Pythagorearn

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 #1
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To find the number of ways of arranging the numbers 1, 2, 3, 4, 5, 6 in a row such that the product of any two adjacent numbers is even, we need to consider the properties of even and odd numbers.

 

For the product of two numbers to be even, at least one of the numbers must be even.

 

Now let's look at the arrangement:

 

1. If we start with an even number (2, 4, or 6), then the number adjacent to it can be either an even or an odd number.


2. If we start with an odd number (1, 3, or 5), then the number adjacent to it must be even.

 

Let's analyze these cases:

 

Case 1: Starting with an even number (2, 4, or 6):


- There are 3 options for the first number (2, 4, or 6).


- For each of these choices, there are 5 options for the second number (either an even number or an odd number).


- For each of the second numbers, there are 4 remaining numbers for the third position, and so on.

 

Therefore, the total number of arrangements in this case is \( 3 \times 5! \).

 

Case 2: Starting with an odd number (1, 3, or 5):


- There are 3 options for the first number (1, 3, or 5).


- For each of these choices, there are 3 options for the second number (only even numbers are allowed).


- For each of the second numbers, there are 4 remaining numbers for the third position, and so on.

 

Therefore, the total number of arrangements in this case is \( 3 \times 3 \times 4! \).

 

Adding the arrangements from both cases:

 

\[ Total \, arrangements = (3 \times 5!) + (3 \times 3 \times 4!) \]

 

Let's calculate the total number of arrangements.

 

To calculate the total number of arrangements:

 

\[ Total \, arrangements = (3 \times 5!) + (3 \times 3 \times 4!) \]

 

\[ Total \, arrangements = (3 \times 120) + (3 \times 3 \times 24) \]

 

\[ Total \, arrangements = 360 + 216 \]

 

\[ Total \, arrangements = 576 \]

 

So, there are 576 ways to arrange the numbers 1, 2, 3, 4, 5, 6 in a row such that the product of any two adjacent numbers is even.

28.04.2024
 #1
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There are two cases to consider:

 

Case 1: Siblings of one pair sit in the same row

 

There are 3 ways to choose which pair of siblings will have siblings sitting in the same row (either the first row, the second row, or both).

 

Once a pair is chosen, there are 2 ways to arrange the siblings within that pair (who sits in the first seat and who sits in the second).

 

For the remaining two pairs, there are only 4 choices for the first seat in the row they occupy (since two seats are already taken by siblings from the chosen pair). Then there are 3 choices for the second seat (since one child has already been seated), for a total of 4 * 3 = 12 ways to arrange the children in that row.

 

Therefore, the total number of arrangements for Case 1 is 3 (ways to choose the pair) * 2 (arrangements within the pair) * 12 (arrangements in the other row) = 72.

 

Case 2: Siblings of each pair sit in different rows

 

There are 6 choices for the child in the first seat of the first row. It doesn't matter which sibling takes it, so suppose child A sits there (denoted by A).

 

Then there are 4 choices for the second seat (child B, child C, child D, or child E). The last seat in the first row cannot be filled by a sibling of child A, so it must be filled by one of the remaining two children (child D or child E).

 

This leaves a pair of siblings (let's say child B and child C) and one person (child D or child E) for the second row. So the only order that will work is child B - empty seat - child C. There are 2 choices for which sibling (B or C) sits in the first seat of the second row.

 

Therefore, the total number of arrangements for Case 2 is 6 (choices for child in first seat of first row) * 4 (choices for second seat) * 2 (choices for who sits first in the second row) = 48.

 

Adding the ways from both cases, there are 72 (Case 1) + 48 (Case 2) = 120 ways to seat the siblings.

28.04.2024
 #1
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To solve this problem, let's break it down step by step:

 

1. **Arrange the pairs of siblings in one row**: There are 3 pairs of siblings, so we need to arrange them in one row. This can be done in \(3!\) ways (3 factorial) because there are 3 pairs and the order matters.

 

2. **Arrange the pairs in the first row**: We have three pairs of siblings, and they can't sit next to each other in the same row. We can start by placing the first pair in one of the three seats, then place the second pair in one of the remaining two seats, and finally, place the third pair in the last remaining seat.

 

This can be done in \(3 \times 2 \times 1 = 3!\) ways.

 

3. **Arrange the pairs in the second row**: After arranging the pairs in the first row, there are three pairs left to be seated in the second row. Since no siblings can sit next to each other, the first pair can be seated in any of the three available seats.

 

Then, the second pair can be seated in one of the remaining two seats, and the last pair will take the remaining seat. This can be done in \(3 \times 2 \times 1 = 3!\) ways.

 

Now, to find the total number of arrangements, we multiply the number of arrangements for each step:

 

Total arrangements = (Arrangements of pairs in one row) * (Arrangements of pairs in the first row) * (Arrangements of pairs in the second row)

 

So, the total number of arrangements is:

 

\[3! \times 3! \times 3! = 6 \times 6 \times 6 = 216\]

 

Therefore, there are 216 ways to seat three pairs of siblings from different families in two rows of three chairs, if siblings may not sit next to each other in the same row.

28.04.2024
 #1
avatar+415 
+1

Let's denote the length of the rectangle as "l" and the width as "w". We are given that the perimeter (total length of all sides) is 40 units and the diagonal is 16 units.

 

Perimeter: Perimeter of a rectangle is calculated by adding the lengths of all its sides. In this case, Perimeter = 2l + 2w = 40.

 

Diagonal: The diagonal of a rectangle divides it into two congruent right triangles. We can use the Pythagorean theorem to relate the lengths of the sides of the rectangle and the diagonal.

 

The Pythagorean theorem states: a^2 + b^2 = c^2, where a and b are the lengths of the shorter sides (length and width in this case), and c is the length of the hypotenuse (diagonal). In this case, we are given c (diagonal) as 16.

 

We can approach this problem in two ways:

 

Method 1: Solve for l and w directly

 

From the perimeter equation, we can rewrite it to isolate "l": l = (40 - 2w) / 2.

 

Substitute this expression for "l" in the Pythagorean theorem equation: [(40 - 2w) / 2]^2 + w^2 = 16^2.

 

Solve this equation for "w" (tedious but solvable). Once you find "w", you can plug it back into the equation for "l" to find the length.

 

Method 2: Exploit the relationship between l and w using the diagonal

 

We know the diagonal is 16, and a rectangle's diagonals cut each other in half, creating right triangles with legs equal to half the length and half the width.

 

Therefore, in each right triangle formed by the diagonal, one leg (half the length) is 8 (half of 16).

 

Since a rectangle has opposite sides equal, the other leg of the right triangle (half the width) must also be 8.

 

Now we know that half the width (w/2) is 8. Solve for the actual width (w) by multiplying by 2: w = 16.

 

Finding the Area

 

Once you have the width (w = 16 from Method 2), you can find the length (l) using the perimeter equation (l = (40 - 2 * 16) / 2 = 4).

 

Now that you know both the length (l = 4) and width (w = 16), the area of the rectangle can be calculated using the formula: Area = l * w = 4 * 16 = 64 square units.

 

Therefore, the area of the rectangle is 64 square units.

28.04.2024