Permutations with Repetition Allowed

Let $$E$$ denote the condition that the number of $$X$$'s is even, and $$O$$

denote the condition that the number of $$Y$$'s is odd.

Each of the $$7$$ positions can be filled with $$X,Y,Z$$

so the total number of strings is $$3^7$$

We use the parity filter.

The required count is

$$N=\frac14\sum_{x=\pm1}\sum_{y=\pm1}(1+x)(1-y)(x+y+1)^7$$

Evaluating the four cases:

For

$$x=1,\ y=1$$

$$(1+x)(1-y)(x+y+1)^7=0$$

For

$$x=1,\ y=-1$$

$$(1+x)(1-y)(x+y+1)^7$$

$$=(2)(2)(1)^7$$

$$=4$$

For

$$x=-1,\ y=1$$

the contribution is $$0$$

For

$$x=-1,\ y=-1$$

the contribution is $$0$$

Hence

$$N=\frac14\left(3^7+1\right)$$

$$=\frac14(2187+1)$$

$$=\frac{2188}{4}$$

$$=547$$

Therefore the number of passcodes is

$$547$$

Final Answer :

$$547$$