GMATBLACKBELT wrote:

I get 5 as well.

7!/4!3! = 35 number of ways w/ no constraints.

Now we have VWXYZAB

There are 5!/4!1! ways to arrange VWXYZ into groups of 4.

So its just 5.

Now can someone show me a method of getting the number of ways w/ A and B using combinations??? I cant figure it out. (Not using the above method to find out that its 30).

Ben and Ann are among 7 contestants from which 4 semifinalists are to be selected. Of the different possible selections, how many contain

both Ben and Ann?

if B and A are in it, then we only have to choose 2 of 5 people.

1*1*5C2 / 7c4

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