Team needs exactly 4 executives and 3 associates
To find the number of teams that can be formed, we need the number of executives and associates
Statement 2nC3 = 10
n!/(n-3)!*3! = 10
n(n-1)(n-2) = 60 = 5*4*3
n = 5
This means number of associates, A is 5-2 = 3
But we don't the number of executives
Insufficient
Eliminates B,D
Statement 1
E-2 = A
But we don't know the value of E or A
Insufficient
Eliminates A
Combined
A = 3
E-2 = 3 => 5
We can now calculate the number of teams using 5C4 *3C3
Suffcient
Answer C
Bunuel
A team is composed of exactly 4 executives and 3 associates. How many such teams can be formed?
(1) If two fewer executives were available for selection there would be an equal number of executives and associates available for selection.
(2) If two more associates were available for selection exactly 10 different teams of 3 associates could be selected.