combinatorics - In how many ways can $8$ LEGO pieces to two types be arranged in one row?

Given the following $8$ LEGO bricks:

• $3$ pieces of length $6$


• $5$ pieces of the length $8$

In how many ways can they be ordered in one layer (one row)?

Total number of ways to order the bricks (permutations): $8!$

But we get some repeated permutations since we have duplicated elements, in e.g:

$S_1=\{6,6,6,8,8,8,8,8\}$

$S_2=\{6,6,6,8,8,8,8,8\}$

So how can I account for the repeated ones?

Update

Permutations, bricks with length $6$: $3!$

Permutations, bricks with length $8$: $5!$

Since their mutual order doesn't matter, we might remove them?

${\displaystyle \frac{8!}{3!\cdot5!}}$

Is this correct?

Answer

Quite Easy.

Just consider the fact that all pieces of same dimensions are exactly similar; and that you can't distinguish them.

If you want to arrange n objects in a row such that $ r_1 $are of 1 kind, $r_2 $ are of other kind,....and $r_n $ are of another similar kind, then you can arrange these n objects in $$ \frac {n!}{(r_1)! \cdot (r_2)! \cdot \cdot \cdot (r_n)!} $$

This is because, you want to avoid those cases when similar kind objects are inter-arranged; i.e. assume that your 3 pieces are named as p1,p2 and p3 and other 5 as q1,q2,q3,q4 and q5 (though all p's and q's are similar)
Now, if you just take 8!, p1,p2,p3,q1,q2,q3,q4,q5 and p1,p3,p2,q1,q2,q3,q4,q5 will be considered as different, and counted twice; although they are similar.
So, you avoid this over-counting by preventing inter-arrangements of p1,p2,p3 in 3! and of q1,q2,q3,q4,q5 in 5! ways, and divide 8! by 3! and 5!

So, here n=8, $r_1$=3 and $r_2$=5
$$ \frac {8!}{3! \cdot 5!} $$
is your answer.