a) If you forget about the best and the worst paper and only consider in how many ways you can arrange the rest, you see that tere are 6! ways to do that.
Then, you can obtain any "legal" way of arranging the 8 papers by putting the two remaining papers inbetween the 6 (or to the left or the right of all the 6 papers), using the rule that there is only room for one paper for each position.
If you have 6 papers then there are 7 places where you can put down an additional paper. So, there are 7*6 ways to pout the 2 remaining papers in. The total number of ways is thus:
6!*6*7 = 6*7!
ANother way to do this problem is by counting the total number of ways to arange 8 papers withoiyut any estrictions and then subtract the number of illegal arrangements. The total number of ways is 8!. To compute the total number of illegal ways, you can glu together the two papers so that it becomes one big paper, with one next to the other. There are two ways to do that. Then you effectively 7 papers and you can arrange them in 7! ways. Multiplying by two to take into accoiunt the oder in which you can glue them gives you 2*7! illegal ways. The total number of ways is thus:
8! - 2*7! = 8*7! - 2*7! = 6*7!
a)in how many different ways can 8 exam papers be arranged in a line so that the best and worst papers are not together?
b)in how many ways can 9 balls be divided equally among 3 students?
c)There are 10 true-false questions. By how many ways can they be answered?
d)lim(x tends to 0)x^x
6 answers
b) You can distribute the balls as follows. You put the 9 balls in some arbitrary order and then let student 1 take the first 3, student 2 the next 3, and student 3 the last 3. Then all possible ways to distribute the balls can be realized this way.
There are 9! possible ways to line up the 9 balls. But permuting the balls inside a group that goes to a particular student does not change the way the balls are distributed to the students.
There are 3!^3 ways to change the order of the balls in the three groups tat go to te students. So, for each possible way of distributing the balls to the student there are 3!^3 permutatons of the 9 balls. So, if the total number of ways to distribute the balls to the students in N, then N*3!^3 is the total number of ways to permute the 9 balls, which should be 9!. So, we have:
N = 9!/(3!^3)
There are 9! possible ways to line up the 9 balls. But permuting the balls inside a group that goes to a particular student does not change the way the balls are distributed to the students.
There are 3!^3 ways to change the order of the balls in the three groups tat go to te students. So, for each possible way of distributing the balls to the student there are 3!^3 permutatons of the 9 balls. So, if the total number of ways to distribute the balls to the students in N, then N*3!^3 is the total number of ways to permute the 9 balls, which should be 9!. So, we have:
N = 9!/(3!^3)
c) Let's write all the possibilities down. Any possible way to answer to the the questions is a string of ten letters
T F T T F....
where T stands for True and F for False. Now, you can generate all possible strings by expanding the product:
(T + F)(T + F)...(T + F) = (T + F)^10
where you expand out the product woithout changing the order of the T's and F's. So, in case of two questions, you would get:
(T + F)^2 = TT + TF + FT + FF
To count the total number of possibilities we need to count the number of terms in this expansion. We can do that simply by substituting
T = 1 and F = 1. So, we find thatthe toital number of possibilities is
(1+1)^10 = 2^10
T F T T F....
where T stands for True and F for False. Now, you can generate all possible strings by expanding the product:
(T + F)(T + F)...(T + F) = (T + F)^10
where you expand out the product woithout changing the order of the T's and F's. So, in case of two questions, you would get:
(T + F)^2 = TT + TF + FT + FF
To count the total number of possibilities we need to count the number of terms in this expansion. We can do that simply by substituting
T = 1 and F = 1. So, we find thatthe toital number of possibilities is
(1+1)^10 = 2^10
c) Let's write all the possibilities down. Any possible way to answer to the the questions is a string of ten letters
T F T T F....
where T stands for True and F for False. Now, you can generate all possible strings by expanding the product:
(T + F)(T + F)...(T + F) = (T + F)^10
where you expand out the product woithout changing the order of the T's and F's. So, in case of two questions, you would get:
(T + F)^2 = TT + TF + FT + FF
To count the total number of possibilities we need to count the number of terms in this expansion. We can do that simply by substituting
T = 1 and F = 1. So, we find thatthe toital number of possibilities is
(1+1)^10 = 2^10
T F T T F....
where T stands for True and F for False. Now, you can generate all possible strings by expanding the product:
(T + F)(T + F)...(T + F) = (T + F)^10
where you expand out the product woithout changing the order of the T's and F's. So, in case of two questions, you would get:
(T + F)^2 = TT + TF + FT + FF
To count the total number of possibilities we need to count the number of terms in this expansion. We can do that simply by substituting
T = 1 and F = 1. So, we find thatthe toital number of possibilities is
(1+1)^10 = 2^10
d)
If Lim x -->0 f(x) = L
then
Lim x---> x ---> g[f(x)] = g(L)
if g(x) is continuous at x = L.
Let's start with the known limit:
Lim x -->0 x Log(x) = 0
If we now take g(x) = exp(x) and use that exp(x) is continuous at x = 0, we get:
Lim x -->0 x^x = 1
If Lim x -->0 f(x) = L
then
Lim x---> x ---> g[f(x)] = g(L)
if g(x) is continuous at x = L.
Let's start with the known limit:
Lim x -->0 x Log(x) = 0
If we now take g(x) = exp(x) and use that exp(x) is continuous at x = 0, we get:
Lim x -->0 x^x = 1
Thanks that was really helpful. But the solution to the limit problem is still not clear to me.
1 answer