Anuj Gupta
2

The lock has 5 digits, and there are 10 digits to choose from. The number of ways the codes can be formed with all digits being different is in ^{10}P_{5} ways.

Using the permutation formula which is ^{n}P_{r} = \frac{n!}{(n-r)!}

^{10}P_{5} = \frac{10!}{(10-5)!} = \frac{10!}{5!} = 7 \cdot 8 \cdot 9 \cdot 10= 30,240 ways.

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How many $4$-digit lock combinations are possible if each digit in ...

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I found an answer from www.quora.com

A lock contains 5 digits in its code. How many codes can be formed ...

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