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(a) There are two ways to create a queue with 3 people and 2 blockers:

The first person is a blocker, and the other two people are in any order.

The second person is a blocker, and the other two people are in any order.

The first case can be achieved in Q(2,2) ways, and the second case can be achieved in Q(2,1) ways. Therefore, the total number of ways to create a queue with 3 people and 2 blockers is 2∗Q(2,2)+2∗Q(2,1).

(b) To create a queue with n people and k blockers, we can either:

Have the first person be a blocker, and the other n−1 people be in any order. This case can be achieved in k∗Q(n−1,k) ways.

Have the second person be a blocker, and the other n−1 people be in any order. This case can be achieved in (k−1)∗Q(n−1,k) ways.

...

Have the $k$th person be a blocker, and the other n−1 people be in any order. This case can be achieved in 1∗Q(n−1,k) ways.

Have the $k + 1$st person be a blocker, and the other n−1 people be in any order. This case can be achieved in 0∗Q(n−1,k) ways.

Note that the last case is impossible, because there would be more blockers than people.

Therefore, the total number of ways to create a queue with n people and k blockers is:

k * Q(n - 1, k) + (k - 1) * Q(n - 1, k) + ... + 1 * Q(n - 1, k)

This can be simplified to:

(k * Q(n - 1, k)) + ((k - 1) * Q(n - 1, k)) + ... + (Q(n - 1, k) + Q(n - 1, k - 1) + ... + Q(n - 1, 0))

The last term in this sum is equal to Q(n−1,k−1), because it is the number of ways to create a queue with n−1 people and k−1 blockers. Therefore, the total number of ways to create a queue with n people and k blockers is:

(k * Q(n - 1, k)) + ((k - 1) * Q(n - 1, k)) + ... + (Q(n - 1, k - 1))

This can be rewritten as:

k * Q(n - 1, k) + (n - k + 1) * Q(n - 1, k - 1)

Therefore, we have shown that for n≥2 and k≥1,

Q(n, k) = k * Q(n - 1, k) + (n - k + 1) * Q(n - 1, k - 1)

i.

There are 52 cards in a deck, so there are 51 pairs of adjacent cards. Each pair has an equal probability of being aces, which is 4/52 * 3/51 = 12/221. So, the expected number of pairs of adjacent cards that are both aces is:

Expected number = 51 * (12/221) = 40/221

Therefore, the expected number of pairs of adjacent cards that are both aces is 40/221.

ii.

There are 12 face cards in a deck, so there are 11 pairs of adjacent face cards. Each pair has an equal probability of being face cards, which is 12/52 * 11/51 = 22/85. So, the expected number of pairs of adjacent cards that are both face cards is:

Expected number = 11 * (22/85) = 2/5

Therefore, the expected number of pairs of adjacent cards that are both face cards is 2/5.

To find the values of x such that f(x)=x, we need to solve each of the following equations:

2x+1=x

7−x=x

10−2x=x

10−x=x

Solving the first equation, we get x=−1​.

Solving the second equation, we get x=3.5​.

Solving the third equation, we get x=5​.

Solving the fourth equation, we get x=5​.

Therefore, the values of x such that f(x)=x are −1, 3.5, and 5.

The floor function rounds a number down to the nearest integer, so the equation ⌊t⌋=3t+4 means that the integer part of t is equal to 3t+4. This can happen only if t is at least −2, since the integer part of any number less than −2 is −3.

If t is at least −2, then the integer part of t is equal to t itself. Therefore, we have the equation t=3t+4. Solving for t, we get 2t=4, so t=2​.

We can find the first few terms of the sum to see if there is a pattern:

f(1) = -1 f(2) = -1 f(3) = -1 f(4) = -2 f(5) = -2 f(6) = -2 f(7) = -3 f(8) = -3 f(9) = -3

We see that the function values repeat in a cycle of length 3: −1, −1, −2, −2, −2, −3, −3, −3. Since there are 1000 terms in the sum, we can divide 1000 by 3 to find that there are 333 complete cycles of length 3, plus one additional term. The sum of the terms in each complete cycle is −1−1−2=−4, so the sum of the first 333 cycles is −4⋅333=−1332. The additional term is f(1000)=−3, so the total sum is −1332−3=∗∗−1335∗∗.

Therefore, the answer is -1335.

Let the length of the bottom piece be x. The length of the top piece is then 50-x.

We know that the top of the pole is now at a point that is 40 units from the base of the pole, so x+40=50

Solving for x, we get x=10

The difference between the lengths of the two pieces is then 50-10 = 40 units.

Therefore, the answer is 40

There are 2 ways to start with the letter A: either from the first A in the array, or from the second A. To reach the second A, we must move down once. Once we have reached the first or second A, we can move in any direction to reach the letter R. There are 3 ways to do this: we can move down, right, or down and then right. Once we have reached the letter R, we can again move in any direction to reach the letter C. There are 3 ways to do this: we can move down, left, or left and then down. Once we have reached the letter C, the last step is to move up to reach the letter H, and there is only 1 way to do this. Therefore there are 2×3×3×1=18​ ways to spell the word "ARCH" in the array provided.

We can break down the problem into smaller parts. First, what is the remainder when 5 is divided by 7? It is 5. So we know that 51≡5(mod7).

What is the remainder when 52 is divided by 7? It is 4. So we know that 52≡4(mod7).

What is the remainder when 53 is divided by 7? It is 6. So we know that 53≡6(mod7).

We can continue in this way to find the remainders of 54, 55, and so on. We will find that the remainders repeat in a cycle of 6: 5, 4, 6, 2, 3, 1. This means that the remainder when 5n is divided by 7 is the same as the remainder when n(mod6) is divided by 7.

Since 207≡3(mod6), we know that the remainder when 5207 is divided by 7 is the same as the remainder when 3 is divided by 7. This is 3​.

The domain of g(x) is all real numbers x such that the expression x+1x2−2​ is defined and evaluates to a value in the domain of f(x).

The expression x+1x2−2​ is undefined when x=−1. Also, the expression x+1x2−2​ is defined and evaluates to 0 when x=−2. Therefore, the intersection of the domain of x+1x2−2​ with the domain of f(x) is the interval (−∞,−2)∪(−1,∞). However, the expression f((x2−2)/(x+1)) is also undefined when x=1. Therefore, the domain of g(x) is the interval (−∞,−2)∪(−1,1)∪(1,∞).

The width of this interval is 18−(−8)+1+1=18​.