In how many ways can 3 distinct numbers fill 6 blanks (each repeated exactly twice)?

Question

I have a basic knowledge of permutations and combinations but using different approaches to this problem lead me to different answers. Basically, there are 6 blanks and 3 distinct digits a,b, and c. All three of them occupy 2 blanks. For example, some possible combinations are:

1. a a b b c c
2. a b c a b c
3. a b c c b a

Please mention the approach used and why the approach is logical and the others are not.

Edit: My Approaches:-

1. We can first fill the 2 a's in any of the 6 blanks using 6C2 (as both are same). Then we can fill the 2 b's in any of the the remaining 4 blanks using 4C2. Then we can multiply them (there is no need to fill up c's as 2C2 is 1). The answer comes out to be 90.
2. We can take the a's ,b's and c's as a1, a2, b1, b2, c1, c2 and find the number of permutations. That comes out to be 6P6 = 720. Since we a1 and a2 are same(similarly b1 and b2), we divide by 2 to get 360 (avoiding repetition).

Edit2: Upon seeing the comments, I realized that in approach2:

1. a1 a2 b1 b2 c1 c2
2. a2 a1 b1 b2 c1 c2
3. a1 a2 b2 b1 c1 c2
4. a2 a1 b2 b1 c1 c2
5. a1 a2 b1 b2 c2 c1
6. a2 a1 b1 b2 c2 c1
7. a2 a1 b2 b1 c2 c1
8. a1 a2 b2 b1 c2 c1

are same. So, I should divide by 8 to get 90. So, is 90 the correct answer?

Answer 1

Your second approach is what needs to be corrected. First approach is correct.

Imagine that you have two systems that you put the variables: $$ a_1,a_2,b_1,b_2,c_1,c_2 $$ through:

1. Permute all six variables $\implies 6!$
2. Flip a coin for each pair $x_1,x_2$ (independently) deciding whether to swap $x_1,x_2$ or not $\implies 2\cdot 2\cdot 2$

So the answer is: $$ \frac{6!}{2\cdot2\cdot2} $$ which gives the same answer you get by using combinations: $$ 6C2\cdot 4C2\cdot 2C2 $$