In the above sum there is no need to consider the red one as the four 1's are already mapped by blue and green group. So the answer will be:
$$F = C{\overline D}{\overline E}+{\overline A}B{\overline C}E+{\overline A}BD{\overline E}$$
In the above sum there is no need to consider the red one as the four 1's are already mapped by blue and green group. So the answer will be:
$$F = C{\overline D}{\overline E}+{\overline A}B{\overline C}E+{\overline A}BD{\overline E}$$