It reminds me a logic which
leads to a small math-concept at the end.
Let us consider that some
persons are standing in a row. Some clues are given and we need to find the
number of persons.
Problem-1)Find the number of persons?
Clue:
"In the row, a person called 'A' is standing second from left and fourth
from right".
This clue is enough to answer the question. If we count from left,
A is second from
left=> there are two persons up to A
A is fourth from right
=> there are three persons after A (as it is four including A)
So total number of persons
= 2+3 = 5
It can be shown like
this:
* A * * *
Now let us go to a little complicated problem by introducing some more info.
Problem-2)Find the number of persons?
First Clue: “person 'A' is fourth from
left”
Second Clue: “person 'B' is fifth from
right”
Are these clues enough to
get the answer? Obviously not...
Third Clue: There is one person in between
A and B
Is it okay now? Can we solve the puzzle? Let us see...
"A is fourth from
left" => there are four persons up to A (including A) from left-end.
"B is fifth from
right" => there are five (including B) up to the right-end.
"There is one person
in between A and B" => This guy shall be added to the count
So total number of
persons = 4+5+1 = 10
It can be shown like
this:
* * * A * B * * * *
But is it okay?
No, there is some ambiguity
in the third clue. Ambiguity is about the order that A and B stand in the row. If we count from left, we are not sure whether A comes before B or the other-way-round. In the scenario shown above (let us call it scenario-I), A comes before B. If B comes before A, then another scenario (call it scenario-II) may arise as
shown below. Can't it satisfy all the three clues?
* B * A * *
Here again, "A is fourth from left" and "B is fifth from right" and "There is one person in between A and B".
And here there are only
six persons in total.
My point is: For the second problem, the
given three clues are not enough to answer the question straight away. This type of concept is
useful in "Data Sufficiency"
type of problems given in competitive exams.
Second problem takes me to "set theory" concept:
Scenario-I of second problem leads us to disjoint sets. One set with 4-persons (includes A) and other set with
5-persons (includes B) and there is one person not belonging to these two sets.
We can consider third set for this guy. These three sets are disjoint to one
another.
n(X) = 4
n(X) = 4
n(Y) = 5
n(Z) = 1
Total number of persons = n(XUYUZ) = n(X)+n(Y)+n(Z)
= 4+5+1 = 10
"n(XUYUZ) = n(X)+n(Y)+n(Z)" -this formula holds good for disjoint sets.
Scenario-II leads us to two intersecting
sets. One set with 4-persons (includes B as well) and
other set with 5-persons
(includes A as well) and there is an intersection of these two sets with 3-persons in
common (A, B and the other guy).
n(X) = 4
n(Y) = 5
n(X∩Y)
= 3
Total number of persons =
n(XUY) = n(X)+n(Y)-n(X∩Y) = 4+5-3 = 6"n(XUY) = n(X)+n(Y)-n(X∩Y)" -this formula holds good for intersecting sets.