Justingavriel1233

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 #1
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Mar 16, 2023
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We can solve this problem using the Principle of Inclusion-Exclusion (PIE). 

First, let's find the total number of ways the children can choose their ice cream flavors. Each child has 3 options, so there are 3^6 = 729 possible outcomes.

Next, let's find the number of ways no flavor is selected by exactly two children. To do this, we can use the complement rule and subtract from the total number of outcomes the number of ways all flavors are selected by at most one child. 

There are C(6,0) ways to choose no child for the first flavor, C(5,0) ways to choose no child for the second flavor, and C(4,0) ways to choose no child for the third flavor. So the number of ways all flavors are selected by at most one child is:

C(6,0) * C(5,0) * C(4,0) = 1

Therefore, the number of ways no flavor is selected by exactly two children is:

3^6 - 1 = 728

However, this includes the cases where some flavor is selected by exactly three children or by all six children. We need to subtract these cases to avoid double-counting. 

Let A be the set of outcomes where flavor 1 is selected by exactly three children, and let B and C be the sets of outcomes where flavors 2 and 3, respectively, are selected by exactly three children. We can use the formula for the size of the union of three sets to find the number of outcomes in which some flavor is selected by exactly three children:

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|

Each of |A|, |B|, and |C| can be found using the following reasoning: There are C(6,3) ways to choose 3 children to select the given flavor, and for each choice, there are 2 options for the remaining 3 children (they can choose either of the other 2 flavors). So:

|A| = C(6,3) * 2^3 = 160
|B| = C(6,3) * 2^3 = 160
|C| = C(6,3) * 2^3 = 160

To find |A ∩ B|, we need to choose 3 children to select flavor 1 and 3 children to select flavor 2, and then the remaining child can select either flavor 1 or flavor 2. So:

|A ∩ B| = C(6,3) * C(3,3) * 2 = 40

Similarly, |A ∩ C| = 40 and |B ∩ C| = 40.

To find |A ∩ B ∩ C|, we need to choose 3 children to select each of the 3 flavors. This can be done in C(6,3) ways:

|A ∩ B ∩ C| = C(6,3) = 20

Substituting these values into the formula for the size of the union of three sets, we get:

|A ∪ B ∪ C| = 160 + 160 + 160 - 40 - 40 - 40 + 20 = 360

Therefore, the number of ways some flavor is selected by exactly two children is:

728 - 360 = 368

So there are 368 ways the children can choose their ice cream flavors so that some flavor of ice cream is selected by exactly two children.

Mar 16, 2023
 #1
 #1
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(a) The probability that Valeria has two heads on her first turn is the probability that one of the coins is tails and the other two are heads. There are 3 ways that this can happen: the first coin is tails and the other two are heads, the second coin is tails and the other two are heads, or the third coin is tails and the other two are heads. Each of these three outcomes has probability (1/2) * (1/2) * (1/2) = 1/8. Therefore, the probability that Valeria has two heads on her first turn is 3/8.

(b) The probability that Valeria has three heads on her first turn is (1/2) * (1/2) * (1/2) = 1/8, since each coin has a probability of 1/2 of coming up heads and all three coins are flipped.

(c) The probability that Valeria has two heads after her second turn is the probability that she had one tail and two heads on her first turn, and then flipped the tail on her second turn. There are three ways that she could have had one tail and two heads on her first turn: the first coin was tails and the other two were heads, the second coin was tails and the other two were heads, or the third coin was tails and the other two were heads. In each of these cases, the probability of flipping the tail on the second turn is 1/2. Therefore, the probability that Valeria has two heads after her second turn is 3/8 * 1/2 = 3/16.

(d) The probability that Valeria has one head after her second turn is the probability that she had two tails and one head on her first turn, and then flipped one of the tails on her second turn. There are three ways that she could have had two tails and one head on her first turn: the first coin was heads and the other two were tails, the second coin was heads and the other two were tails, or the third coin was heads and the other two were tails. In each of these cases, the probability of flipping one of the tails on the second turn is 1/2. Therefore, the probability that Valeria has one head after her second turn is 3/8 * 1/2 = 3/16.

(e) The probability that Valeria has one head after her third turn is the probability that she had two tails and one head on her first turn, flipped the head and one of the tails on her second turn, and then flipped the remaining tail on her third turn. There are three ways that she could have had two tails and one head on her first turn, and we can consider each case separately:

- If the first coin was heads and the other two were tails, then the probability of flipping the head and one of the tails on the second turn is 1/2, and the probability of flipping the remaining tail on the third turn is 1/2. Therefore, the probability of this sequence of flips is (1/2) * (1/2) * (1/2) = 1/8.
- If the second coin was heads and the other two were tails, then the probability of flipping the head and one of the tails on the second turn is 1/2, and the probability of flipping the remaining tail on the third turn is 1/2. Therefore, the probability of this sequence of flips is (1/2) * (1/2) * (1/2) = 1/8.
- If the third coin was heads and the other two were tails, then the probability of flipping the head and one of the tails on the second turn is 1/2, and the probability of flipping the remaining tail on the third turn is 1/2. Therefore, the probability of this sequence of flips is (1/2) * (1/2) * (1/2) = 1/8.

Adding up the probabilities for each case, we get a total probability of 3/8 * 1/8 = 3/64.

(f) The probability that Valeria has three heads after her fourth turn is the probability that she had one tail and two heads on her first turn, flipped the tail on her second turn, flipped the two tails on her third turn, and flipped the remaining head on her fourth turn. There is only one way that she could have had one tail and two heads on her first turn, and the probability of this is 3/8. The probability of flipping the tail on the second turn is 1/2. The probability of flipping the two tails on the third turn is 1/4, since there are two tails that need to be flipped and each has a probability of 1/2 of coming up tails. Finally, the probability of flipping the remaining head on the fourth turn is 1/2.

Mar 13, 2023
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We can begin by applying the Cauchy-Schwarz inequality to the three pairs of terms under the square roots:

√(a^2 + ab + b^2) * √(a^2 + ac + c^2) ≥ a^2 + ab + ac + c^2
√(a^2 + ab + b^2) * √(b^2 + bc + c^2) ≥ ab + b^2 + bc + c^2
√(a^2 + ac + c^2) * √(b^2 + bc + c^2) ≥ ac + bc + c^2

Adding these three inequalities, we get:

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ (a^2 + ab + ac + b^2 + bc + c^2 + ac + bc + c^2)^(1/2)

Simplifying the right-hand side, we get:

(a^2 + ab + ac + b^2 + bc + c^2 + ac + bc + c^2)^(1/2) = (a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)^(1/2)

Now, we can use the AM-GM inequality on the last three terms inside the square root:

2ab + 2ac + 2bc ≥ 3(2(abacbc)^(1/3)) = 3√(a^2b^2c^2)

Substituting this into the previous equation, we get:

(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)^(1/2) ≥ (a^2 + b^2 + c^2 + 3√(a^2b^2c^2))^(1/2)

Now, we can use the AM-QM inequality on a^2, b^2, and c^2:

(a^2 + b^2 + c^2)/3 ≥ (abc)^(2/3)

Substituting this into the previous equation, we get:

(a^2 + b^2 + c^2 + 3√(a^2b^2c^2))^(1/2) ≥ (4/3)(a^2 + b^2 + c^2)^(1/2)

Therefore, we have:

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ (4/3)(a^2 + b^2 + c^2)^(1/2)

Finally, we can use the fact that (a^2 + b^2 + c^2) ≥ ab + ac + bc (which follows from the inequality (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0) to get:

(4/3)(a^2 + b^2 + c^2)^(1/2) ≥ (4/3)(ab + ac + bc)^(1/2) ≥ (4/3)(3abc)^(1/2) = 2(abc)^(1/2)

Therefore, we have:

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ 2(abc)^(1/2) = √(4(ab)(ac)(bc)) = √(3(ab) + 3(ac) + 3(bc))

Dividing both sides by √(ab) + √(ac) + √(bc), we get:

(√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2))/(√(ab) + √(ac) + √(bc)) ≥ √(3)

Therefore, we have shown that

√(a^2 + ab + b^2) + √(a^2 + ac + c^2) + √(b^2 + bc + c^2) ≥ √(3)(√(ab) + √(ac) + √(bc))

Equality occurs when a, b, and c are equal, or when two of them are equal and the third is zero.

Mar 13, 2023
 #1
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(a) To place 8 counters in the 4 x 4 grid so that each row contains exactly two counters, we can proceed as follows:

- Choose any two squares in the first row and place counters in them. There are (4 choose 2) = 6 ways to do this.
- Choose any two squares in the second row that do not already have counters and place counters in them. There are (2 choose 2) = 1 way to do this.
- Choose any two squares in the third row that do not already have counters and place counters in them. There are (2 choose 2) = 1 way to do this.
- Place the remaining two counters in the fourth row. There is only one way to do this.

Therefore, the total number of ways to place 8 counters in the 4 x 4 grid so that each row contains exactly two counters is 6 x 1 x 1 x 1 = 6.

(b) To place 12 counters in the 4 x 4 grid so that each column contains exactly three counters, we can proceed as follows:

- Choose any three squares in the first column and place counters in them. There are (4 choose 3) = 4 ways to do this.
- Choose any three squares in the second column that do not already have counters and place counters in them. There are (3 choose 3) = 1 way to do this.
- Choose any three squares in the third column that do not already have counters and place counters in them. There are (2 choose 3) = 0 ways to do this, since there are only two squares left in the third column.
- Place the remaining three counters in the fourth column. There are (3 choose 3) = 1 way to do this.

Therefore, the total number of ways to place 12 counters in the 4 x 4 grid so that each column contains exactly three counters is 4 x 1 x 0 x 1 = 0. There is no way to do this.

Mar 13, 2023
 #1
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To form a right isosceles triangle, we need to select three points such that two of them lie on one of the grid's horizontal lines and the third lies on a line that is perpendicular to that line and passes through the midpoint of the first two points.

There are 5 horizontal lines and 5 vertical lines we can choose from. Let's consider the cases where the two points are on the first horizontal line:

- If we choose the two points that are closest to each other, they are one unit apart, and there are two points on the second horizontal line that would form a right isosceles triangle with them.
- If we choose two points that are two units apart, there are no points on the second horizontal line that would form a right isosceles triangle with them.
- If we choose two points that are three units apart, there is only one point on the second horizontal line that would form a right isosceles triangle with them.

Therefore, for each horizontal line, we have a total of 2 + 0 + 1 = 3 ways to select two points that are one, two, or three units apart. For each of these pairs of points, there is exactly one point on a vertical line that would form a right isosceles triangle with them. Therefore, the total number of ways to select three points that form a right isosceles triangle is:

5 horizontal lines x 3 ways to choose the pair of points on each line x 1 way to choose the third point on a perpendicular line = 15.

Therefore, there are 15 ways to select three points from the given grid that form a right isosceles triangle.

Mar 13, 2023