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What are you confused on. I try help if I know it.

No I didn't learn that at school.

But I have a simple idea of it and now I understand it.

Could I have a question on it please?

:)

Wow....that 's pretty neat, Alan.....I've never looked at that in this way before.....!!!

Notice that Alan has pointed out the fact that the final "landing points"  have coordinates whose sum MUST be even!!!!  It is impossible to land on a point whose coordinate sum is ODD!!!...Try it for yourself....I dare you!!!

Notice that, along the x-axis "row," there are 11 possible landing points.

And, on  each successive  "row" above the x axis, we have (11-n) possible landing points where n is the y value associated with that "row"......so 10 + 9 + 8 +........+2 + 1 = 55 possible landing points above the x axis. And, by symmetry, we have the same number below the x axis.

So, there are  2(55) + 11  = 121 possible "landing points."

Also notice that we could have gotten the same answer by just "squaring" the 11 points on each side of the.....well.....square !!!!  [Hey....I like to do things the hard way  !!!]

Thanks, Alan.....that's a very clear "visual" answer  !!!!

[I'm recommending this one for the Daily Wrap. ]

30 because 6 5 times is 30

I assume you want this.....???

Ln 82  = 4.4067192472642531

The larger ones weigh.....

820(1 +.175)  = 963.5 g

I'll have a go at this one.....!!!

Since the perimeter is 64, the width and the length must be (1/2) of this = 32

And if each is an integer quantity....we have the following possibilites

31 + 1  ... 30+  2 ...  29 + 3....etc.

But notice, when we get to 16 + 16, we can stop....because all combinations after this are just "reversals" of the sums that came before 16 + 16 !!!

So

31 - 16 + 1   = 16  noncongruent rectangles.....!!!!

Thanks.....I get lucky every now and then on these "counting" problems..........LOL!!!

We can choose 1 of the 3 triplets and from the other 11 players, any 5 of them.

So......C(3,1) * C(11,5)  = 1386 different teams...

This is correct. I entered this answer. And I understand fully. Thanks CPhil!!

See if this image helps:

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I'll give this one a try  ....though it doesn't appear to be too easy!!!

Notice that we could have these 2 configurations right off the bat......

yyyy zzzz xxxx      and  zzzz xxxx yyyy

Next, we could have this

yyyz  xzzz xxxy  .......but....since each single letter within each grouping can appear  in one of four positions,

we have 4 x 4 x 4 = 64 possiblities

Next, we could have

yyzz xxzz xxyy ....notice that within each subgroup, we have (4!)/(2! * 2!)  = 6 recognizable "words"

So  6 x 6 x 6 = 216 possibilities

We could also have

yzzz xxxz xyyy...which is again, 64 possibilities

So...if I haven't missed any combos, the total possible "words" = 2 + 2(64) + 216 = 346

Could somebody else check this....(Melody, Alan, heureka ??)

What do you mean?

Oh! Thanks CPhill! It is greatly appreciated

This formula is rare. I have yet to find it after searching several statistical and combinatorics sites. I am sure it is on the web somewhere. How do you search for a formula that does not have a name?

This formula was given to me years ago by a classmate. I am unsure if he found it in a publication or independently discovered it –he definitely had the math skills to do that. I wrote a Fortran program to test it. All the tests demonstrated (not proved) it distributes the counts such that N contains at least one (or more) k elements.

I’ll continue to look for the theory behind this formula, either on the web or by dissecting it. It looks very similar to the Binomial CDF that should give a starting point.

How do you know which number is N and which one is k. Should N be the smallest one, it was for your other example.

Yes, the (k) has to be equal to or greater than the (N). The (i) functions as the k in the binomial.

Thanks for your help :)

Since 5 evenly divides 60, it already IS to the "nearest 5m^2."

Yes, I guess they do.  I forgot about the zeros! 

In both cases these points will still lie on the circle of radius 5/sphere of radius 7.

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Alan...in the first one, don't (±5, 0) and (0,±5) also "work ??"

And in the second....don't  (±7, 0, 0), (0,±7, 0 ) and (0, 0, ±7) also do the job??

Each point is connected to 8 others, so, 9*8.  But this counts each line twice, so the number of lines is 9*8/2 = 36

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(a) All combinations of (±3,±4) and (±4,±3) satisfy the equation.  Geometrically, these points lie on the circumference of a circle of radius 5.

(b) All combinations of (±2, ±3, ±6), (±2, ±6, ±3), (±6, ±2, ±3), (±6, ±3, ±2),(±3, ±2, ±6) and (±3, ±6, ±2) satisfy the equation.  Geometrically, these points lie on a sphere of radius 7.  See image below:

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Well since "C" is in a given position, and we're going in a clock-wise manner, we have 4 ways to choose the next letter, 3 ways to choose the letter after that, etc.

So  4 x 3 x 2 x 1= 4!  = 24 arrangements

4! ways 

You cannot learn every thing at once MG :)

Letters just stand for numbers so they behave exactly the same way.

$$\\\mbox{Hopefully you already know how to add fractions}\\\\
\frac{2}{3}+\frac{4}{5}\\\\
=\frac{2*5}{3*5}+\frac{4*3}{5*3}\\\\
=\frac{10}{15}+\frac{12}{15}\\\\
=\frac{10+12}{15}\\\\
=\frac{22}{15}\\\\\\
\mbox{Now add pronumerals (letters) the same way}\\\\
\frac{2}{a}+\frac{4}{b}\\\\
=\frac{2*b}{a*b}+\frac{4*a}{b*a}\\\\
=\frac{2b}{a*b}+\frac{4a}{a*b}\\\\
=\frac{2b+4a}{ab}\\\\$$