Kenneth Rosen Edition 6th Exercise 5.1 Question 23c (Page No. 345)

Question

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?

Answer 1

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 - $9*1*1* \frac{3!}{2!}=27$

Comments

you mean to say there is no need to do 3C2 when we are considering all possible permutations of such string that contain exactly two 4's.?

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Yes exactly.
Either combination or permutation is required. Both will work same in this case.

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Everytime we are choosing positions for both 4's we are omitting the combination with an extra 4. (i know i'm being captain obvious).