Bertie

Just come in on this conversation, here's my take on it.

First method: First ball can go into any of the three boxes, second ball can go into any of the three boxes etc, total number of arrangements is 3*3*3*3*3*3 = 729.

Second method: All six b***s could go into a single box, or they could be split five in one box and one in another, or they could be split 4 - 2 - 0, or 4 -1 -1 and so on, seven possible splits.

The number of possibilities for each split as follows.

6 - 0 - 0: 3 ways of doing this, (the 6 can be in any one of the three boxes).

5 - 1 - 0: 36 ways. There are 6 possible orders for the 5-1-0, and 6 possible choices for the odd ball.

4 - 2 - 0: 90 ways. Again 6 possible orders and 6C2=15 choices for the 2.

4 - 1 - 1: 90 ways. 6 possible orders, (3 possibilities for the 4, but double this because the two 1's are different), then 6C4=15 choices for the 4.

3 - 3 - 0: 60 ways. 3 possible orders (positions for the 0), and 6C3=20 choices for either 3.

3 - 2 - 1: 360 ways. 6 possible orders, 6 choices for the 1 and 5C2=10 coices for the 2 or 3.

2 - 2 - 2: 90 ways. 6C2=15 choices for the first 2, 4C2=6 choices for the second 2.

Total = 3 + 36 + 90 + 90 + 60 + 360 + 90 = 729. 

You might get different answers.

Show us your calculation and we will find where your mistake is.

Attractively ?

Add them up and divide the result by 6.

For the first relationship, cot inverse and tan inverse (to borrow the notation), have different principal ranges for negative values of x. The principal range for the inverse cotangent is pi/2 to pi while for the inverse tangent it is -pi/2 to 0. What that means is that cot inverse will return an angle in the second quadrant while tan inverse returns an angle in the fourth quadrant. The two angles differ by pi, so pi has to be added to the negative angle returned by tan inverse to achieve equality with cot inverse. (So should there be a plus sign rather than a negative sign beteween the two terms on the RHS ?)

(Incidently, not everyone uses the same principal range for cot inverse. For some, including Wolfram I think, the principal range is -pi/2 to 0, the same as for tan inverse. In that case the pi would not be needed.)

With the cos inverse and sec inverse though the principal ranges are the same, pi/2 to pi, so no additional pi is needed.

Actually Melody, my first method is exactly equivalent to Alan's method.

The equation ds/dt = (D/s)(dD/dt) can be written as (s/D)(ds/dt) = dD/dt, which, in Alan's notation is

v.cos(theta)=95, (with or without the negative sign)

Hi Melody,

Like you  I'm struggling with the physics of this problem, I don't fully understand Alan's diagram, how the radar gun measures the component of the cars speed at right angles to the direction of the radar beam, and how that relates to the diagram. The logic as to why the speed of the car should be greater than that shown by the radar gun is nice though.

When I first looked at the problem, I tried two different methods of solution. In both cases I put the right angle at the road as you did, so the line of the radar beam is down the hypotenuse of the triangle.

First method.

Call the length of the hypotenuse D and the distance that the car is down the road s, then D^2 = s^2 + 30^2, and differentiating that, 2D(dD/dt) = 2s(ds/dt) so that ds/dt = (D/s)(dD/dt).

Initially we have s = 120, D^2 = 120^2 + 30^2, so D = 123.69 and dD/dt = -95, so ds/dt = 123.69.(-95)/120 = -97.92.

Second method.

Consider the movement of the car over 100th of a second. The hypotenuse decreases by 0.95 ft, and will now have a length approximately 123.69 - 0.95 = 122.74, and that translates to a distance of the car on the road as s^2 = 122.74^2 - 30^2 = 14165.885, so s = 119.02. That seems to imply that the car has moved a distance of 120 - 119.02 = 0.9795 ft which in turns implies a speed of 97.95 fps.

If instead of using 100th of a second, a limiting process were used, I would expect the second result to agree with the first, (though they are pretty close anyway).

The method is to look at the final digit, it has to be a 1 or a 9. (Choose an intermediate digit like 5 for example  and some of those to the left will be bigger, and some smaller.)   Similarly there will be two choices for the digit in the eighth position, two choices for the digit in the seventh position and so on.  There will only one choice left for the first digit, so 2 to the eight or 256 possibles

The method requires you to multiply the first equation by some number, the second equation by some number, in order that either the x coefficients are the same or that the y coefficients are the same. Sometimes, you get lucky and a pair of coefficients, either the x's or the y's are already equal in which case the jobs already done, or it's sufficient to multiply just one of the equations by a number in order to make a pair of coefficients equal.

With these equations, you have a choice. Multiply the first equation by 3 to get 6x + 9y = -3, multiply the second equation by 2 to get 6x + 8y = -8, and now subtract one equation from the other. The x's cancel out. That allows you to find the value of y, and by substituting that into one of the earlier equations allows you to find the value of x.

Alternatively, multiply the first equation by 4 to get 8x + 12y = -4, the second equation by 3 to get

9x + 12y = -12,  and subtract one from the other. That gets you the value of x, and by substituting that into one of the earlier equations, the value of y.

You should do it both ways, as an exercise, to show that you get the same results, either way.

There will be sets of equations where the x's or y's are opposite in sign in which case the equations have to be added rather than subtracted.

I'm going to cheat slightly and refer to Alan's rather nice diagram. (I drew it on paper).

The angle that OB makes with the negative y-axis is 15 deg which means that the angle AOB is 120 deg, which in turn means that the angle OBQ is 60 deg.

OB and BQ being the same length implies that the triangle OBQ is equilateral in which case the length of OQ is 6 and it's at an angle of 45 deg to the negative y-axis.

That means that Q has polar co-ordinates (6, -45deg).