# 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

## (6) Comment(s)

Pritish
()

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

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

then the number thus formed, will be 354 or 534 or 594 or 954 nut none of them is multiple of 4, as abc should be multiple of 4

and same goes for the c=8

Saikiran
()

a, b, c are distinct.. it was never mentioned they are 3 digit number.. so think in this way, the solution given will be correct :)

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 $a,b=1,7$