Need help on probability question?

With 4 items we have 4! ways of arranging them

i.e. 4 x 3 x 2 x 1 = 24

However, because the digit 8 is repeated we have twice the amount

so, 8₁348₂ is the same number as 8₂348₁

The 8 can be positioned in 2! = 2 ways

Hence, 4!/2! => 24/2 = 12

Note: if we had 8 appearing three times we would divide by 3!, because they can be arranged in 3 x 2 x 1 ways

We would then have 4!/3! => 24/6 = 4

Extending to say 23256676, we have 2 appearing twice and 6 appearing three times so we'd have:

8!/2!3! => 3360 ways

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Actually, you can use formula.

For permutation in a line of n objects in which r objects are identical.
The formula is:
Number of permutations = ₙPₙ/r!

In the question, n = 4 and r = 2.
Hence, number of four-digit numbers = ₄P₄/2! = 4!/2! = 4 × 3 = 12

If there is no repeating numbers it's just n!
1

12, 21

123, 132, 213, 231, 312, 321

1234, 1324, 2134, 2314, 3124, 3214, 1243, 1342, 2143, 2341, 3142, 3241, 1423, 1432, 2413, 2431, 3412, 3421, 4123, 4132, 4213, 4231, 4312, 4321

Now if you replace all the 2s with 1s you will notice that half as many non duplicate solutions.