Logic Puzzle Answer

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Photography Class - Answer

The pairs who were successful getting pictures were:

Perry and Quinn photographed the anteater.

Perry and Toby photographed the bear.

Rex and Scott photographed the crocodile.

Scott and Toby photographed the deer.

Quinn and Rex photographed the emu.

Solution

There are ten possible pairs. Each person went out four times, each time with a different partner. Each person got a photograph twice and got nothing twice.

First, to determine which pairs photographed an animal and which didn't, make a 5x5 matrix with row and column headings of Perry, Quinn, Rex, Scott and Toby. Since we don't always know what animal was photographed, assign the animals a number, 1 through 5.

Everybody snapped two animals. So you can assign animal 2 to P [(P, RST) = 2] and you can assign animal 3 to Q [(Q, RST) = 3] in the matrix.

Then you are left with assigning animal 4 to S and animal 5 to T [(R, S) = 4 and (S, T) = 5.]

Since S has both animals 4 and 5, S cannot have animals 2 and 3. That means that we can assign 2 to R and 3 to T.

The pairs who were successful were (P, Q), (P, T), (Q, R), (R, S) and (S, T).

Next, we need to determine which pair photographed which animal. We know that

• The Emu is animal 1 or 3.
• The Anteater is not 2 and 5.
• The Bear is not 4 and 5. Therefore, the bear is 2 or 3.

Consider these cases separately: (Emu, Bear) = (1, 2), (1, 3) and (3, 2). The first two cases lead to contradictions. Therefore, (Emu, Bear) = (3, 2) which leads to Crocodile = 4.

Then, since Toby did not photograph the Anteater, the Anteater cannot be 2 or 5. So the Anteater must be 1.

Then the Deer is 5.

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