# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. $a, b, c$ are three distinct integers from 2 to 10 (both inclusive). Exactly one of $ab, bc$ and $ca$ is odd. $abc$ is a multiple of 4. The arithmetic mean of $a$ and $b$ is an integer and so is the arithmetic mean of $a, b$ and $c$. How many such triplets are possible (unordered triplets)
 ✔ A. $4$ ✖ B. $5$ ✖ C. $6$ ✖ D. $7$

Solution:
Option(A) is correct

Exactly one of $ab, bc$ and $ca$ is odd

⇒ Two are odd and one is even

$abc$ is a multiple of 4

⇒ the even number is a multiple of 4

The arithmetic mean of $a$ and $b$ is an integer

⇒ $a$ and $b$ are odd

and so is the arithmetic mean of $a, b$ and $c$.

⇒ $a+ b + c$ is a multiple of 3

$c$ can be 4 or 8.

$c = 4; a, b$ can be $3, 5$ or $5, 9$

$c = 8; a, b$ can be $3, 7$ or $7, 9$

Four triplets are possible

## (4) Comment(s)

Shubham Kumar
()

If both a and b are odd then $ab$ and $ca$ are odd but given is that one of $ab$, $bc$ and $ca$ are odd. How is this possible?

Manohar Tangi
()

5th way,c=8,and a,b=7,5

Awai
()

The Correct Ans is (B) 5

Ravi
()

one more possibility is $c=4$ and $a,b=1,7$