anonymoushehe
Active Member
![Screenshot 2024-08-25 at 4.13.34 pm.png](/data/attachments/43/43962-5aae514db38e16f33bb1d542cf9079da.jpg)
The answer was E, but I'm a bit confused.
what i did was calculate the number of combinations without E, so I got 5C5 = 1.
But i needed to determine the total number of distinct combinations but i thought that doing 8C5 = 56 would be wrong because wouldnt that assume that each letter is distinct even though we have three E's?? so we would end up repeating some of the combinations containing E more than once??