The worlds first riddle!

Mark, don't forget; you got an invite to join the party. Whereas, I had to gatecrash! :wink:


SHEAINSE


SWECIAGLEHS



My last four letters denote a conservative. I am an 'established record.'


I was a famous poet. Remove one of my letters & you have something often held by candlelight.


Remove my first three letters & you have a flower. I am an extremely unfortunate occurrence.

[size=8]he's insane
weigh in scales[/size]

Remove my first three letters & you have a flower. I am an extremely unfortunate occurrence.

DIS ASTER

My last four letters denote a conservative. I am an 'established record.'

Score - Core

Mark:

SURVEY
C(211,3) = 1,543,465



We seek the number of non-negative integer solutions to the equation MP+FP+MS+FS=208, where each two-letter variable represents one of the four categories of respondents. The number of such solutions is C(208+4-1,4-1)=C(211,3).


CCOSUTTS
cut in costs


PRCIUCTE
cut in price


SHEAINSE
he's insane

SWECIAGLEHS
weigh in scales


Whim wrote, "I have been giving answers in other topic. Have you missed that?"

No, I was following, following Mark that is. Either he had already answered or I didn't know what you were talking about. However, it is always good to see the riddles you come up with. Oh, and answers too of course. Such as:

Whim:

"Remove my first three letters & you have a flower. I am an extremely unfortunate occurrence."

DIS ASTER



My last four letters denote a conservative. I am an 'established record.'

Score - Core

(That is good. I had: His tory)





A strange request by a recently sun burnt mathematician who was almost eaten by a flock of Tiger Sharks, involved the two nephews of the mathematician. The two nephews were to divide the mathematician's sports card collection, with the only requirement being that each nephew get at least one card.

If there were k different cards in the collection, with no card appearing more than once, how many different divisions were possible between the two nephews



RSINAGIINGN


SIGLAHNDT

2^k divisions
but to have at least one card
(2^k)-2 divisions

[size=8]singing in the rain
land in sight[/size]

Usamashaker:

2^k divisions
but to have at least one card
(2^k)-2 divisions


This problem can be viewed as determining the number of subsets from a k-element set and adjusting for the requirement that each nephew get at least one card. There are 2^k different subsets, including the empty set. We must eliminate both the empty set and the set itself as potential subsets, for use of either of these violates the restriction that each nephew get at least one card. Therefore, there are 2^k - 2 ways to divide the cards.

usamashaker made it look easy. Good to see you buddy.



Mark:

RSINAGIINGN
singing in the rain

SIGLAHNDT
land in sight



Shari walks from her apartment to the library every day. Her apartment is 7 blocks north and 12 blocks east of the library. Shari walks along city streets that are laid out in a grid system.

(a) Given access to all the streets in the system, how many different 19-block routes could Shari use to get from home to the library

(b) The east/west street running outside Shari's apartment is under repair. She cannot walk west on that street for the first 3 blocks outside her apartment. Now how many different 19-block routes could Shari use to get from home to the library

(c) Again given access to all the streets in the system, how many different 21-block routes could Shari use to get from home to the library



BFLLOOWEORSM


SCOIMDEE

Flowers in Bloom....
Come inside

[size=7]SHARI/LIBRARY
(a) C(19,7) = 50,388
(b) She'll never be able to get back up to this street; so it doesn't matter how many of it's blocks are out. The reduces to C(18,6) = 18,564
(c) This was interesting.
I got 3,452,870.[/size]

Shari:

Flowers in Bloom.
BFLLOOWEORSM

SCOIMDEE
Come inside


Mark:

SHARI/LIBRARY
(a) C(19,7) = 50,388

(b) She'll never be able to get back up to this street; so it doesn't matter how many of it's blocks are out. The reduces to C(18,6) = 18,564

(c) This was interesting.
I got 3,452,870.

Now this is how I have it. I am not saying I am right, experience has often shown otherwise.

For each of the C(19,7) routes Shari could follow from her home to the library, there are 20 intersection points of E/W with N/S streets. At each intersection point, she could go in one of 4 directions one block and return to the same point. : C(19,7)*20*4




Referring to the letters in the word ENTEROBACTERIACEAE


(a) How many arrangements are there for the letters in this word

(b) How many arrangements exist if each cannot begin nor end with E



CLLOAVMEBSR


LSTIANNDE


WSIANKTINEGR

What about the intersections with fewer than four directions?

Here's my method:
Calculate the number of ways of getting to an intersection by having gone backward (N or E) at the final step. This is the sum of the ways of getting directly to the intersections that are just south and just west of the current intersection. Multiply that sum by the number of ways of getting directly from that intersection to the goal. Do this for each intersection and sum the products. I'd bet almost anything the method works. I verified it by hand on a few small cases. Try your method on some small cases.

[size=8]ENTEROBACTERIACEAE
(a) 18! / (5! * 2! * 2! * 3! * 2!) = 1,111,523,212,800

(b) [18! / (5! * 2! * 2! * 3! * 2!)] - 2* [17! / (4! * 2! * 2! * 3! * 2!)] + [16! / (3! * 2! * 2! * 3! * 2!)] = 566,658,892,800

lambs in clover
stand in line
sinking in water[/size]

C…"given access to ALL the streets in the system, how many different 21-block routes could Shari use to get from home to the library?"


"city streets that are laid out in a grid system."


"What about the intersections with fewer than four directions?"


I am being brave here by saying; my answer does not have any intersections with less than four directions.

However, Shari is one foxy woman, so she might have tricked me.

Oh! We can venture beyond the rectangle? Your answer seems correct then. I assumed the system was the 19x7 block grid.

Oh well. I had fun solving the over-constrained problem.

"I had fun solving the over-constrained problem."

Thank you for the answer, I will add it to the mix in future.



QUETSRTIOYN


FLIEVAER


The RCAAUGIHNT

[size=8]tryin' question
live in fear
caught in the rain[/size]

Mark:

ENTEROBACTERIACEAE
(a) 18! / (5! * 2! * 2! * 3! * 2!) = 1,111,523,212,800

(b) [18! / (5! * 2! * 2! * 3! * 2!)] - 2* [17! / (4! * 2! * 2! * 3! * 2!)] + [16! / (3! * 2! * 2! * 3! * 2!)] = 566,658,892,800


And that is a lot of ways.


CLLOAVMEBSR
lambs in clover

LSTIANNDE
stand in line

WSIANKTINEGR
sinking in water


QUETSRTIOYN
tryin' question


FLIEVAER
live in fear


The RCAAUGIHNT
caught in the rain




Referring to the letters in the word ENTEROBACTERIACEAE (cont).


How many arrangements can be made if no two consonants can be adjacent to each other



If the only distinction we can make is between vowels and consonants, how many arrangements can be made



TWAISMTEE


ACSCUENNT

[size=8]ENTEROBACTERIACEAE
(a) C(11,8) * [10! / (5! * 3!)] * [8! / (2! * 2! * 2!)] = 4,191,264,000
(b) 18! / (10! * 8!) = 43,758

wastin' time
cunnin' ascent / sunnin' accent[/size]

Mark:


ENTEROBACTERIACEAE
(a) C(11,8) * [10! / (5! * 3!)] * [8! / (2! * 2! * 2!)] = 4,191,264,000

(b) 18! / (10! * 8!) = 43,758


The guy is a genius.



TWAISMTEE
wastin' time


ACSCUENNT
cunnin' ascent / sunnin' accent

(Both good, I had sun (in) accent)




Shari is taking a True/False test in her automobile mechanics class. There are 10 unique items on the test. Using a numbered answer sheet, students are requested to circle either True or False corresponding to each item.

(a) How many different responses could Shari submit on the 10-item answer sheet, assuming she responds either True or False to each item



(b) If we allow for the possibility that Shari might not circle either True or False on any or all items (i.e., she could leave items blank), how many different answer-sheet responses could she submit




(c) Shari's friend Mark took the True/False test before Shari did. Mark told Shari there were 5 True items and 5 False items. If Shari heeded Mark's advice and circled 5 True and 5 False on the answer sheet, how many different answer-sheet responses could she submit



HSEHADOLICNKES


NSHEOCWKS