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Thanks in advance for any response!!!

why is the below method not correct?
A _ _ , A has to be one of the letters. For the remaining 2 position it will be --> 4c1*3c1=12.

Since, the correct answer is 6, the possible groups are:
ABC, ABD, ABE, ACD, ACE, ADE.
Each group can again be arranged in 3! ways. So, What would the approach/solution be if the question asked for number of arrangements?
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Itisallinurhead
Thanks in advance for any response!!!

why is the below method not correct?
A _ _ , A has to be one of the letters. For the remaining 2 position it will be --> 4c1*3c1=12.

Since, the correct answer is 6, the possible groups are:
ABC, ABD, ABE, ACD, ACE, ADE.
Each group can again be arranged in 3! ways. So, What would the approach/solution be if the question asked for number of arrangements?

Hi Itisallinurhead

Here for the question being asked, in the 12 options possible when
you use 4c1*3c1 - AED and ADE are both possible(but since we are
asked to find the total possible groups) it would be wrong!

If the question read -
"total number of arrangements possible if A is one of the letters "
One of the letters will be A.
The other two letter can be chosen as 4c2 = 6
There are 3!(6) ways of arranging them.

Total arrangments possible are 1*6*6 = 36

Hope this helps you!
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Bunuel
How many different groups of three distinct letters are possible if the three letters are chosen from the letters A, B, C, D, and E and A must be one of the letters selected?

A. 60
B. 10
C. 6
D. 4
E. 3

First, we know this will be a combination problem, since the order of selection doesn’t matter.

Since A must be selected, we are actually choosing only 2 letters from the 4 letters: B,C,D,E. Thus, the selection can be made in 4C2 = (4 x 3)/2 = 6 ways.

Answer: C
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