GingerAle

Pending review....

Ginger speaks: Yep I screwed up. I neglected the caveat of “at least” in the question.

Ginger Troll speaks: I didn’t post a wrong answer. I correctly answered the wrong question. A home run on an error is still a home run.

GA

How many different three-word code phrases in this system use 8 letters?

The posted example is inconsistent with the description preceding it. 

This is a valid example: <<  >>>  <> 

Assumptions: a single bit of “<” or “>” can represent two words. Two bits can represent four words, etc. 

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Solution:

Treat as a binary code of 8 bits partitioned into sets of three (3). The value of each partition is an exponent of (2). Multiply these partitions, then multiply by (6) or (3) for the unique permutations of each partition. 

8 bits partitioned into sets of (3) where the number of bits is the exponent of (2).

(2^6)*(2^1)*(2^1) =256 | 256*3 =768

(64) words * (2) words * (2) words * (3) unique permutations.

(2^5)*(2^2)*(2^1) = 256 | 256*6 =1536

(32) words * (4) words * (1) word * (6) unique permutations.

(2^3)*(2^3)*(2^2) = 256 | 256*3=768

(2^4)*(2^3)*(2^1) = 256 | 256*6 =1536

(2^4)*(2^2)*(2^2) = 256 | 256*3=768

5376 Total Phrases

GA

Solution:

There are (10C3) 120 ways to choose where to place the three false responses. This leaves (2⁷) 128 ways to respond to the remaining questions with true or false. 

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^{How many ways are there to answer a 10-question true/false test, where at least 3 of the questions have been answered with a false?}

There are (10C3) 120 ways to choose where to place the three false responses. This leaves (2⁷) 128 ways to respond to the remaining questions with true or false. 

 (10C3)   = 120   *       128                   15360

 This continues with

(10C4)   = 210    *         64                   13440

(10C5)   = 252    *         32                     8064

(10C6)   = 210    *         16                     3360

(10C7)   = 120    *           8                       960

(10C8)     = 45    *           4                       180

(10C9)     = 10    *           2                         20

(10C10)     = 1    *           1                           1

Sums        968             256                 35185

Analysis:  There are 35185 ways to answer 10 question T/F with 3 or more pre-answered questions. Total includes multiple redundancies. At least 64 permutations will not exist in the totals.

GA

Laughter is usually among the best of medicines; although, I suppose it could top-off someone on the edge.  Anyway, you’ve had a few maga-doses and it’s not topped you yet, so keep borrowing from the borrowed-time zone –the interest rates are cheap, and though the days are getting shorter, you’ll have more of them, and you’ll make it up in volume.  

What’s the “other way” that might be better? 

GA

^{Just picture all the 2's and 4's with extra dots turning them all into 6's, say.}

^{To all intents and purposes they're the same, same as the 1's, 3's and 5's.}

Twos, fours, and sixes, Oh MY! This sounds like analysis using magical incantations and charms. “Snake eyes and boil, slick lies and toil, these ducks are now pigs.”  The ducks aren’t really pigs, but some members of the viewing audience may think they are.  This is entertaining, but not as much as using telekinetic powers to roll snake eyes or not snake eyes.

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DC Thomas, my cat, demonstrated his telekinetic powers for rolling snake eyes. This was very impressive to watch. He rolled snake eyes on about 60% of the attempts. The dice seemed to roll with normal expectation when he wasn’t using his telekinetic powers. This went on for a while, until Mr. Peabody, my dog, caught him switching out the loaded dice. The fight was on...

DC knows he’ll get caught when he uses loaded dice playing craps, or embezzles Monopoly money when he’s the banker. He does this just to pissoff Mr. Peabody –he likes to fight. Mr. Peabody always obliges. Long ago, I started securing the fragile artwork, and put the veterinarian on speed-dial, whenever we play games.

GA

Hi Melody,

The (4!) is in there for the same reason it’s in your solution. Four of the dice are indistinguishable from each other, so their permutations are factored out of the (10!).

To clarify, though the four dice appear to be distinguishable from each other because 2, 4, and 6 are distinguishable, they are treated as an indistinguishable group, that is, they are only distinguishable from the 1, 3, and 5. So this question has four groups: three groups of 2 indistinguishable element, each consisting of 1s, 3s, and 5s, and one group four that is an indistinguishable collective of NOT 1s, 3s, or 5s.

GA

Here’s a “sensible way” that hides the demonstration of the nuanced details.

\(\dfrac {\binom{10}{2}\cdot \binom{8}{2}\cdot \binom{6}{2}\cdot (3^4)}{6^{10}} = \dfrac{175}{6912} \approx 2.532\% \)

GA