Assume that there are 8 positions in a row. The first position can be occupied by any of 8
people. Then the second position can be occupied by any of 4 people who has the different
sex with the person in the first position; and so on. Hence it follows from the generalized
basic principle of counting that there are 8 · 4 · 3 · 3 · 2 · 2 · 1 · 1 = 1152 possible ways.
(d) In how may ways can 8 people be seated in a row if there are 5 men and they must sit
together?
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Solution. This question can be solved following the same reasoning as the question 7(c).
We can first ask 3 women to sit and there are 3! possible ways. Since the 5 men must sit
together, they have the following four choices for their positions.
1 1 1 1 1 0 0 0 or 0 1 1 1 1 1 0 0 or 0 0 1 1 1 1 1 0 or 0 0 0 1 1 1 1 1,
where “1” denotes men and “0” represents women. Furthermore, 5 men can permutate their
positions. Therefore, there are totally 3! · 4 · 5! = 2, 880 possible ways.
(e) In how may ways can 8 people be seated in a row if there are 4 married couples and each
couple must sit together?
Solution. Firstly, 4 groups of couples can permutate their positions and there are 4! possible
ways. Then within one group, the husband and the wife also can exchange their positions.
Therefore, there are totally 4! · 24 = 384 possible ways.
11. (a) In how may ways can 3 novels, 2 mathematics books, and 1 chemistry book can be arranged
on a bookshelf if the books can be arranged in any order?
Solution. It is a permutation of 6 books. Therefore, there are 6! = 720 possible ways.
(b) In how may ways can 3 novels, 2 mathematics books, and 1 chemistry book can be arranged
on a bookshelf if the mathematics books must be together and the novels must be together?
Solution. The position of 3 categories of books can be permutated and there are 3! possible
ways. Furthermore, the position of books within one group can also be permutated and there
are 3!, 2! and 1! possible ways for novels, mathematics books and chemistry book, respectively.
Therefore, there are totally 3! · 3! · 2! · 1! = 72 possible ways.
(c) In how may ways can 3 novels, 2 mathematics books, and 1 chemistry book can be arranged
on a bookshelf if the the novels must be together but the other books can be arranged in any
order?
Solution. This question can be solved following the same reasoning as the question 7(c).
We can first arrange 3 other books and there are 3! possible ways. Since the 3 novels must
be arranged together, they have the following four choices for their positions.
1 1 1 x x x or x 1 1 1 x x or x x 1 1 1 x or x x x 1 1 1,
where “1” denotes novels and “0” represents other books. Furthermore, the positions of 3
novels can be permutated. Therefore, there are totally 3! · 4 · 3! = 144 possible ways.
19. (a) From a group of 8 women and 6 men a committee consisting of 3 men and 3 women
is to be formed. How many different committee are possible if 2 of the men refuse to serve
together?
( )( )
Solution. There are 83 43 possible committees that do not contain either of the 2 men, and
( )( )( )
there are 83 21 42 possible committees that contain exactly 1 of them. Therefore, there are
(8)(4) (8)(2)(4)
totally 3 3 + 3 1 2 = 896 possible committees.
(b) From a group of 8 women and 6 men a committee consisting of 3 men and 3 women is
to be formed. How many different committee are possible if 2 of the women refuse to serve
together?
3
( )( )
Solution. There are 63 63 possible committees that do not contain either of the 2 women,
( )( )( )
and there are 63 21 62 possible committees that contain exactly 1 of them. Therefore, there
(6)(6) (6)(2)(6)
are totally 3 3 + 3 1 2 = 1000 possible committees.
(c) From a group of 8 women and 6 men a committee consisting of 3 men and 3 women is to
be formed. How many different committee are possible if 1 man and 1 woman refuse to serve
together?
( )( )
( )( )
Solution. There are 73 53 possible committees in which neither feuding party serves, 72 53
( )( )
possible committees in which the feuding woman serves, and 73 52 possible committees in
( )( ) ( )( ) ( )( )
which the feuding man serves. Therefore, there are totally 73 53 + 72 53 + 73 52 = 910 possible committees.
21. Consider the grid of points shown below. Suppose that starting at the point labeled A you can
go one step up or one step to the right at each move. This is continued until the point labeled
B is reached. How many different paths from A to B are possible?
Solution. Each path is a linear arrangement of 4 r’s and 3 u’s (r for right and u for up).
For instance the arrangement r, r, u, u, r, r, u specifies the path whose first 2 steps are to the
right, next 2 steps are up, next 2 are to the right, and final step is up. Therefore, there are
7!
3!4! = 35 different paths from A to B are possible.
23. A psychology laboratory conducting dream research contains 3 rooms, with 2 beds in each
room. If 3 sets of identical twins are to be assigned to these 6 beds so that each set of twins
sleeps in different beds in the same room, how many assignments are possible?
Solution. Each set of twins can sleep in the different rooms and there are 3! possibilities.
Furthermore, each person of the twins can select different beds within one room. Therefore,
there are totally 3! · 23 = 48 possible assignments.
24. Expand (3x2 + y)5 .
Solution. Using the binomial theorem, we have
()
()
()
()
(3x2 + y)5 = y 5 + 51 (3x2 )1 y 4 + 52 (3x2 )2 y 3 + 53 (3x2 )3 y 2 + 54 (3x2 )4 y 1 + (3x2 )5
= y 5 + 15x2 y 4 + 90x4 y 3 + 270x6 y 2 + 405x8 y + 243x10 .
25. The game of bridge is played by 4 players, each of whom is dealt 13 cards. How many bridge
deals are possible?
( 52 )
Solution. There are 13,13,13,13
possible bridge deals.
28. If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What
if each school must receive 2 teachers?
Solution. Assuming teachers are distinct, then there are 48 possible divisions since each
teacher
( 8 ) can8! be divided to any of 4 schools. If each school must receive 2 teachers, there are
2,2,2,2 = 24 = 2520 possible divisions.
30. Delegates from 10 countries, including Russia, France, England, and the United States, are
to be seated in a row. How many different seating arrangements are possible if the French
and English delegates are to be seated next to each other, and the Russia?
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Solution. We first arrange seats for delegates from other 6 countries and there are 6! possible
arrangements. Then we arrange two adjacent seats for delegates from France and England.
There are 2! · 7 = 14 possible arrangements. Finally, we arrange non-adjacent seats for
delegates from Russia and the United States and keep
(8) the adjacent seats for the French and
English delegates at the same time. There are 2! · 2 = 56 possible arrangements. Hence,
it follows from the generalized basic principle of counting that there are totally 6! · 14 · 56 =
564, 480 possible arrangements.
Theoretical Exercises
2. Two experiments are to be performed. The first can result in any of m possible outcomes.
If the first experiment results in outcome number i, then the second experiment can result in
any of ni possible outcomes, i = 1, 2, . . . , m. What is the number of possible outcomes of the
two experiments?
∑
Solution. The number of possible outcomes of the two experiments is m
i=1 ni .
5. Determine the number of vectors (x1 , . . . , xn ), such that each xi is either 0 or 1 and
n
∑
xi ≥ k.
i=1
( )
Solution. There are nj different 0 – 1 vectors whose sum is j, since any such vector can be
characterized
∑ by( a) selection of j of the n indices whose value are then set equal to 1. Hence
there are nj=k nj vectors that meet the criterion.
(
) ( )( ) ( )(
)
( )( )
n+m
n m
n
m
n m
=
+
+ ··· +
.
r
0
r
1
r−1
r
0
(n+m)
Proof. Consider
groups of size r. As
r
(n)( m ) a group of n men and m women. There are
there are i r−i groups of size r that consist of i men and r − i women, we see that
8. Prove that
)
(
) ∑
r ( )(
n
m
n+m
.
=
i
r−i
r
i=0
10. For a group of n people, suppose that we want to choose a committee of k, k ≤ n, one of whom
is to be designated as chairperson.
(a) By focusing( first
) on the choice of the committee and then on the choice of the chair, argue
that there are nk k possible choices.
( )
Solution. We first choose the committee and there are nk possible choices. Since (each
)
member in the committee of k people can be designated as the chair, there are totally nk k
possible choices.
(b) By focusing first on the choice(of the
) nonchair committee members and then on the choice
n
of the chair, argue that there are k−1
(n − k + 1) possible choices.
5
( n )
Solution. We first choose the nonchair committee members and there are k−1
possible
(choices.
) Then we can choose a chair from the remaining n − k + 1 people. Hence, there are
n
k−1 (n − k + 1) possible choices.
(c) By focusing first on the choice
( of )the chair and then on the choice of the other committee
members, argue that there are n n−1
k−1 possible choices.
Solution. We first choose a chair from n people and there are n possible choices. Then we
can choose
( ) the other k − 1 committee members from the remaining n − 1 people. Hence, there
are n n−1
k−1 possible choices.
(d) Conclude from parts (a), (b), and (c) that
( )
(
)
(
)
n
n
n−1
k
= (n − k + 1)
=n
.
k
k−1
k−1
Solution. The parts (a), (b) and (c) describe three different methods to form a committee
of k people, one of whom is the chair. Although the order of the selection is different, three
methods will give the same result, which implies the above three equalities.
( )
(e) Use the factorial definition of m
r to verify the identity in part (d).
Solution. We can verify the identities
( )
n
k
=
k
(
)
n
(n − k + 1)
=
k−1
(
)
n−1
n
=
k−1
13. Show that for n > 0,
in part (d) as follows:
k!n!
n!
=
(n − k)!k!
(n − k)!(k − 1)!
(n − k + 1)n!
n!
=
(n − k + 1)!(k − 1)!
(n − k)!(k − 1)!
n!
n(n − 1)!
=
(n − k)!(k − 1)!
(n − k)!(k − 1)!
( )
n
∑
n
= 0.
(−1)i
i
i=0
Proof. Using the binomial theorem to expand (−1 + 1)n , we have the following:
( )
n ( )
n
∑
∑
n
i n−i
i n
0 = (−1 + 1) =
(−1) 1
=
(−1)
,
i
i
n
i=0
i=0
which gives you the result.
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