- We have
ways to choose three
numbers from ten
numbers
. And the biggest sum
of these three
numbers is 9+8+7=24 , and the smallest sum of these three numbers
is 0+1+2=3 , thus we have 22 choice of the sum of the
three
numbers.
According to the Pigeonhole principle, there are at least
different ways to choose three numbers
such that they have the same sum. - We'll draw the transition table for the
FSM.
- (a).
- The transition table for the FSM is following
where
is the starting state,
and
is the final state.
- (b).
- The transition table for the FSM is following
where
is the starting state,
and
is the final state.
- (c).
- The transition table for the FSM is following
where
is the starting state,
and
is the final state.
- In order to form a number greater than 4,000,000 , the
most
significant digit should be
. When we choose
4 as
the most significant digit, we have
ways to choose the other digits.
When we choose 5 as the most significant digit, we have
ways to choose the other digits.
Hence we have
ways to form a number
greater than 4,000,000. - According to Course notes 1 page 23 (just
before section 2.8.2),
we have
numbers between
0000 and 9999 , whose
four digits form a nondecreasing sequence. - We'll use following
combinatorial arguments
- (a).
- The left handside says that we want to choose
r+k balls
from n+k balls. This can be done in
ways.
This may also be done by first splitting the n+k
balls into two
groups A and B each having
k and n balls,
respectively, and then picking (k-i) balls from
group A and
(r+i) balls from group B for
. For each
this can be done in
ways. However, since
we get
- (b).
- A committee of k may be chosen from
n people
in
ways and its chairman may be chosen in
ways. Thue
is the total number of
ways one
may choose a committee of k people out of
n, with a chairman.
For
the total number of ways of doing this is
therefore
. This is the
left hand side.
The same committee selection with a chairman can also accomplished by
first selecting a chairman, which can be done in
ways,
and then selecting an arbitrary subset of the remaining
(n-1)
people as members of the committee, which can be done in
ways. Thus
- Let x be any irrational number and let
be
jx-N(jx) where
N(jx) is the integer closest to
jx for
. Each
is an irrational number between
-1/2 and 1/2 . We will assume that
n is any even number;
the case where n is odd is messier.
Consider the n open intervals
and
for
:
If for some
,
or
then
we are done. If for none of the j's,
is
in
or
then all of the n
s should
be distributed among the
remaining (n-2) intervals. Since there are
n-2 intervals and
n numbers
, by the Pigeonhole
principle tells there should
be an interval,
for some
j,
contain a
and a
with r
<
s. Clearly then,
and therefore
.
- Let us define a random variable X
as the score Linda gets.
And we also define random variables
as the
scores Linda
gets for each true/false and multiple choice questions, respectively.
We have
Assuming independence we get,
- If we choose five books satisfying that no two books are
adjacent, that means these five books has already divided the shelf
into six parts
, which each part
satifying following constraints
And we also assume
Hence we have
And the number of solution to the above equation is
. Thus there are
ways to
choose
five books so that no two adjacent books are chosen.
-
- (a).
- If he has just one fruit a day, it will cost him seven days
to eat the whole fruits. The number of ways for him to consume is
.
- (b).
- If it will at most cost him seven days to consume the
fruits, then we will have six divisor to divide the fruits. The
number of ways to permutate the fruits and divisors is
.
- (c).
- If all the fruits are identical, the number of ways to
permutate the fruits and divisors is
.
- The number of ways to get thirteen cards is
.
- (a).
- We have only one ways to get all thirteen hearts, hence
the probability is
.
- (b).
- We have four ways to get thirteen cards of the same suit,
hence the probability is
.
- (c).
- We have
ways to get seven
spades and
six clubs, hence the probability is
.
- (d).
- We have
ways to
get seven cards of one suit and six cards of the other, hence the
probability is
.
- (e).
- We have
ways to get six cards
of one suit, six cards of a second
suit, and one card of the other.
- (f).
- We have
ways to get four cards
of one
suit and six cards of a second, two cards of the third, and one card
of the fourth, hence the probability is
.
-
- (a).
- Assume the number of bit string of length n not
containing three consecutive zeros is f(n) . And we
can add a zero
or a one in the end of a bit string of length n not
containing
three consecutive zeros to get a bit string of length of
n+1 .
If the last three bit is 100 , we can't add a zero to it.
We have
- (b).
- The initial condition is
- (c).
- We have
- (d).
- Lets assume
.
substituting the values for f , we get
- Let A denote the event that all three chips
drawn are white,
and let
denote the event that
the ith urn is sampled.
And according to Bayes theorem, we have
and how to draw three chips without replacement is really an example
of Hypergeometric distribution.
Hence
- First we shall declare there is a typing error in this problem,
it should be
Since the chips are drawn with replacement and we are asked the
probability of a white chip drawn for the first time, it's actually a
Geometric distribution, with
.
- Let A denote the event that only one color is
left, and let
B denote that only black chips are left, and let
W denote that
only white chips are left. We then have
In order to make only white chips left, we must draw at least
b
chips and at most b+w chips from the urn. It's a
Hypergeometric
distribution.
Symmetrically, we have
Hence we have