Henry Ernest Dudeney/Puzzles and Curious Problems/284 - Lamp Signalling/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $284$

Lamp Signalling
Two spies on the opposite sides of a river devised a method for signalling by night.
They each put up a stand, like the diagram, and each had three lamps which could show either white, red or green.
Dudeney-Puzzles-and-Curious-Problems-284.png
They constructed a code in which every different signal meant a sentence.
Note that a single lamp on any one of the hooks could only mean the same thing,
that two lamps hung on the upper hooks $1$ and $2$ could not be distinguished from two on, for example, $4$ and $5$.
However, two red lamps on $1$ and $5$ could be distinguished from two on $1$ and $6$,
and two on $1$ and $2$ from two on $1$ and $3$.
Remembering the variations of colour as well as of position, what is the greatest number of signals that could be sent?


Solution

$471$.


Proof

With $3$ lamps all of one colour, we can make $15$ distinct messages for each colour.

Because there are $3$ colours, that makes $3 \times 15 = 45$ messages.


With $2$ of one colour and $1$ of another, we can make the same number: $15$ distinct messages.

Each of these admits $3$ variations of colour order, making $45$.

There are $6$ ways of choosing $1$ lamp of one colour and $2$ of another colour.

Hence there are $6 \times 45 = 270$ messages made with $3$ lamps of $2$ colours.


With $1$ lamp of each colour we can make $6 \times 15 = 90$ messages.


With $2$ lamps of one colour we can make $7$ messages.

Because there are $3$ colours, that makes $3 \times 7 = 21$ messages.


With $1$ lamp of one colour and $1$ lamp of another colour, we can get $14$ messages.

There are $6$ ways of choosing $1$ lamp of one colour and $1$ lamp of another colour.

Hence there are $6 \times 14 = 42$ messages made with $2$ lamps of $2$ colours.


With $1$ lamp only, you can make only $1$ message.

As there are $3$ colours, that makes $3$ more messages.


Hence the total:

$45 + 270 + 90 + 21 + 42 + 3 = 471$

$\blacksquare$


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