(1) Tabulating the problem to see it more clearly.
For a finite sequence of non zero numbers,
the
number of variations in sign is defined as the
number of pairs of consecutive terms of the sequence for which the
product of the two consecutive terms is negative.
What is the
number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?
(2) Converting to a more compact version.
For a finite sequence of non zero numbers,
V =
number of pairs of consecutive terms in {1, -3, 2, 5, -4, -6} when
product of the two consecutive terms is negative.
What is
V=?
The effort is spent in figuring out what V is. Do not get confused with
consecutive terms and
consecutive integers. Clearly the author of the problem wants you to confuse the concept of consecutive integers and terms, however, resist the temptation.
In Set S = {a, b, c}, the terms a and b are consecutive terms, as well as b and c. However, a and c are not consecutive terms.
As such, consecutive terms for {1, -3, 2, 5, -4, -6} are:
{
1, -3, 2, 5, -4, -6} = 1* -3 = -3 (negative)
{1,
-3, 2, 5, -4, -6} = -3 * 2 = -6 (negative)
{1, -3,
2, 5, -4, -6} = 2* 5 = 10 (positive)
{1, -3, 2,
5, -4, -6} = 5 * -4 = -20 (negative)
{1, -3, 2, 5,
-4, -6} = -4 * -6 = 24 (positive)
Note that you do not even need to multiply the numbers, however, you need to realize what happens when you multiply a negative times a positive or vice versa.
As a result of the analysis above, you can conclude that you would have three (3) negative pairs.