THE BARRELS OF BALSAM.
(
Combination and Group Problems)
A merchant of Bagdad had ten barrels of precious balsam for sale. They
were numbered, and were arranged in two rows, one on top of the other,
as shown in the picture. The smaller the number on the barrel, the
greater was its value. So that the best quality was numbered "1" and the
worst numbered "10," and all the other numbers of graduating values.
Now, the rule of Ahmed Assan, the merchant, was that he never put a
barrel either beneath or to the right of one of less value. The
arrangement shown is, of course, the simplest way of complying with this
condition. But there are many other ways--such, for example, as this:--
1 2 5 7 8
3 4 6 9 10
Here, again, no barrel has a smaller number than itself on its right or
beneath it. The puzzle is to discover in how many different ways the
merchant of Bagdad might have arranged his barrels in the two rows
without breaking his rule. Can you count the number of ways?
Answer:
This is quite easy to solve for any number of barrels--if you know how.
This is the way to do it. There are five barrels in each row Multiply
the numbers 1, 2, 3, 4, 5 together; and also multiply 6, 7, 8, 9, 10
together. Divide one result by the other, and we get the number of
different combinations or selections of ten things taken five at a time.
This is here 252. Now, if we divide this by 6 (1 more than the number in
the row) we get 42, which is the correct answer to the puzzle, for there
are 42 different ways of arranging the barrels. Try this method of
solution in the case of six barrels, three in each row, and you will
find the answer is 5 ways. If you check this by trial, you will discover
the five arrangements with 123, 124, 125, 134, 135 respectively in the
top row, and you will find no others.
The general solution to the problem is, in fact, this:
n
C
2n
-----
n + 1
where 2n equals the number of barrels. The symbol C, of course, implies
that we have to find how many combinations, or selections, we can make
of 2n things, taken n at a time.
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