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 #1
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We can solve this problem by considering two cases:

 

Case 1: One family has siblings in each row

 

There are 3 families (let's call them A, B, and C). In this case, we need to place one child from each family in each row. There are 3 choices for the first seat in the first row (child from any family).

 

Once the first seat is filled, there are 2 choices left for the second seat (child from one of the remaining families). The third seat is automatically filled by the remaining child.

 

Similarly, for the second row, there are again 3 choices for the first seat, and so on. Therefore, the number of arrangements for this case is:

 

Arrangements (Case 1) = 3 * 2 * 3 * 2 = 36

 

Case 2: Two families have siblings in one row

 

There are again 3 choices for which family will have siblings in the same row (A, B, or C). Without loss of generality, let's say family A has siblings in the same row.

 

There are now 2 choices for which child of family A sits in the first seat of the first row. The other child of family A must sit in the second seat.

 

The remaining child from family B can sit in the third seat. Now, for the second row, there are 2 choices for the first seat (child from either family B or C). Following the same logic as before, the number of arrangements for this case is:

 

Arrangements (Case 2) = 3 * 2 * 1 * 2 * 2 = 24

 

Total Arrangements

 

To get the total number of arrangements, we add the number of arrangements from both cases:

 

Total Arrangements = Arrangements (Case 1) + Arrangements (Case 2)

 

Total Arrangements = 36 + 24

 

Total Arrangements = 60​

21 avr. 2024
 #3
 #2
avatar+1537 
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For question 1:

 

Here's how to find the number of positive five-digit integers containing the digit grouping "24" at least once:

 

Total Possible Numbers:

 

There are 9 choices (excluding 0) for each digit in a five-digit number (1, 2, 3, 4, 5, 6, 7, 8, 9).

 

So, the total number of possible five-digit positive integers is 9 * 9 * 9 * 9 * 9 = 59,049.

 

Unwanted Scenarios (Numbers without "24"):

 

No "24" group: There are 8 choices for the digit in the place where "24" could be (all digits except 2 and 4). For the remaining 4 digits, there are 9 choices each. So, there are 8 * 9 * 9 * 9 = 5832 cases where "24" doesn't appear in any of the five digits.

 

"24" only appears once, but not next to each other:

 

Choose the digit that will be between "2" and "4": There are 7 choices (all digits except 0, 2, and 4).

 

Choose the positions for "2" and "4" (not next to each other): There are 4 choices for the first position (remaining slots after fixing one digit). There are 3 choices for the second position (remaining slots after fixing the first and the middle digit). So, there are 4 * 3 = 12 ways to arrange "2" and "4" with another digit in between.

 

Choose the remaining 2 digits: There are 9 choices each.

 

Total unwanted cases with "24" not next to each other: 7 * 12 * 9 * 9 = 5832.

 

Total Numbers with "24" at least once:

 

We want to find the number with "24" at least once. So, subtract the unwanted cases from the total:

 

Total numbers - Unwanted cases = Numbers with "24"

 

59,049 - (5832 + 5832) = 47,385

 

Therefore, there are 47,385 positive five-digit integers containing the digit grouping "24" at least once.

14 avr. 2024
 #2
avatar+1537 
0

We can solve this problem using the properties of isosceles triangles and the Pythagorean theorem.

 

Isosceles Triangle Properties:

 

Since AB = AC in an isosceles triangle ABC, angles B and C are congruent.

 

Dividing the Base:

 

The height from A to BC is the perpendicular bisector of the base (divides it into two segments of equal length) in an isosceles triangle.

 

Segment Lengths:

 

Given that BC = 8 cm and BD = 3 cm, we know DC = BC - BD = 8 cm - 3 cm = 5 cm.

 

Finding the Height:

 

We can use the Pythagorean theorem on right triangle ABD (since the height is perpendicular to the base).

 

Pythagorean Theorem:

 

a^2 + b^2 = c^2 (where a and b are the lengths of the legs and c is the length of the hypotenuse)

 

In this case:

 

a (length of leg BD) = 3 cm

 

c (length of hypotenuse AB) = unknown (represents the height of the triangle)

 

b (length of leg AD) = unknown

 

Solving for Height (c):

 

b^2 = c^2 - a^2 (rearranging the equation)

 

b^2 = c^2 - 3^2 (substitute known values)

 

Since AD is half of the base (property of isosceles triangle):

 

AD = 1/2 * BC = 1/2 * 8 cm = 4 cm

 

Substitute the value of AD (b) in the equation:

 

4^2 = c^2 - 3^2

 

Simplify:

 

16 = c^2 - 9

 

Add 9 to both sides:

 

25 = c^2

 

Take the square root of both sides (remember to consider both positive and negative since squaring either results in 25):

 

c = ± 5

 

Positive Height:

 

Since the height represents a distance, we take the positive value:

 

c (height of the triangle) = 5 cm

 

Therefore:

 

Length of segment DC = 5 cm

 

Height of the triangle = 5 cm

14 avr. 2024