CPhill

Nauseated.....how does  C(2N, N) = (2N)!/N!  ???

My guess of 70 unique routes wth a 4 x 4 square is actually the same as your formula of C(8, 4)  = 70

I had forgotten that, in Pascal's tirangle, the summing of the middle two terms in the 7th row would actually produce the 5th term in the 8th row, ie., C(8,4).....thus......your result appears justified.....

But  (2*4)! / 4!  = 8! / 4!   = 1680 routes  ????  Shouldn't there be an additional (2N - N)! in the denominator??

Am I missing something...or.....is premature CDD actually here to stay???

BTW.....I like your logic......!!!

Choose a boy to "anchor" the circle......he can sit in any seat

To his right, any of the 4 girls can be chosen

To her right, any of the 3 other boys can be chosen

Etc.

So..... the total ways  = 4! x 3!  = 24 x 6  = 144 ways

?????

Disregard my previous answer....I want to think about this one a little more....!!!

I'll try this one.....here goes nothing......!!!

We have 4 ways of choosing a boy to sit in any of the chairs.

And we have 4 ways to select any girl who sits to the right of him......3 ways to seat the next boy to the right of her......3 ways to seat the next girl to the right of the second boy, etc.

So....the total number of ways is just 4! x 4!   = 576 ways

Don't know if I'm correct, but that's my two cents worth.......

This is easiest calculated by first choosing the ranks, and then choosing the cards within those ranks...

So.....

From 13 ranks, we want to choose any 2  =  C(13,2)

And from the two ranks, we want to choose any 2 cards from each = [C(4,2)]^2

And we want to choose one card from any of the remaining 11 ranks.....

So we have

C(13,2)[C(4,2)}^2 * C(11,1)C(4,1)  = 123,552

Here is a method that is similar to Melody's...it converges pretty fast....even if our intial guess is not too good.

This method is known as "Guess - Divide - Check "

Step 1  : Let's say that we guess the square root of 6 as 2

Step 2 : Divide 6 by 2  = 3

Step 3 :  Now add 2 +3  and divide by 2 = 2.5

Step 4: Now divide 6 by 2.5 = 2.4

Step 5: Now add 2.5 and 2.4 and divide by 2 = 2.45

Step 6 :  Divide 6 by 2.45 = 2.448

Step 7 : Add 2.448 and 2.45 and divide by 2 = 2.449

Step 8: Divide 6 by 2.449 = 2.449......note this is pretty close to the actual square root of 6 !!!

This isn't hard to do.....once you get the "rhythm" down.....!!!!......with a better initial guess.....the procedure may have even converged much more quickly!!!

 P.S.  -  The method might be better termed as "Guess - Divide - Average - Divide - Average - Divide......."

-4 is even because, when divided by 2, there is no remainder.....

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"or do even and odd numbers have a minimum value of 0??"

I think you're asking if negative numbers have parity....("eveness" and "oddness").....yes they do.....just like the positive integers....!!!

I'll give this one a shot.......

"Anchor" John and Sam in any two opposite positions.   Now, Stacy and Roger can occupy any of the other 4 diametrically opposite positions in 2 ways each, and for each of these arragements the other children can be further arranged in 6! ways.

So we have  4 x 2 x 6!   = 5760 ways

Hey...Alan and I agree......whether that makes us correct - or not -might be another story..!!! 

To elaborate on Melody's answer....we can have either math book at both ends...so that's two ways to arrange those.

And, in the "middle,' the other 4 books can be arranged in 4! = 24  ways.

So   ..   2 x 4!  = 2 x 24  = 48 ways 

Also....ASA  ....Angle-Side-Angle....

The one that we have to be careful with is SSA....This may or  may  not produce congruent triangles.....as one of my former teachers might say.....Turning this around backwards may prevent you from making an "A*S" of yourself.....!!!

Or, as "Nauseated" might say.....turning it around backwards will help prevent CDD....!!!!