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MATH 236: Discrete Mathematics with Applications
TUTORIAL 3: 28 February, 2013
1. Let cm,n be the number of onto functions from a set of m ≥ 1 elements to a
set of n ≥ 1 elements.
(1) Find cm,1 .
(2) Find cm,n when m < n.
(3) Find cm,2 when m ≥ 2.
(4) Find cm,m .
Solution. (1) cm,1 means n = 1 so that all elements must map onto 1
element, hence there is only 1 onto map. Thus cm,1 = 1.
(2) An onto map from a smaller set to a bigger set will be one to many, so
that this is not a function. Thus cm,n = 0 whenever m < n.
(3) Assume that X and Y = {a, b} are two set with |X| = m and |Y | = 2.
If f : X → Y is an onto map then {f −1 (a), f −1 (b)} is a partition of X, where
f −1 (a) 6= ∅, X and f −1 (b) = f −1 (a). So the number of onto maps is also the
number of nonempty subsets A of X with A 6= X. Hence cm,2 = 2n − 2.
(4) Each onto map between two sets of the same size is also a one-to-one
map so that cm,m = m!.
2. For integers m and n with 1 ≤ n ≤ m, the Stirling number of second kind
denoted by Sm,n is defined as the number of ways a set of size m can be partitioned into n subsets. If X is a set of size m and Y be a set of size n. Express
the number of onto functions from X to Y using Stirling number of second kind.
Solution. Partition S into n parts, then map each part onto an element of
Y. This map will be a bijection. The total number of maps will be the (number
of ways the elements of X can be partitioned into n parts) × (the number of
permutations on Y )=Sm,n · n!.
3. How many 3-digit numbers have distinct and non-zero digits?
Solution. Each 3-digit number has the form abc, where a, b, c are pair-wise
distinct and nonzero, so that there are 9 possibilities for a, 8 possibilities for b,
and 7 possibilities for c. In total, there are 9 · 8 · 7 numbers.
4. (1) How many odd integers are there from 10000 to 99999 ?
(2) How many integers from 10000 to 99999 have distinct digits?
(3) How many odd integers from 10000 to 99999 have distinct digits?
Solution. (1) Each odd integer from 10000 to 99999 has the form abcde,
where 1 ≤ a ≤ 9, e ∈ {1, 3, 5, 7, 9} and 0 ≤ b, c, d ≤ 9. Thus there are 9 · 10 · 10 ·
10 · 5 = 45000 odd integers.
(2) We must have that a 6= 0, and a, b, c, d, e are pair-wise distinct, so that
there are 9 · 9 · 8 · 7 · 6 integers with distinct digits.
(3) We must have that a 6= 0, and e ∈ {1, 3, 5, 7, 9} and a, b, c, d, e are pairwise distinct, so that there are 8 · 8 · 7 · 6 · 5 odd integers with distinct digits.
1
5. (1) How many ways can the letters of the word ALGORITHM be arranged
in a row?
(2) How many ways can the letters of the word ALGORITHM be arranged
in a row if the letters RIT must remain together (in order) as a unit?
Solution. (1) There are 9 letters so that there are 9! ways. This is the
number of permutations on a set of size 9.
(2) ‘RIT’ behaves like one letter so that the number of letters will be 7. Thus
there are 7! ways.
6. (1) How many ways can three of the letters of the word ALGORITHM be
selected and written in a row?
(2) How many ways can six of the letters of the word ALGORITHM be
selected and written in a row if the first two letters must be OR?
Solution. (1) Each way of writing three letters in a row is the number of
9!
= 9 · 8 · 7.
3-permutation on a set of size 9 so that this number is P (9, 3) = (9−3)!
(2) As the first 2 letters are fixed, the other 4 letters are chosen from the
7!
= 7 · 6 · 5 · 4 ways.
remaining 7 letters, so that there are P (7, 4) = (7−4)!
7. How many different ways can three of the letters of the word SAMPLE be
chosen and written in a row
(1) with repetition of letter allowed?
(2) without repetition of any letters?
(3) without repetition of any letters and containing letter A?
(4) with repetition of letters allowed and containing letter A?
Solution. (1) There are 6 possibilities for each letter so that there are 63
ways.
(2) There are 6 possibilities for a, 5 possibilities for b, and 4 possibilities for
c, so that there are 6 · 5 · 4 = 120 ways.
(3) there are 3 different positions for A, and when A is chosen, there are
5 · 4 = 20 choices for the remaining two letters so that there are 3 · 20 = 60 ways.
(4) There are 53 ways of choosing 3 letter without A and so there are 63 −53 =
91 ways of choosing 3 letters and writing in a row and containing A.
8. A committee of five: a president, secretary and 3 committee members, is
to be chosen from a group of 15 people. In how many ways can the committee
be chosen?
Solution. We can choose the president and secretary first and then choose
3 committee members from the remaining 13 people. There are
13
13
P (15, 2) ·
= 15 · 14 ·
= 15 · 14 · 13 · 12 · 11/6 = 60060
3
3
ways.
We can also choose 3 committee members and then choose the
president
and secretary from the remaining 15 − 3 = 12 people. There are 15
3 P (12, 2) =
15·14·13
·
12
·
11
=
60060
ways.
1·2·3
9. A committee of four is to be chosen from a club whose membership comprises
12 men and 10 women.
2
(1) In how many ways can the committee be chosen?
(2) In how many ways can the committee be chosen if it must include at
least 2 men?
Solution. (1) There are 12 + 10 = 22 people and so there are 22
4 ways.
(2) We can have 2 men, 2 women, 3 men, 1 woman or 4 men. Hence there
are
12 10
12 10
12 10
+
+
2
2
3
1
4
0
ways.
10. How many different five digit numbers can be constructed from the digits
1, 2, 3, 3, 3?
Solution. There is
P (5; 1, 1, 3) =
5!
= 20
1!1!3!
numbers.
11. (1) How many ways are there to arrange the letters of the word HULLABALOO?
(2) How many of these arrangements contain the three letter LAB next to
each other in order?
(3) How many of these arrangements in (a) have no adjacent Ls?
Solution. (1) There are 10 letters with 3 letters ‘L’, 2 letters ‘A’, 2 letters
‘O’ and 1 letter ‘H’, 1 letter ‘U’, and 1 letter ‘B’. There are
P (10; 3, 2, 2, 1, 1, 1) =
10!
3!2!2!1!1!1!
ways.
(2) the word LAB counts as one and so there are only 8 letters and so there
8!
are P (8; 2, 2, 1, 1, 1, 1) = 2!2!1!1!1!
ways.
(3) We first arrange 7 letters H, U, A, B, A, O, O and then insert 3 letters
L into 8 positions between these 7 letters. Hence there are P (7; 2, 2, 1, 1, 1) 83
ways.
12. Determine the number of (staircase) paths in the xy-plane from (1, 2) to
(9, 6) where each such path is made up of individual steps going one unit to the
right (R) or one unit upward (U).
Solution. There are 9 − 1 = 8 steps right R0 s and 6 − 2 = 4 step ups U 0 s.
Each path corresponds to a list of 8 ‘R’s’ and 4 ‘U’s’. Hence there are
P (8 + 4; 8, 4) =
12!
8!4!
paths.
13. Provide a combinatorial
argument to show that
n
2
(1) 2n
=
2
+
n
.
2
2
(2) n m+n
= (m + 1) m+n
m
m+1 .
3
(3)
m n
m
n
m n
=
+
+ ··· +
.
0
r
1
r−1
r
0
Solution. (1) The left side, namely 2n
2 , counts the number of ways of
choosing 2 members from a set of 2n people. This can be also counted as
follows: Think of the 2n people consisting of n men and n women. There are
different types of committees that can be chosen: those consisting of 2 women,
those with 1 women and 1 men, and those with no women. Now applying the
Addition Rule.
(2) Suppose we have a squad of m + n soccer players from which we wish
to select a team of m + 1 players including the goalie. This can be counted
as follows: On the one hand we first choose the m team players, excluding the
goalie, from the m + n players in the squad, and then from the remaining n
players choose a goalie. On the other hand, we could have counted as follows:
First choose the m + 1 team players from the m + n players in the squad and
then from these m + 1 team
players, choose a goalie.
(3) The quantity m+n
is the number of ways of choosing a committee of r
r
members from a group of m + n people. This can be counted as follows: Think
of the m + n people as consisting of m men and n women. There are r + 1
distinct types of committees that can be selected: For i = 0, 1, · · · , r, let type
i consist of all those committees that consist of i men and r − i women. Now
applying the Multiplication Rule and Addition Rule.
14. Using the Binomial theorem to find the coefficient of:
(1) x8 y 12 in the expansion of (2x2 − y 2 )10
2
2
x 8
) .
(2) xy2 in the expansion of ( xy 2 − 2y
Solution. (1) The coefficient of
m+n
r
(2x2 )4 (−y 2 )6
in the expansion of
(2x2 − y 2 )10
is
10
4
=
10 · 9 · 8 · 7
= 210.
1·2·3·4
4
6
Thus the coefficient of x4 y 12 in the expansion of (2x2 − y 2 )10 is 10
4 2 (−1) =
210 · 16 = 3360.
2
2
(2) We first need to write xy2 in the form ( xy 2 )k ( xy )8−k . Observe that
(
y 2 k x 8−k
x
) ( )
= ( )8−3k .
x2
y
y
Hence 8 − 3k = 2 and then k = 2. Therefore
x2
y2 2 x 6
=
(
) ( ) .
y2
x2
y
4
The coefficient of
coefficient of
x2
y2
y2 2
x 6
x2 ) (− 2y )
in the expansion of
y2
x2
is
8 −1 6
7
8·7
(
) = 7 =
.
2
2
2
16
5
−
x 8
2y )
is
8
2
and so the