Fragen 431
Antworten 16


We can divide this problem into two cases:


Case 1: Each row has exactly one child from each family.


Choose one child from each family for the first row: there are 3 choices


Arrange the remaining two children in the first row (siblings can't be together): there are 2!=2 ways.


Arrange the second row similarly: 3⋅2=6​ ways total.


Case 2: One row has two children from the same family.


There are two subcases depending on the arrangement of siblings in the rows:


Subcase 2a: The first child in each row is from the same family.


Choose one pair of siblings: 3 choices


Arrange the siblings within the pair: 2 ways.


Arrange the remaining 4 children (2 from another pair and 2 from the third pair) in the second row: 4! ways.


Overcount: we've counted the arrangement as if sibling order matters within a pair (which it doesn't) twice (once for each sibling in the first row). So, we divide by 2!⋅2!.


Total: 2!⋅2!3⋅2⋅4!​=36 ways.


Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.


Choose one pair to have their children occupy the third seats in each row: 3 choices.


Arrange the remaining 4 children in the first row: 4! ways.


Overcount: similar to subcase 2a, we divide by 2!⋅2! to account for sibling order not mattering.


Total: 2!⋅2!3⋅4!​=36 ways.


Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case:

6 + 36 + 36 = 78​.


This gives us a final answer of 78.


Since a rotation about point O maps A to B, B to C, and C to D, we know the following:


Angle AOD: This is given as 24 degrees.


Full Rotation: A full rotation around a point is 360 degrees.


Possible Rotations:


There are three possibilities for the rotation that maps A to B, B to C, and C to D, resulting in three possible measures for angle AOB:


Case 1: Single Rotation of 24 Degrees


In this case, the rotation about O maps A to B by 24 degrees clockwise, B to C by 24 degrees clockwise, and C to D by 24 degrees

clockwise, resulting in a total rotation of A to D of 24 + 24 + 24 = 72 degrees.


Since the full rotation is 360 degrees, the remaining angle for AOB must be 360 - 72 = 288 degrees.


However, angle measures are typically represented between 0 and 180 degrees. We can achieve this by subtracting a multiple of 180 (full rotations) from 288:


AOB = 288° - 180 = 108.


Case 2: Double Rotation (360 + 24 Degrees)


Here, the rotation about O maps A to B by a full rotation (360 degrees) followed by a 24-degree clockwise rotation. This effectively brings B back to its original position and then continues the rotation to C and D.


The total rotation for AOD in this case becomes 360 + 24 + 24 = 408 degrees.


Similar to case 1, we can adjust this to fit the 0-180 degree range:


AOB = 408° - 360° = 48°


Case 3: Triple Rotation (2 * 360 + 24 Degrees)


This case involves two full rotations followed by a 24-degree clockwise rotation from A to B. Again, the first two rotations effectively bring B back to its original position.


The total rotation for AOD becomes 2 * 360 + 24 = 744 degrees.


Adjusting for the 0-180 degree range:


AOB = 744° - 2 * 360° = 124°




Therefore, the three possible degree measures for angle AOB, considering rotations between 0 and 180 degrees, are:


108° (Case 1)


48° (Case 2)


124° (Case 3)


Understanding the Transformations:


Tasha's Transformation: Moves the point sqrt(2) units to the right.


Richard's Transformation: Rotates the point clockwise about point O by 90 degrees.


Round 1:


A' is sqrt(2) units to the right of A (Tasha's translation).


A" is obtained by rotating A' by 90 degrees clockwise around O (Richard's rotation).


Round 2:


A1 is obtained by rotating A by 90 degrees clockwise around O (Richard's rotation).


A2 is sqrt(2) units to the right of A1 (Tasha's translation).


Finding A"A2:


To find the distance between A" and A2, we can consider the following:


A" and A1 are the same distance away from O (since both are obtained by a 90-degree rotation from A and they share the same original distance from O).


A'A2 forms a right triangle with A"A1 as the hypotenuse (due to the 90-degree rotations).


Pythagorean Theorem:


We can use the Pythagorean theorem to find A"A2:


a^2 + b^2 = c^2




a = A'A2 (we know this is sqrt(2) from Tasha's translation)


b = A"A1 (distance between A" and A1, which is the same as the distance between A and O, which is sqrt(3))


c = A"A2 (what we want to find)


Substituting Values:


(sqrt(2))^2 + (sqrt(3))^2 = (A"A2)^2


2 + 3 = (A"A2)^2


5 = (A"A2)^2


Taking the Square Root (consider both positive and negative):


A"A2 = ±√5


Since distance cannot be negative, we take the positive square root:


A"A2 = √5


Therefore, the distance between A" and A2 is √5 units.


This is a series where the numerator follows the Fibonacci sequence and the denominator is a geometric sequence with common ratio 1/10. To find the sum of such a series, we can use a technique involving manipulation and summation of geometric series.


Here's how to solve it:


Step 1: Splitting the Series


We can represent the series as the sum of two series:


S = (1/10^2 + 1/10^3 + 1/10^4 + ...) + (2/10^4 + 3/10^5 + 5/10^6 + ...)


Step 2: Recognizing Geometric Series


The first series is a geometric series with first term (a = 1/10^2) and common ratio (r = 1/10). The second series is another geometric series with first term (a = 2/10^4) and common ratio (r = 1/10).


Step 3: Find the Sum of Each Series


The formula for the sum (Sn) of a finite geometric series is:


Sn = a(1 - r^n) / (1 - r)




a is the first term


r is the common ratio


n is the number of terms


Step 4: Apply the Formula to Each Series


First Series:


S1 = (1/10^2) * (1 - (1/10)^n) / (1 - 1/10)


Second Series:


S2 = (2/10^4) * (1 - (1/10)^n) / (1 - 1/10)


Step 5: Note on Infinite Series


Since we're dealing with an infinite series (n tends to infinity), both (1/10)^n terms in the numerators approach zero. Therefore, we can simplify the expressions:


S1 ≈ (1/10^2) * (1 - 0) / (1 - 1/10) = 1/100


S2 ≈ (2/10^4) * (1 - 0) / (1 - 1/10) = 1/50


Step 6: Sum the Simplified Series


The total sum (S) of the original series is the sum of S1 and S2:


S = S1 + S2 ≈ 1/100 + 1/50 = 3/100




Therefore, the sum of the series is 3/100.


The strategy here is to calculate the probability of the event's complement (i.e., the probability that the coin is tossed ten times or fewer) and subtract this from 1.


Favorable Outcomes:


There are two favorable outcomes:


HHH appears in exactly 10 flips: There are 3 choices for the position of the first heads (leaving 9 flips remaining), then 2 choices for the second heads (leaving 8 flips remaining), and 1 choice for the third heads (leaving 7 flips remaining).


So, there are 3⋅2⋅1=6 successful outcomes where HHH appears in exactly 10 flips.


TTT appears in exactly 10 flips: This follows the same logic as scenario 1, so there are also 6 successful outcomes.


Total Favorable Outcomes:


There are a total of 6 + 6 = 12 successful outcomes where the coin is tossed exactly 10 times.


Total Outcomes:


Since the coin is fair, there are 210=1024 total possible outcomes for 10 flips (heads or tails for each flip).


Probability of Favorable Outcomes:


The probability of needing exactly 10 flips is the number of successful outcomes divided by the total number of outcomes:


P(exactly 10 flips) = 12 / 1024


Complement's Probability:


We want the probability of needing more than 10 flips. This is the complement of the event where the coin is tossed ten times or fewer.


P(more than 10 flips) = 1 - P(exactly 10 flips)


Final Answer:


Substitute the probability of needing exactly 10 flips:


P(more than 10 flips) = 1 - (12 / 1024)




P(more than 10 flips) = (1024 - 12) / 1024 = 1012 / 1024


Both the numerator and denominator have a common divisor of 4, so we can simplify:


P(more than 10 flips) = 253 / 256


Therefore, the probability of needing more than 10 flips is 253/256.