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
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!
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)
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
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
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
Thanks that was really helpful. But the solution to the limit problem is still not clear to me.
a) To solve this problem, we need to consider the different possible positions for the best and worst papers not being together. Let's denote the best paper as "B" and the worst paper as "W".
Step 1: Place the best paper (B) in any position on the line. There are 8 possible positions for this.
Step 2: Now, there are two scenarios to consider:
a) If the best paper is placed at either end of the line, there are 6 remaining positions for the worst paper (W).
b) If the best paper is placed at any other position, there are 7 remaining positions for the worst paper (W).
Step 3: For each scenario, calculate the number of ways to arrange the remaining papers (6 papers in scenario a, and 7 papers in scenario b).
Step 4: Add up the number of ways for each scenario.
The answer will be: (2 * 6!) + (6 * 7!).
b) To determine the number of ways to divide 9 balls equally among 3 students, we can use a concept called "stars and bars" or the "balls and urns" method.
Imagine we have 9 indistinguishable balls and 2 dividers (represented by bars). We need to distribute the balls among 3 students, and since the balls are indistinguishable, the only thing that matters is the number of balls each student receives.
Arrange the 9 balls and 2 dividers in a line, for a total of 11 objects. The first student will get the number of balls to the left of the first divider, the second student will get the balls between the two dividers, and the third student will get the balls to the right of the second divider.
Using the stars and bars method, calculate the number of ways to arrange these 11 objects: (11 choose 2).
The answer will be: (11 choose 2) = 55.
c) For each of the 10 true-false questions, there are 2 possible answers (true or false). Since each question is independent of the others, we can multiply the number of possibilities together.
For each question, there are 2 possible answers.
Therefore, the total number of ways to answer all 10 questions is: 2^10 = 1024.
d) To evaluate the limit as x tends to 0 of x^x, we can start by noticing that this expression is indeterminate when x = 0^0.
To further explore the limit, we can take the natural logarithm (ln) of both sides, as the natural logarithm is a continuous function. This will help us simplify the expression: ln(x^x) = x * ln(x).
Now, we can rewrite the limit as x tends to 0 as: lim(x tends to 0) [x * ln(x)].
Applying L'Hôpital's Rule, differentiate both the numerator and denominator with respect to x: lim(x tends to 0) [(1 * ln(x) + x * (1/x))].
Simplifying further, we get: lim(x tends to 0) [ln(x) + 1].
As x approaches 0, ln(x) approaches negative infinity. Therefore, the limit as x tends to 0 of x^x is 0.