The number of ways we can choose three different numbers in order for a unique code for this case is 2730
We can use combinations for this case,
Total number of distinguishable things is m.
Out of those m things, k things are to be chosen such that their order doesn't matter.
This can be done in total of [tex]^mC_k = \dfrac{m!}{k! \times (m-k)!}[/tex] ways.
If the order matters, then each of those choice of k distinct items would be permuted k! times.
So, total number of choices in that case would be:
[tex]^mP_k = k! \times ^mC_k = k! \times \dfrac{m!}{k! \times (m-k)!} = \dfrac{m!}{ (m-k)!}\\\\^mP_k = \dfrac{m!}{ (m-k)!}[/tex]
This is called permutation of k items chosen out of m items (all distinct).
Given that:
In a lock, the order of the way numbers are arranged matters.
Because of no repetition allowed, we can simulate this situation as:
The number of ways we can choose 3 numbers out of 15 numbers with order = The number of different ways we can choose three different numbers in order for a unique code.
And we have:
The number of ways we can choose 3 numbers out of 15 numbers with order = [tex]^{15}P_3 = \dfrac{15!}{(15-3)!} = \dfrac{15 \times 14 \times 13 \times 12!}{12!} = 2730[/tex]
Thus, the number of ways we can choose three different numbers in order for a unique code for this case is 2730
Learn more about combinations and permutations here:
https://brainly.com/question/16107928
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