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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?
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1. 

they have exactly two digits that are 4's.

So choices for remaining digits will be 9 only. (Exclude 4).

2. You are selecting 2 places for 4's so remaining one place will automatically contain the last digit. Now why are you permuting them again?

Solution - $9*1*1* \frac{3!}{2!}=27$

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