GingerAle

MIRB, The counts (not the actual sequence) correspond to row #8 of Pascal’s triangle (with zero starting the count, so it’s actually the ninth row). If zero were included in the count then row #9 would give the counts for each number.

Again, despite Mr. BB’s insistence, there is nothing coincidental about these numeric relationships. All of these are mathematically related.  The set theory gives the solution to the sum, which is what the question asks for. The binomial distribution gives the leading coefficient for the count of each sequence (via a generating function). Still, further logic and mathematics are required to produce the numeric sequences. There are 362,880 ways to arrange those numbers, but only 256 of them fall within the criteria of the question.   

GA

Well, Mr. BB, like usual... you’re an arrogant idiot. 

Heureka’s post details the 256 individual solutions to this problem, and they link unambiguously and decidedly to the count obtained via the set theory solution—that is, they are NOT incidental nor are they coincidental.   The counts relate to the binomial distribution in Pascal’s triangle. (See below.)

Your solution looks good, EP.

Condensed Form

\(\dfrac{1}{2} * \large (\Large \rho \normalsize \text{(two balls of same color bag A)} + \Large \rho \normalsize \text{(two balls of same color bag B)}\large)\\ \text { }\\ \large \dfrac{1}{2} \small \dfrac{\dbinom{3}{2}+\dbinom{2}{2}}{\dbinom{5}{2}} +\large \dfrac{1}{2} \small \dfrac {\dbinom{6}{2}+\dbinom{3}{2} }{\dbinom{9}{2}} =\Large \dfrac{9}{20} = \ 0.45 \)

GA

It looks like the above answer is from Mr. BB or one of his students.

For this question, it’s easier to count the negations, i.e when there are no matches.

\(\small \text {There are six ways to roll the first die. (6)}\\ \small \text {There are five ways to roll the second die for it NOT to match the first. (5)}\\ \small \text {There are four ways to roll the third die for it NOT to match the first or second. (4) }\\ \small \text {So, there are } 6*5*4 = 120 \text { ways to roll three dice with NO matches. }\\ \small\text {And there are }6^3 \text {ways to roll all combinations. (216) }\\ \large 1- \dfrac {6*5*4}{216} = \dfrac{5}{9} \approx 44.44 \% \)

GA

The last digit has to be either a one (1) or nine (9), then are two (2) options (>,<) for the remaining eight (8) digits, so there are 2^8 =256 arraignments where each digit is either greater than or less than the preceding digits.

GA

In this scenario, there are only six games. Your team obtains the fourth win in game six. This means they must have won three of the previous five games.  This gives 5C3 = 10 for the number of way this could happen.

GA

You’re right! I withdraw my critical assessment.  

The knot does equate to a straight line; changing from the circular permutation P=(n-1) to the linear P=n!

The caveat for reflections (/2) being equivalent for a linear permutation seems unusual, but there isn’t a reason you couldn’t do it. When I scratch my left ear while looking in a mirror I see me scratching my right ear in Alice’s universe, but were both content.

Your answer matches because of coincidence, not because of logical analyses.   

You said the passengers for the bus are standing in a line, that’s not a circle. If they were standing in a circle, and the bus door prevented them from rotating, then your solution is wrong because you have to count the bus door as a passenger, and you didn’t do that. 

Your methodology solves a string of beads in a line, not attached at the end, or a bracelet (ends attached) and each rotation counting as a unique arrangement –this isn’t normally done in circle combinations.  

Unless the people line up in a circle, it’s not the same.

For a bracelet without a knot, there are (4-1)! / 2 arrangements. One (1) is subtracted to fix the position of a bead, because the beads can be rotated such that one particular bead can be placed in a start position without rearranging them. The division by two (2) is because the bracelet can be flipped.

A bracelet with a knot prevents rotation. Treat the knot as an additional bead, fixed in position, giving (5-1)! / 2 = 12 arrangements. 

GA

Yes, Lancelot uses his opposable thumbs for multiple purposes, including thumbs up or down. Nauseated, however, may give paws for dewclaw consideration, but his central tendency is toward his central digit with adjustments for any ambiguity.  

GA