Download MATH 351 Fall 2009 Homework 1 Due: Wednesday, September 30

MATH 351 Fall 2009 Homework 1
Due: Wednesday, September 30
Problem 1. How many different letter arrangements can be made from the letters
BOOKKEEPER.
This is analogous to one of the problems presented in class, its also very similar to
the ”flag-code” problem. The solution is as follows. If each of the repeated letters
was distinct B1 O1 O2 K1 K2 E1 E2 P1 E3 R1 then there would be 10! ways to arrange these
letters. However we need to divide by 2!2!3! since each of the 3! permutations of the
E’s, the 2! permutations of the O’s and 2! of the K’s that give us the same word.
Thus the total number of words is:
10!
=.
3!2!2!
We also can think of this as 10 objects with 1 of type B, 3 of type E, 2 of type K,
and 2 of type 0, and 1 of type R. As verified in class the number of distinct orderings
from this set is exactly the same as the number above. We also called this number
10
.
1, 3, 2, 2, 1
Problem 2. There are 130 students at a school and three dormetories A, B, and C
with capacities 30, 35, and 65 respectively.
(a) How many ways are there to fill up the dormitories?
We have to split 130 distinct students into three distinct subgroups (dormitories A,
B and C) with fixed sizes (30, 35, and 65 respectively). This is exactly counted by:
130!
130
=
30, 35, 65
30!35!65!
(b) Suppose that of the 130 students, 60 are men and 70 are women and that A is
an all men’s dorm, B is an all women’s dorm and C is co-ed. How many ways
are there to fill up the dormitories?
1
60
To count this, first choose the 30 men for dorm A from the 60 possible men by
.
30
70
Second choose the 35 women from the 70 women, that will go into dorm B by
.
35
The remainder will then be in dorm C. By the basic principle of counting the total
possible placings of the students is the product:
60!70!
60
70
.
=
30
35
30!30!35!35!
Problem 3. In how many ways can ten people be seated in a row if:
(a) there are no restrictions on the seating arrangement?
This is a simple permutation of 10 distinct objects. Thus the solution is 10! =.
(b) persons A and B must sit next to each other?
If A and B must sit together we treat them as one ”person” and order them with the
remaining 8 people in 9! different ways. Then for each of those orderings we either
have A first or B first or 2 possible arrangements. By the basic counting principle we
have a total of 2 × 9! = orderings.
(c) there are 5 men and 5 women, and no 2 men or 2 women can sit next to each
other?
First choose if they sit mwmwmwmwmw or wmwmwmwmwm. Once this is chosen
then order the men in their seats. Since there are five of them there are 5! arrangements. Similarly the women can then be seated in 5! different ways. Thus the total
number of arrangements is 2 × 5! × 5! =.
(d) there are 5 men and they must sit next to each other?
2
If the 5 men must sit next to each other then if the seates are ordered #1 − #10
then the first man can be in seat #1 − #6. Choose this seat (6 ways). Then order
the 5 men (5!). Then order the 5 women in the remaining seats (5!). Thus the total
number of orderings is 6 × 5! × 5! =.
(e) there are 5 married couples, and each couple must sit together?
To count this treat each couple as an object and order them in 5! ways. Then once
their two seats are chosen there are 2 ways the first couple can sit (mw or wm). Then
there are 2 ways in which the second couple can sit and so forth. Giving a total of
5!25 = arrangements of the people.
Problem 4. How many integers greater then 5400 have both of the following
properties?
(a) The digits are distinct.
(b) The digits 2 and 7 do not occur.
To do this problem we split it into cases. First consider the numbers with 8 digits.
in this case we must order the digits 1, 3, 4, 5, 6, 8, 9, and 0. However when we pick
the digit for the first entry we only have 7 choices since 0 cannot be the first digit.
Then once we have chosen that digit we have to order the remaining 7 digits in any of
the 7! possible permutations. Thus the total number is 7 × 7! = 35280. In a similar
fashion there are 7 × 7! = 35280 seven digit numbers, 7 × 7 × 6 × 5 × 4 × 3 = 17640
six digit numbers of this type and 7 × 7 × 6 × 5 × 4 = 5880 five digit numbers. The
four digit numbers must be handled in cases since they must be bigger then 5400.
First count the numbers that are 6000 or bigger. To do this there are 3 choices for
the first digit (6,8, or 9) then 7 choices for the second, 6 for the third, and 5 for the
fourth position, giving 3 × 7 × 6 × 5 = 630 numbers of this type. Secondly consider
the numbers that begin with a 5. Then there are four choices for the second digit
(4,6, 8, or 9 ) that can be picked for the second digit. Then the third digit must be
chosen from the remaining 6 digits and then there are 5 choices for the fourth digit
Giving 4 × 6 = 120 numbers of this type. Thus the total number of integers satisfying
conditions (a) and (b) is 120 + 630 + 5880 + 17640 + 35280 + 35280 = 94830.
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Problem 5.
(a) Suppose that Michigan table tennis team of 16 players splits into pairs for
practicing. How many different splittings are possible?
One way to think of this is that you have to take 16 people and envision placing two
at each of 8 tables numbered 1-8. This is splitting 16 people into 8 distinct groups of
2 people each. The number of ways to do this is:
16!
16
.
=
2, 2, 2, 2, 2, 2, 2, 2
2!2!2!2!2!2!2!2!
However we have over counted since if Ms A and Mr. B are at table one or at table
2 makes no difference. Thus we must divide by 8!. Giving pairings.
(b) Suppose that Michigan table tennis team of 12 players competes with MSU
team of 12 players. How many different splittings are possible (each pair
contains a player of each team)?
This is actually easy to count because we envision the Michigan players lined up.
Then we start with the first one in line and pick a MSU player to be paired with him.
There are 12 ways to do this. Next we go to the second Michigan player and we have
11 possible choices of MSU players to play him, and so on. So a total of 12! possible
pairing are possible.
Problem 6. A classroom has 2 rows of 8 seats each. There are 14 students, 5 of
whom always sit in the front row and 4 of whom always sit in the back row. In how
many ways can the students be seated?
Consider the probblem by breaking it into pieces. First suppose the ”back-row sitters”
walk into the room. Then there are 8 choices of where the first one to walk in can sit.
7 for the second, 6 for the third and 5 for the fourth. Next the ”front-row-sitters”
walk in. There are 8 choices for the first 7 for the second, ...4 for the 5th. Now the
five remaining students walk in one at a time. There are 7 seats remaining and the
first of these students that walks in has 7 choices of where to sit, the second has six
choices, .... the fifth has 3 choices of where to sit. Viewing the problem in this way
tells us that the answer should be:
(8 · 7 · 6 · 5)(8 · 7 · 6 · 5 · 4·)(7 · 6 · 5 · 4 · 3·) = 28449792000
4
Problem 7. From a group of 8 women and 6 men a committee consisting of 3 men
and 3 women is to be formed. How many dierent committees are possible if:
(a) 2 of the men (Mr. X and Mr. Y) refuse to serve together;
We count this by first considering all the possible committees with out any restriction
and then subtracting the number of committees
with Mr. A and Mr. B. The total
6
number of ways to choose 3 men from 6 is
and the total ways of choosing 3
3
8
women from 8 is
. Thus the total number of committees with 3 men and 3
3 6
8
women is the product
= 1120. Now if we count the number of committees
3
3
that have both Mr. A and Mr. B then we must choose the one
remaining man from
4
8
the other 4 or
and then choose the 3 women from 8
. Giving a total of
1
3
8
4·
= 224. Thus subtracting the two gives 1120 − 224 = 896. (Note one could do
3
this problem by considering the possible committees with neither A nor B and then
as a separate case consider the committees with exactly one of A or B and then add
the cases together)
(b) 2 of the women (Ms. A and Ms. B) refuse to serve together;
This proceeds in an analogous fashion to part (a) and we subtract the committees
with both Ms. A and Ms. B from the total (1120 as counted in part (a)). The number
of
that Ms. A and Ms B can serve together is by first choosing
ways
the three men
6
6
and then choosing one more woman from the other 6 or
= 6. Thus the
3
1
6
total number of committees with both Ms. A and Ms. B is 6 ·
= 120. Thus, the
3
total number of committees when they DO NOT serve together is 1120 − 120 = 1000.
(c) 1 man (Mr. X) and 1 woman (Ms. A) refuse to serve together?
5
Again we proceed in a similar fashion and count the committees when Mr. X and Ms.
A serve
5,
together. To do this we first choose the 2 other men from the remaining
5
7
or
= 10 and then choose the other 2 women from the remaining 7 or
= 21.
2
2
Thus the total number of ways in which they can serve together is the product of
these two numbers of 210. Now the total number of committees in which they do not
serve together is our total number of committees with no restrictions (1120) minus
the committees where they do serve together (210). Thus the answer is 910.
Problem 8. Let r, n, and m be positive integers, and let r < n and r < m. Then
the following identity holds:
X
r n+m
n
m
=
.
r
i
r−i
i=0
Prove it combinatorially by considering a group of n men and m women, and determining (in two different ways) how many groups of size r are possible.
To prove this we follow the suggestion and consider a group of n men and m women,
and count the number of committees of r people we can choose from the total (n+m)
people. Clearly one way to count this is the left hand side (LHS) of the equation we
wish to verify.
Now consider counting the same thing but breaking it into cases were we consider
the committees that have no men, exactly 1 man, exactly 2 men, ...., all r are men.
To do this consider the arbitrary case where there are i men in the committee
and r-i
n
women. Thus we must choose the i men from the total n men or
and we must
i
m
choose the r-i women from the total m women or
Thus the total number of
r−i
committees
with r people with i of them men and r-i of them women is the product,
n
m
. Now since we want to count all the possible committees we must add
i
r−i
all the cases where i = 0, i = 1, ....i = r. This is exactly the right hand side of the
equation we wish to verify.
Since both the left and the right hand side count the same thing, they must be equal.
Problem 9. In the expansion of (z + 8y)7 , what is the coefficient of z 4 y 3 ?
6
Recall the binomial theorem:
n
(a + b) =
n X
n
i
i=0
ai bn−i .
Using this,
7
(z + 8y) =
7 X
7
i=0
i
z i (8y)7−i .
7 4
If I want the coefficient of z y then i must consider the term
z (8y)3 . Thus the
4
7
coefficient of that term is
(8)3 = 17920
4
4 3
Problem 10. There are 12 people to attend a dinner party and sit at a round table.
(a) How many ways can these 12 people be arranged at the table? ( supposing that
any seating arangement is equivalent to another if they are rotations of each
other)
Since any arrangement of the people at the table can be rotated so that a certain
distinguished person (call him Bob) is sitting at the head of the table. We can count
just the arrangements where Bob is sitting at the top of the table. To do this we
order the remaining 11 people around the table in 11! ways.
(b) Suppose there are 6 men and 6 women, How many seating arrangements are
there if no two men are to sit together.
In the same fashion we suppose Bob is sitting at the top. Then the person sitting to
he left must be a woman and there are 6 choices for that seat. Then we go the her
left and there must be a man (other then bob so there are 5 choices for this seat. We
continue around the table giving a total of 6! · 5! = 86400 arrangements around the
table.
(c) How many arrangements are there if the host and the hostess are to sit opposite
eachother, but the rest of the guests may sit anywhere?
7
Again we proceed in the same fashion and suppose that the Host is at 12 o’clock and
that the hostess is seated at 6 o’clock. Thus we must just order the remaining 10
people and there are 10! = 3628800 ways to do so.
Problem 11 How many integer solutions of:
x1 + x2 + x3 + x4 = 30
satisfy, x1 ≥ 2, x2 ≥ 0, x3 ≥ −5, and x4 ≥ 8?
To solve this problem we re-scale the variables by letting y1 = x1 − 2, y2 = x2 ,
y3 = x3 + 5, and y4 = x4 − 8. Solving for each xi and substituting into the equation
above we obtain:
(y1 + 2) + y2 + (y3 − 5) + (y4 + 8) = 30,
where each yi is now greater then or equal to 0. Manipulating the constants gives us:
y1 + y2 + y3 + y4 = 25
. As per definition in class, the number of integer solutions to this equation is given
by:
28 · 27 · 26
25 + 4 − 1
28
= 3276.
=
=
4−1
3
3!
8