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

combinatoricspermutations

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?

Best Answer

We have eight positions to fill with three pieces of length six and five pieces of length eight.

Notice that the sequence is completely determined by choosing which three of these eight positions are filled with pieces of length six. In how many ways can you choose three of the eight positions to fill with pieces of length six?

$$\binom{8}{3}$$

I suggest that you verify that this works by listing the distinguishable sequences for a simpler problem such as three pieces of length six and two pieces of length eight.

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