How many strings of three decimal digits can be formed such that they have exactly two digits that are 4's.
My approach as to select 2 positions for these 4's in $\binom{3}{2}$ ways and the last bit will have 10 choices.Now I can permute the string formed in $\frac{3!}{2!}=3$ ways.
So total such strings should be $\binom{3}{2}$ * 10*$\frac{3!}{2!}$ = 90.
But the answer is 27.
How?