perms and combs q help (1 Viewer)

[anonymoushehe]

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??

 

[Constantspy977]

View attachment 43959
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??

Use complementary events and find out the probability of NOT getting an E when choosing 5 letters. Whatever you get, just do 1 minus that. so it would be 5/8 for the first pick, 4/7 for the second, then 3/6, 2/5, 1/4 multiply all of these and you get 1/56, so the probability of getting at least 1 E is 55/56.