THE GLASS BALLS.
(
Combination and Group Problems)
A number of clever marksmen were staying at a country house, and the
host, to provide a little amusement, suspended strings of glass balls,
as shown in the illustration, to be fired at. After they had all put
their skill to a sufficient test, somebody asked the following question:
"What is the total number of different ways in which these sixteen balls
may be broken, if we must always break the lowest ball that remains on
any string?" Thus, one way would be to break all the four balls on each
string in succession, taking the strings from left to right. Another
would be to break all the fourth balls on the four strings first, then
break the three remaining on the first string, then take the balls on
the three other strings alternately from right to left, and so on. There
is such a vast number of different ways (since every little variation of
order makes a different way) that one is apt to be at first impressed by
the great difficulty of the problem. Yet it is really quite simple when
once you have hit on the proper method of attacking it. How many
different ways are there?
Answer:
There are, in all, sixteen balls to be broken, or sixteen places in the
order of breaking. Call the four strings A, B, C, and D--order is here
of no importance. The breaking of the balls on A may occupy any 4 out of
these 16 places--that is, the combinations of 16 things, taken 4
together, will be
13 x 14 x 15 x 16
----------------- = 1,820
1 x 2 x 3 x 4
ways for A. In every one of these cases B may occupy any 4 out of the
remaining 12 places, making
9 x 10 x 11 x 12
----------------- = 495
1 x 2 x 3 x 4
ways. Thus 1,820 x 495 = 900,900 different placings are open to A and B.
But for every one of these cases C may occupy
5 x 6 x 7 x 8
------------- = 70
1 x 2 x 3 x 4
different places; so that 900,900 x 70 = 63,063,000 different placings
are open to A, B, and C. In every one of these cases, D has no choice
but to take the four places that remain. Therefore the correct answer is
that the balls may be broken in 63,063,000 different ways under the
conditions. Readers should compare this problem with No. 345, "The Two
Pawns," which they will then know how to solve for cases where there are
three, four, or more pawns on the board.
