The worlds first riddle!

[size=8]TEST
(a) 2^10 = 1,024
(b) 3^10 = 59,049
(c) C(10,5) = 252

shockin' headlines
show in necks / showin' necks[/size]

Perhaps, this will be more to your taste!


An unlimited supply of dominoes is available for building a rectangle that measures exactly 2 units high and n units long. Each domino is a 2-by-1 rectangle.

Can you write a recursive description for R(n), the number of different ways to build a 2-by-n rectangle with dominoes


FLBIGIHROT


USTANDP

[size=8]R(n) is the Fibonacci sequence.
R(0) = R(1) = 1
R(n) = R(n-1) + R(n-2), n > 1

To see the recursive nature:
Given n,
- if the first domino is vertical, the number of solutions is R(n-1)
- if the first two dominoes are horizontal (one on top of the other), the number of solutions is R(n-2)


Perhaps that should be FLBIGIHRDT for
bird in flight

standin' up[/size]

Mark:

TEST
(a) 2^10 = 1,024

There are two choices at each of the 10 answer-blank positions on the answer sheet.


(b) 3^10 = 59,049

There are now three choices at each of the 10 answer-blank positions on the answer sheet.


(c) C(10,5) = 252

Among the 10 answer blanks, select 5 and circle True. The 5 remaining blanks must be False.



HSEHADOLICNKES
shockin' headlines

NSHEOCWKS
show in necks / showin' necks

Yippy! I think I win for the very first time. SHOCK (in) NEWS. :wink:



R(n) is the Fibonacci sequence.
R(0) = R(1) = 1
R(n) = R(n-1) + R(n-2), n > 1

To see the recursive nature:
Given n,
- if the first domino is vertical, the number of solutions is R(n-1)
- if the first two dominoes are horizontal (one on top of the other), the number of solutions is R(n-2)

Mark once again makes it look easy, for the rest of us:


Let us look at some of the initial values for R(n).

R(1) represents the number of 2-by-1 rectangles we can create with 2-by-1 dominoes. There is just one way to do this, so R(1)=1.

R(2) represents the number of 2-by-2 rectangles we can create with 2-by-1 dominos. There are two way to do this, with a pair of dominoes both horizontal or both vertical, so R(2)=2.

Now let's see how we can build R(3) from previous cases. From the 2-by-2 rectangles (i.e., those two in the R(2) case), we can append one vertical 2-by-1 domino. From the 2-by-1 rectangles (i.e., the one in the R(1) case), we can append two horizontal 2-by-1 dominoes. It seems, then, that R(3) is just R(2)+R(1), or, equivalently, R(3)=3.

This pattern will continue, for to get R(n) we need to just look at those in the R(n-1) case and append one vertical domino, or look at those in the R(n-2) case and append two horizontal dominoes.

Therefore, R(n)=R(n-1)+R(n-2), with R(1)=1 and R(2)=2.


Perhaps that should be FLBIGIHRDT for
bird in flight

(No perhaps about it)

USTANDP
standin' up


Now, if I have the math right: That's one to me and 647 to you. Boy, it sure is getting to close to call.


Payback time!


In the expansion of (r + s + t + u + v) ^15, determine the number of different ways a coefficient of 15 appears among the collected terms




DSOIWTN


WTRAETAEDR


OPAIINLTS

[size=8]sitting down
treading water
paint in oils[/size]

How does he do that?

SLOPSATCE


LILANSTE

[size=8]lost in space
last in line[/size]

I haven't forgotten about the expansion problem, but I'm quite busy with work these days.

[size=8]EXPANSION
That would be all terms of the form (x^14)y. There are P(5,2) = 20 such terms.[/size]

Mark:

EXPANSION
That would be all terms of the form (x^14)y. There are P(5,2) = 20 such terms.


We want to determine the number of times K=15 when we examine the collected terms of the form Kr^Rs^St^Tu^Uv^V where R+S+T+U+V=15 over the nonnegative integers.

We know that K=(15!)/(R!S!T!U!V!), the multinomial coefficient. For K to be 15, it must be the case that R!S!T!U!V!=14! [Why is that?] For that to be true, two conditions must be met: (i) We must have exactly one of R, S, T, U, or V equal to 14 and (ii) the remaining 4 values must sum to 1.

There are 5 ways for the first condition to be met: R or S or T or U or V must be 14. There are 4 ways for the second condition to be met: one of the four remaining values from among R,S,T,U,V must be 1 and the other three must be 0.

By the multiplication principle, there are 5*4=20 total ways for a coefficient of 15 to appear. Which is exactly what Mark told us.


DSOIWTN
sitting down


WTRAETAEDR
treading water


OPAIINLTS
paint in oils


SLOPSATCE
lost in space


LILANSTE
last in line


Fantastic!!!



"I'm quite busy with work these days."

That is excellent news, I am pleased business is picking up, may it continue to pre downturn levels and beyond.





A shelf is to contain nine different books, six different paperback books and three different hardback books. If the paperback books must be shelved in pairs (that is, exactly two paperback books must be adjacent to each other), how many ways can the nine books be shelved


MAWCHASINHE


TMAEXRIMEUSMT

Washin' machine

[size=8]maximum interest[/size]

[size=8]BOOKS
Taking "exactly two" to mean no fewer and no more, there are four ways to separate the paperback pairs:
hphphp
phhphp
phphhp
phphph
There are 6! ways to order the paperbacks and 3! ways to order the hardbacks.
Total = 4 * 6! * 3! = 17,280[/size]

Shari:

WOW IRAHS KCAB


MAWCHASINHE
Washin' machine


Mark:

BOOKS
Taking "exactly two" to mean no fewer and no more, there are four ways to separate the paperback pairs:
hphphp
phhphp
phphhp
phphph
There are 6! ways to order the paperbacks and 3! ways to order the hardbacks.
Total = 4 * 6! * 3! = 17,280

Somewhat surprisingly there are.


TMAEXRIMEUSMT
maximum interest




Ten dogs come upon eight biscuits, and dogs do not share biscuits!

(a) In how many different ways can the biscuits be consumed by the dogs, assuming the dogs are distinguishable but the biscuits are not


(b) In how many different ways can the biscuits be consumed by the dogs, assuming both the dogs and the biscuits are distinguishable



CHADIONGS


PBIASCSKUIETT


SALSVATEEIOKN

[size=8]dog in chains
biscuit in pasket?
seeking salvation[/size]

i had......... dong in chai's

but I knew that wasn't right.

ahhh poop. mark got it.

DP wrote, "i had......... dong in chai's

[size=8]right in the middle[/size]

DOG BISCUITS
(a) C(17,9) = 24,310
(b) C(17,9) * 8! = 980,179,200

Mark:


DOG BISCUITS
(a) C(17,9) = 24,310

Let d(i) represent the number of biscuits eaten by the ith dog, where 0 ² i ² 10. Then we seek the number on non-negative integer solutions to d(1)+d(2)+…+d(10) = 8.



(b) C(17,9) * 8! = 980,179,200

For each of the 8 biscuits, there are 10 possible dogs that could have eaten the biscuit.





CHADIONGS
dog in chains


PBIASCSKUIETT
biscuit in pasket? (It would be PACKET if the idiot could spell)


SALSVATEEIOKN
seeking salvation (apt)


MRIIDGDHLTE
right in the middle



There is a street of length 4. The street is initially empty. Cars then come along to fill the street until there is no space left that is large enough to park a car. Every car is length 1. Drivers will choose a location to park at random among all possible locations left. No consideration whether good or bad, is given to other cars.

What is the expected number of cars that will be able to park



MODRVIIVEE


APCLTAIONN




As the next post will be the 5,000th may I congratulate all those who have taken part and made a little piece of history.