In the equation A + B + C + D + E = FG, where FG is the two digit number whose value is 10F + G and letters, A, B, C, D, E, F and G each represents different digits and FG is as large as possible.
If a five digit number is made using the digits A, B, C, D, E without repetition then
 

9, 8, 6, 5, 4
(A and B) ……. + 9 + 8 + 6 + 5 + 4 = 32
$|(9+8)-(6+5+4)|=2$  not divisible by 11
$|(9+6)-(8+5+4)|=3$  not divisible by 11
$|(9+5)-(8+6+4)|=4$  not divisible by 11
 $|(9+4)-(8+6+5)|=6$ not divisible by 11
$|(8+6)-(9+5+4)|=4$  not divisible by 11
 $|(8+5)-(9+6+4)|=6$ not divisible by 11
 $|(8+4)-(9+6+5)|=8$ not divisible by 11
 $|(6+5)-(9+8+4)|=10$ not divisible by 11
$|(6+4)-(9+8+5)|=12$  not divisible by 11
$|(5+4)-(9+8+6)|=14$  not divisible by 11
(C) ……..  $$\begin{gathered} \underset{\_}{4}\underset{\_}{8} \\ \underset{\_}{5}\underset{\_}{6} \\ \underset{\_}{6}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8} \\ \underset{\_}{8}\underset{\_}{4} \\ \underset{\_}{9}\underset{\_}{6} \\ \text{Probability = }\frac{6\times 3!}{5!}=\frac{6}{20}=\frac{3}{10} \end{gathered}$$