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The area of the shaded region is the sum of the circle and triangle minus the white region. 

The white region is a 60-degree sector, so its area is \({10 \times {60 \over 360}} = {5 \over 3}\)

This means that the area of the shaded region is \(20 - {5 \over 3 } = {\color{brown}\boxed{55 \over 3}}\)

sorry, but your answer is wrong

so as u said "from white balls numbered from 1 to 12, and one SuperBall is drawn (at random) from red balls numbered from 13 to 20. When you buy a ticket, you choose three numbers from 1 to 12, and one number from 13 to 20".
14.1%

The combined area of the two gray regions is 16*sqrt(3) - 4*pi.

You made a small typo. The answer should be: 1/2

Thanks Justin.

You had basically the right idea too.  We both used complimentary counting.

Here is a very lay definition:

A convergent series is one where the sum of the terms gets closer and closer to a specific constant.  It converges.

A divergent series is one where the sum of the terms diverges off to pos or neg infinity.

If you look at this sight there a a number of examples of divergent and convergent series.

Second/third lines 'Applying this to the triangle BID'.

BID is not a triangle.

6th line, DI = 6.

The question says that DI = 3.

etc.

Your approach to the problem is correct. 

The total number of ways to arrange the 9 students without any restrictions is 9! = 362,880.

To count the number of arrangements that meet the given conditions, we can use the complementary counting principle. First, we count the number of arrangements in which no two sophomore students are standing next to each other, or no freshman students are at the ends of the line. Then, we subtract this count from the total number of arrangements to get the desired count.

As you correctly noted, there is only one way to arrange the students such that there are no two sophomore students standing next to each other and no freshman students at the ends of the line. This gives us 5*4*3!*4! = 28,800 arrangements. 

Therefore, the number of arrangements that meet the given conditions is:

9! - 28,800 = 97920

So, the answer you arrived at, 97920, is correct.

The code you provided calculates the values of the binomial coefficients (n choose 4) for n from 1 to 11 using the Python syntax for binomial coefficients, `(11+n) nCr 4`, and stores them in a list called `a`. 

The print statement then outputs the values of `a` and their sum. The sum of `a`, which is calculated correctly by the code, is 32857.

However, the statement that this is the total number of solutions is not correct without additional context. It is unclear what problem these binomial coefficients are related to, so it is difficult to determine whether the total number of solutions is indeed 32857.

To find the number of paths from A to I, if each step must be in a right-ward direction, we can use the principle of multiplication. 

Starting from A, we have only one option to move to C. From C, we can move to either D or E. From D, we can only move to E, and from E, we can move to either F or G. From F, we can only move to G, and from G, we can move to either H or I. Finally, from H, we can only move to I.

Therefore, the total number of paths from A to I, moving only in a right-ward direction, is the product of the number of choices at each step:

1 (choice at A) x 2 (choices at C) x 1 (choice at D) x 2 (choices at E) x 1 (choice at F) x 2 (choices at G) x 1 (choice at H) = 4

So there are 4 possible paths from A to I, if each step must be in a right-ward direction.

We can solve this problem using the multiplication principle of counting. 

First, we need to choose one American skater to win a medal. There are 'n' ways to do this, where 'n' is the number of American skaters. 

Next, we need to choose two skaters from the remaining non-American skaters to win the other two medals. There are (n-1) ways to do this, since we have already chosen one American skater to win a medal. We can choose the first non-American skater to win a medal in (n-1) ways, and then choose the second non-American skater to win a medal in (n-2) ways (since there is one less non-American skater remaining).

Therefore, the total number of ways to award the medals, given that exactly one American skater wins a medal, is:

n * (n-1) * (n-2)

Note that this expression assumes that there are at least three skaters competing, and that there is at least one American skater. If either of these assumptions is not true, the expression will need to be modified accordingly.

Let X be the random variable that represents the number of 6's rolled when three fair 6-sided dice are rolled. We can define X as the sum of three indicator variables, Xi, where Xi is 1 if the ith die rolls a 6 and 0 otherwise. 

Then, the expected value of X is:

E(X) = E(X1 + X2 + X3)
     = E(X1) + E(X2) + E(X3)

Since each die is fair, the probability of rolling a 6 on any given roll is 1/6. Therefore, the expected value of each Xi is:

E(Xi) = 1/6 * 1 + 5/6 * 0
      = 1/6

Substituting this into the equation for E(X), we get:

E(X) = E(X1) + E(X2) + E(X3)
     = 1/6 + 1/6 + 1/6
     = 3/6
     = 1.5

Therefore, the expected number of 6's rolled when three fair 6-sided dice are rolled is 1.5.

Jeff can choose his lunch in 24 ways if he has to choose one soup, one sandwich, and one dessert.

To see why, we can use the multiplication principle of counting. Jeff has 3 choices of soup, 4 choices of sandwich, and 2 choices of dessert. To determine the total number of ways he can choose his lunch, we multiply the number of choices for each category: 

3 soups x 4 sandwiches x 2 desserts = 24

Therefore, Jeff can choose his lunch in 24 ways if he has soup, a sandwich, and dessert.

(a) What is the probability that Valeria has two heads on her first turn?

:-  Assuming that Valeria is flipping two coins on her first turn, the probability of getting two heads is calculated as follows:

The probability of getting a head on a single coin flip is 1/2, since there are two possible outcomes (heads or tails) that are equally likely.

To calculate the probability of getting two heads, we need to multiply the probability of getting a head on the first flip by the probability of getting a head on the second flip, since these are independent events. Thus:

P(two heads) = P(head) x P(head) = (1/2) x (1/2) = 1/4

Therefore, the probability that Valeria has two heads on her first turn is 1/4 or 0.25.

Thanks Justin :)

Are you looking for help Rob

or are you just posting a fun puzzle question?

Either way it is fine but you should make it clear whether or not you are actually looking for help.   

Hi Guest and Justin,

Thanks for answering Justin, just thought I'd give it a go myself.  :)

A debate team consists of 5 freshmen and 4 sophomores.  (Everyone is distinguishable.)  In how many ways can they stand in line, so at least two of the sophomores are standing next to each other, and at least two of the freshmen are at the ends of the line?

The only way you can have  2 freshman at the end and NOT 2 sophomore together is

S F S F S F  S        F F 

So the number of ways this can happen is     5*4  * 3!*4! = 20*6*24 = 2880

Now all the possible combinations wiith  FF at the end but no other restrictions is  5*4 * 7! 

So the number of ways that has 2 freshmen at the end AND at least 2 sofomores next to each other is    

\(5*4*7!    \quad  -\quad        5*4  * 3!*4!   \\ =5*4*4!(5*6*7-6)\\ =4*5*4!*6*34\\ =97920\)

I suggest you check what we have both done and decide for yourself which answer (if any) is correct.

listfor(n, 10, 29, (n+1) / n) == (11 / 10, 12 / 11, 13 / 12, 14 / 13, 15 / 14, 16 / 15, 17 / 16, 18 / 17, 19 / 18, 20 / 19, 21 / 20, 22 / 21, 23 / 22, 24 / 23, 25 / 24, 26 / 25, 27 / 26, 28 / 27, 29 / 28, 30 / 29)==3

His first fraction == 11 / 10

Fix one of the scientists at the head of table:

The other 4  can sit either to the left or to the right, or split 2 to the left and 2 to right, or 1 to right and 3 to the left or vise versa (it makes no difference) in this many ways:

4! = 24 ways

The remaining people (2 + 5 + 2 + 3 = 12) can be seated in:

12! =479,001,600 ways - since there is no restriction to their seating.

So, the total number of ways =24 x 479,001,600 = 11,496,038,400 

We can approach this problem by first noting that in order for the product of two adjacent numbers to be even or a multiple of 3, at least one of the numbers must be even or a multiple of 3. Thus, we can group the numbers into two sets: even numbers and multiples of 3, and odd numbers that are not multiples of 3.

There are 5 even numbers (2, 4, 6, 8, 10) and 3 multiples of 3 (3, 6, 9), so there are 8 numbers in the first set. We can arrange these numbers in 8! ways.

There are 2 odd numbers that are multiples of 3 (3, 9) and 3 odd numbers that are not multiples of 3 (1, 5, 7), so there are 5 numbers in the second set. We can arrange these numbers in 5! ways.

Now, we need to consider the placement of the second set of numbers (odd numbers). Since the product of two adjacent numbers must be even or a multiple of 3, the odd numbers can only be placed in the spaces between the even numbers and multiples of 3.

There are 9 spaces between the 8 even numbers and multiples of 3. We need to choose 5 of these spaces to place the 5 odd numbers. We can do this in 9 choose 5 ways (or 9!/5!4!).

Therefore, the total number of arrangements is 8! × 5! × [9 C 5] =609,638,400 .

1 - If each twin gets "0" candies, then there are: [10 + 3 -1] C [3 - 1] ==66 ways

2 - ........................   1  ........................................: [8 + 3 - 1] C [3 - 1] ==45 ways

3 - ........................  2  .........................................: [6 + 3 - 1 ] C [3 - 1] ==28 ways

4 - ........................  3  .........................................: [4 + 3 - 1] C [3 - 1] == 15 ways

5 - ........................  4...........................................: [2 + 3 - 1] C [3 - 1] == 6 ways

6 - ........................  5  ..........................................: [0 + 3 - 1] C [3 - 1] == 1 way

7 - Total number of ways ==[66 + 45 + 28 + 15 + 6 + 1] ==161 ways