[Prof. Templeton]

Long before Professor Templeton started his career at the Redrum University he had a job at Whiskey distillery in Tennessee. The plant made six different grades of whiskey that went into numbered barrels from 1 to 6. The best stuff was labeled “1” and the least in quality was labeled “6” with other grades in between. Now the man in charge of the warehouse was a bit eccentric (aren’t we all ) and had a special way of stacking the graded barrels. The six graded barrels were stacked in two rows of three, one row above the other, such that a barrel was never beneath or to the right of a barrel of lesser value. With the six different grades of barrel there were only five ways this could be done…

123 on top of 456

124 on top of 356

125 on top of 346

134 on top of 256

135 on top of 246

After much thought by the marketing department, it was decided the distillery would double the amount of grades of whiskey that it produced. The warehouse man also decided to double his number of barrels in a row to six in order to accommodate the increase, but he still wanted to keep his rule of never placing a barrel below or to the right of a lesser graded barrel. How many ways now were there to stack the twelve different graded barrels?

[Brandonb]

haven't gone through the distinct possibilities in my mind yet but I would have to say...

10

Edited January 20, 2009 by Brandonb

[Guest]

132

Edited January 20, 2009 by solarfish

[preflop]

since you defined to the right to mean only on the same row ...

for example in the original problem this stack is ok

123

456

(technically 2 is right of 4 - just not in the same row).

i come up with 40

[Guest]

Admittedly, my method wasn't very carefully executed so I'm not too sure, but my guess is 76.

[Prof. Templeton]

None right so far,

solarfish is the closest.

[Guest]

None right so far,

solarfish is the closest.

I did originally have 10 more than that number but managed to talk myself out of it. I assume that was not the right answer either?

[Prof. Templeton]

I did originally have 10 more than that number but managed to talk myself out of it. I assume that was not the right answer either?

Nope. If this was the price is right you wouldn't have made it on the stage.

[Guest]

I count 144...?

[HoustonHokie]

Well, I haven't worked out a lot of math on this, but I do notice something - namely, the number of possible arrangements for the barrels has gone up by a lot, and I suppose the number of "correct" rule-following arrangements has gone up by some related factor. For the original 6 barrels, we are only really moving 4 of them (#1 is always top left and #6 is always bottom right). So the number of permutations for 4 barrels is 24, of which 6 are "correct". Now, with 10 barrels (#1 always top left, #12 always bottom right), there are 3,628,800 possible ways to arrange them. I doubt the answer grows by a factor of 3,628,800/24, but perhaps there's a relationship between these permutations and the answer we're looking for. Am I on the right track?

[Prof. Templeton]

Well, I haven't worked out a lot of math on this, but I do notice something - namely, the number of possible arrangements for the barrels has gone up by a lot, and I suppose the number of "correct" rule-following arrangements has gone up by some related factor. For the original 6 barrels, we are only really moving 4 of them (#1 is always top left and #6 is always bottom right). So the number of permutations for 4 barrels is 24, of which 6 are "correct". Now, with 10 barrels (#1 always top left, #12 always bottom right), there are 3,628,800 possible ways to arrange them. I doubt the answer grows by a factor of 3,628,800/24, but perhaps there's a relationship between these permutations and the answer we're looking for. Am I on the right track?

Combinations. In the first case with six barrels what is C(n,k) and how does it relate to the available configurations?

[HoustonHokie]

Combinations. In the first case with six barrels what is C(n,k) and how does it relate to the available configurations?

I had thought about combinations, but the order is significant, which is what led me to permutations. Sounds like I'm in the right forest, but still barking up the wrong tree. I'll keep thinking.

Edited January 21, 2009 by HoustonHokie

[Guest]

I tried this myself and I got the same answer as solarfish.

Are you sure the answer isn't 132?

[Prof. Templeton]

132

Your answer is correct. Opps!

I tried this myself and I got the same answer as solarfish.

Are you sure the answer isn't 132?

My apologizes to solarfish, his answer is indeed correct. I got back to work and checked his answer with my math and it's spot on. Not sure what I was thinking when I said "No". Sorry. Anywho...

C(2n,k)/n+1

C(12,6)=12!/6!(12-6)!=[12*11*10*9*8*7]/[6*5*4*3*2*1]=924

924/7 = 132

[Guest]

My apologizes to solarfish, his answer is indeed correct. I got back to work and checked his answer with my math and it's spot on. Not sure what I was thinking when I said "No". Sorry. Anywho...

Thank goodness for that, I had given up after my attempt as I thought the puzzle was beyond me!

[Guest]

I agree with solarfishes answer, i figure that no digit can be twice the place from the left, which makes 2,*,*,*,*,* and 1,4,*,*,*,* and 1,2,6,*,*,* and 1,3,6,*,*,* and so on imposible. Also, any posible combination for the top can have only one combination for the bottom since the remanding 6 barrels must be in numerical order. So solarfish's answer would be the maximum number of solutions following this rule. So i am really confused as too how solarfish's answer could be less than the actual number.

????dazed and confused?????

just reading the last comment few it is right

[Guest]

120