mikemcgarry wrote:

Attachment:

chessboard with knight-L tile.JPG

A chessboard is an 8×8 array of identically sized squares. Each square has a particular designation, depending on its row and column. An L-shaped card, exactly the size of four squares on the chessboard, is laid on the chessboard as shown, covering exactly four squares. This L-shaped card can be moved around, rotated, and even picked up and turned over to give the mirror-image of an L. In how many different ways can this L-shaped card cover exactly four squares on the chessboard?

(A) 256

(B) 336

(C) 424

(D) 512

(E) 672Obs.: when I came here to post my solution, I found that it coincides with "thefibonacci" solution.

I will post it anyway, so that the readers have another "wording" (it may help)!

\(?\,\,\,:\,\,\,\# \,\,L - {\rm{shaped}}\,\,{\rm{positions}}\)

Although we will need to separate the problem in 8 configurations... all of them are trivial:

(The time I took to type the solution - drawing included - was approximately 15min.

But only 3min to find the solution to myself - ugly-hand-drawing included.)

Configuration 1: the "head" (guide-point in red) is up, the "tail" to the right.

We have 6 positions for the head in the first left viable column (I have shown the first and last in it)

We have 7 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

(Numbers 6 and 7 are explained in the first drawing. The others below are analogous.)

Configuration 2: the "head" (guide-point in red) is up, the "tail" to the left.

We have 6 positions for the head in the first left viable column (I have shown the first and last in it)

We have 7 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

Configuration 3: the "head" (guide-point in red) is down, the "tail" to the right.

We have 6 positions for the head in the first left viable column (I have shown the first and last in it)

We have 7 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

Configuration 4: the "head" (guide-point in red) is down, the "tail" to the left.

We have 6 positions for the head in the first left viable column (I have shown the first and last in it)

We have 7 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

Configuration 5: the "head" (guide-point in red) is left, the "tail" to the right-down.

We have 7 positions for the head in the first left viable column (I have shown the first and last in it)

We have 6 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

Configuration 6: the "head" (guide-point in red) is right, the "tail" to the left-down.

We have 7 positions for the head in the first left viable column (I have shown the first and last in it)

We have 6 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

Configuration 7: the "head" (guide-point in red) is left, the "tail" to the right-up.

We have 7 positions for the head in the first left viable column (I have shown the first and last in it)

We have 6 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

Configuration 8: the "head" (guide-point in red) is right, the "tail" to the left-up.

We have 7 positions for the head in the first left viable column (I have shown the first and last in it)

We have 6 positions for the head in each row (I have shown the lower left and the lower right).

Multiplicative Principle: 6*7 = 42 possibilities

All cases above are exhaustive (i.e, cover all scenarios) and mutually exclusive (i.e., no double-countings), hence:

\(? = 8*42 = 336\)

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik :: GMATH method creator (Math for the GMAT)

Our high-level "quant" preparation starts here: https://gmath.net