- Assume, there was
units of the isotope at
the hour n.
Then the amount of isotope left at the hour n+1 will be
. Hence we get the recurrence equation,
and the initial condition will specify what
was.
-
Let
be the number of bacteria after n
hours, where
. Then,
since
new bacteria are born at time n, these newly born
ones should be added to those that already existed at time (n-1).
>From this total
is subtracted because
that many died at
time n. Initial conditions should specify
and that
.
- (a). Let
be the number of permutations of a set with
n elements. Clearly,
. Then,
is the required recurrence relation, since the first
element
can be chosen in n ways, and the remaining
n-1 elements can be
arranged in
ways.
(b). Proof by iteration:
,
,
,
,
. The product of the left
hand side
of the above equations is
. The product of the right
hand sides of the above
equations is
. Hence,
Cancelling,
on
either side of the above equation we get
.
- Let
be the number of ways of climbing
n steps.
Then, by the conditions of the problem,
, since the
person could have reached step n either by
directly jumping to it from step n-3, or by jumping to it
from
step n-2 or by stepping on to it from step
n-1. The initial
conditions are,
and
.
- This problem was discussed in the solutions to review 1
problems. It should be clear that
and
. - (a). Let
be the number of
different ways the driver
can pay n = 5m cents using only nickels and dimes.
Then
, and
since the last coin
dropped by the driver could be either a nickel or a dime. This is
the fibbonaci series.
(b). The answer is
for m =
9.
and
.
- Let
be the number of ways of getting stamps valued
n cents. Initial conditions are,
and
. Then,
, since
stamps for n cents may be made up by adding a
10 cent stamp to
stamps for (n-10) cents, a 6 cent stamp to
stamps for
(n-6) cents, or a 4 cent stamp to stamps for
(n-4) cents. - Let
be the
nth Fibonacci number.
For n < 5 it is easy to see that
for the initial
conditions given in the problem. By repeated application of Fibonacci
expansion, we have,
=
=
=
. Thus,
for
also we have
. Hence,
for
all
.
For
we have
. Clearly,
for n = 1,
is divisible by 5.
-
Assume,
is
divisible by 5. If we now prove that
is
divisible by 5,
then we have the result.
by the result proven
above. But,
, since
is a Fibonacci number. Hence,
. Clearly,
is divisible
by 5 and
is divisible by 5
because of the induction
hypothesis. Therefore,
is divisible by 5.
Hence, the result.
- We have,
,
,
, and
- (a).
-
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ALT="tex2html_wrap_inline480" SRC="img61.gif" > ,
and
. You can add
up the result.
- (b).
- 11,100.
- (c).
-
.
- There are 200 integers divisible by 5,
142 divisible
by 7 (because
) and 28
which
are divisible by both 5 and 7 (because
). The answer is 1,000 minus the number
of integers which are divisible by 5 or 7, which is
1000 - (200 + 142 - 28). - This is the same as
(this is the
number of integers that are divisible by 3) minus the number of
integers that are divisible by both 3 and 7, namely
. -
The largest integer n such that
is 31.
The largest integer m such that
is 10.
The number of integers less than 1000 which are both squares and cubes
is just 1, since 1 is the only integer whose square is equal to its
cube. Hence, the answer is (31 + 10 - 1). - Consider
fish and rat as single tokens. The
number of
permutations containing fish is (26 - 4 + 1)! = 23!,
the number
of permutations containing the token rat is similarly,
(26 - 3 +
1)! = 24!, and the number of permumations containing both
fish
and rat is (26 - 7 + 2)! = 21!.
Hence the total number of permutations not containing fish
or
rat is, the total number of permutations of the 26 alphabets
minus the number of permutations containing fish or
rat.
This is 26! - (23! + 24! - 21!).
- Let
for
be the seven sets.
There will be seven singleton terms,
,
doubleton terms
,
triples
,
quadruples and
quintuples. For
j > 5 all intersections of j sets
will be empty. Hence, the
answer is
. - This is
minus the total number of bit strings that
do contain four consecutive 1's. The number of bit strings that
do contain 4 consecutive ones are
Hence, the
answer is
.
Notice, this problem is in general hard to solve for n-bit strings
for arbitrary
.
- If we throw a fair dice
8 times, the size of the sample
space is
. And that two sixes
and two ones and one others,
will appear in
.
Hence the probability of eight throws and get two sixs and two ones
and one others is
. - When we draw balls out
of the bag, with replacement, it's a
Binomial distribution, we have
When we draw balls out of the bag, without replacement. It's
Hypergeometric distribution (according to Theorem 3.6.3),
we have
- We draw cards from the pack without replcement. Again,
Hypergeometric distribution:
- This is also Hypergeometric distribution.
- (a).
- If we get more salmons than trouts, then the number of
salmons is either
. Hence we have
- (b).
- If we get seven males
- This uses Poisson distribution. An average of 3.2
telephone calls per hour to call X, in other word
.
- It's also a Poisson distribution with
.
- In certain distance s feet, we have
.
solve it get
.
It's no solution. So whatever the distance is we are never sure that
we can find a sapling.
- Actually it's a Binomial distribution with
p=1/50 , but we
also can use Poisson distribution to approximate it.
The claim that one out every fifty books brought out is a bestseller means
.
- It's Hypergeometric distribution. According to Theorem 3.6.3
(Larsen' book page 153), we have
- (a).
-
.
- (b).
-
- We can do it directly using Binomial distribution, certainly we
can use Poisson distribution to solve it. we have
.