Answer:
The least possible result is <em>-10</em>.
Step-by-step explanation:
Given the numbers 4, 5 and 6 are to be chosen one of the letters A, B or C.
First of all,
Let A = 4, B = 5 and C = 6
![A(B-C) = 4 \times (5-6) = 4 \times -1 = -4](https://tex.z-dn.net/?f=A%28B-C%29%20%3D%204%20%5Ctimes%20%285-6%29%20%3D%204%20%5Ctimes%20-1%20%3D%20-4)
Let A = 4, B = 6 and C = 5
![A(B-C) = 4 \times (6-5) = 4 \times 1 = 4](https://tex.z-dn.net/?f=A%28B-C%29%20%3D%204%20%5Ctimes%20%286-5%29%20%3D%204%20%5Ctimes%201%20%3D%204)
Let A = 5, B =4 and C = 6
![A(B-C) = 5 \times (4-6) = 5 \times -2 = -10](https://tex.z-dn.net/?f=A%28B-C%29%20%3D%205%20%5Ctimes%20%284-6%29%20%3D%205%20%5Ctimes%20-2%20%3D%20-10)
Let A = 5, B = 6 and C = 4
![A(B-C) = 4 \times (6-4) = 4 \times 2 = 8](https://tex.z-dn.net/?f=A%28B-C%29%20%3D%204%20%5Ctimes%20%286-4%29%20%3D%204%20%5Ctimes%202%20%3D%208)
Let A = 6, B = 4 and C = 5
![A(B-C) = 6 \times (4-5) = 6 \times -1 = -6](https://tex.z-dn.net/?f=A%28B-C%29%20%3D%206%20%5Ctimes%20%284-5%29%20%3D%206%20%5Ctimes%20-1%20%3D%20-6)
Let A = 6, B = 5 and C = 4
![A(B-C) = 6 \times (6-5) = 6 \times 1 = 6](https://tex.z-dn.net/?f=A%28B-C%29%20%3D%206%20%5Ctimes%20%286-5%29%20%3D%206%20%5Ctimes%201%20%3D%206)
Summarizing the above values in the form of a table:
![\begin{center}\begin{tabular}{ c c c c}A & B & C & A(B-C)\\ 4 & 5 & 6 & -4\\ 4 & 6 & 5 & 4\\ 5 & 4 & 6 & -10\\ 5 & 6 & 4 & 10\\ 6 & 4 & 5 & -6\\ 6 & 5 & 4 & 6\end{tabular}\end{center}](https://tex.z-dn.net/?f=%5Cbegin%7Bcenter%7D%5Cbegin%7Btabular%7D%7B%20c%20c%20c%20c%7DA%20%26%20B%20%26%20C%20%26%20A%28B-C%29%5C%5C%204%20%26%205%20%26%206%20%26%20-4%5C%5C%20%204%20%26%206%20%26%205%20%26%204%5C%5C%20%205%20%26%204%20%26%206%20%26%20-10%5C%5C%20%205%20%26%206%20%26%204%20%26%2010%5C%5C%20%206%20%26%204%20%26%205%20%26%20-6%5C%5C%206%20%26%205%20%26%204%20%26%206%5Cend%7Btabular%7D%5Cend%7Bcenter%7D)
So, the least possible result is <em>-10</em>.