Henry Ernest Dudeney/Puzzles and Curious Problems/26 - Cross and Coins/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $26$

Cross and Coins

Take any $11$ of the $12$ current coins of the realm,

and using one duplicate coin, can you place the $12$ coins, one in each division of the cross,

so that they add up to the same value in the upright and in the horizontal?

Solution

The solution given by Dudeney is:

who then goes on to explain:

The illustration shows a solution.

Only one coin -- the farthing -- is repeated, and in both directions the sum is $\pounds 1, 2 \shillings 7 \tfrac 1 2 \oldpence$

Readers may possibly ask how many different arrangements are possible, all correctly adding up to this amount.

Well, the penny must, in every solution, remain in its present position, but the seven coins in the upright may be permuted in $2520$ ways.

(the two farthings being regarded as indistinguishable),

and the four coins in the horizontal may be permuted in $24$ ways.

Therefore, $2520 \times 24 = 60 \, 480$ -- the number of different ways in which the coins may be arranged.

But we can exchange the three coins $2 \shillings$, $6 \oldpence$ and $\tfrac 1 2 \oldpence$ in the horizontal with the $2 \shillings 6 \oldpence$, $\tfrac 1 4 \oldpence$ and $\tfrac 1 4 \oldpence$ in the upright.

This new arrangement will give $5040 \times 12 = 60 \, 480$ solutions also.

so the total number of ways is twice $60 \, 480$ or $120 \, 960$.

Sources

• 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $26$. -- Cross and Coins