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Hence for we get paths, for we get
paths, and for
paths. This gives us
hence the total number of paths is
we get
paths through the second quadrant,
.
Combinatorial Solution 2
Each path that goes through the second quadrant must pass through exactly one of the points
,
, and
.
There is exactly path of the first kind, paths of the second kind, and
paths of the third type. The conclusion remains the same.
Problem 19
Each of 2010 boxes in a line contains a single red marble, and for box in the
, the
position also contains white marbles. Isabella begins at the first box
be the probability that Isabella
and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let
stops after drawing exactly marbles. What is the smallest value of for which
?
Solution
The probability of drawing a white marble from box is drawing a red marble from box is
.
. The probability of
The probability of drawing a red marble at box is therefore
It is then easy to see that the lowest integer value of that satisfies the inequality is
.
Problem 20
Arithmetic sequences value of ?
and and
have integer terms with
for some . What is the largest possible
Solution Solution 1
Since
and
have integer terms with
, we can write the terms of
each sequence as
where and ( Since
) are the common differences of each, respectively.
it is easy to see tha