Is that 8 columns (positions) where a vertical domino can go?Originally posted by joe shmoI'm stuck on finding approach to eliminate undesirable combinations in the problem below, any help would be appreciated.

How many ways can a 8x2 rectangle be tiled with 2x1 dominoes such thatexactlytwo of the dominoes are vertical?

Or 2 columns of 8?

Leftmost domino in positionOriginally posted by joe shmo8 columns where a vertical domino can be placed.

1 then second can go in 5 places (not 3 or 7)

3 then second can go in 3 places (not 5 or 7)

4 then second can go in 2 places (not 6 or 7)

5 then second can go in 2 places (not 7)

6 then second can go in 0 places (not 7 or 8)

7 then second can go in 1 place (only 8)

So I make that 13 in total.

The answer is 10, so you have a few more to find.Originally posted by wolfgang59Leftmost domino in position

1 then second can go in 5 places (not 3 or 7)

3 then second can go in 3 places (not 5 or 7)

4 then second can go in 2 places (not 6 or 7)

5 then second can go in 2 places (not 7)

6 then second can go in 0 places (not 7 or 8)

7 then second can go in 1 place (only 8)

So I make that 13 in total.

Correct me if i'm wrong, but your method is somewhat brute force? Is their a way that is mathematically succinct and generalized to any 2*n x 2 rectangle using combinatorics?

OK. Sober now.Originally posted by joe shmoThe answer is 10, so you have a few more to find.

Correct me if i'm wrong, but your method is somewhat brute force? Is their a way that is mathematically succinct and generalized to any 2*n x 2 rectangle using combinatorics?

The answer is n(n+1)/2 for all n.

(triangle numbers)

The proof is obvious but difficult to put down .....

Start by considering that the left most domino is always odd and right most is always even..

Ok.Originally posted by wolfgang59OK. Sober now.

The answer is n(n+1)/2 for all n.

(triangle numbers)

The proof is obvious but difficult to put down .....

Start by considering that the left most domino is always odd and right most is always even..

n(n+1)/2 is true for n=1

Now to prove if true for n then true for m=(n+1)

When we have 2m columns there is ONE additional place for the right most tile for all n positions of the left most tile. In addition there is one more configuration with left most tile in column (m-1) and right most in position m,

Therefore there are an additional m, or (n+1) configurations.

Total = n(n+1)/2 + (n+1) = {n(n+1)+2(n+1)}/2

= (n+1)(n+2)/2 = m(m+1)/2

QED

Seven. Basically there are four 2x2 squares which each fit two dominoes. Bother must be either vertical or horizontal. So depending on which of those four squares you put the vertical dominoes you can have four possibilities.Originally posted by joe shmoI'm stuck on finding approach to eliminate undesirable combinations in the problem below, any help would be appreciated.

How many ways can a 8x2 rectangle be tiled with 2x1 dominoes such thatexactlytwo of the dominoes are vertical?

You can also have one vertical one on each end. So five.

Or one vertical, two horizontal, one vertical, four horizontal. That's six.

Or one vertical, four horizontal, vertical, two horizontal. Seven.

The answer is 10...so your missing a few.Originally posted by AThousandYoungSeven. Basically there are four 2x2 squares which each fit two dominoes. Bother must be either vertical or horizontal. So depending on which of those four squares you put the vertical dominoes you can have four possibilities.

You can also have one vertical one on each end. So five.

Or one vertical, two horizontal, one vertical, four horizontal. That's six.

Or one vertical, four horizontal, vertical, two horizontal. Seven.

Are there any methods in combinatorics that you know of that can be used to eliminate combinations of the remaining un-chosen columns?; That is removing combinations for which any two of the un-chosen columns are not adjacent?Originally posted by wolfgang59Ok.

n(n+1)/2 is true for n=1

Now to prove if true for n then true for m=(n+1)

When we have 2m columns there is ONE additional place for the right most tile for all n positions of the left most tile. In addition there is one more configuration with left most tile in column (m-1) and right most in position m,

Therefore there are an additional m, or (n ...[text shortened]... configurations.

Total = n(n+1)/2 + (n+1) = {n(n+1)+2(n+1)}/2

= (n+1)(n+2)/2 = m(m+1)/2

QED