# The number lock of a suitcase has 4 wheels each labelled with ten digits i.e. From $0$ to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Last updated date: 28th Mar 2023

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Answer

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Hint: In order to solve such type of question firstly there are number of wheels in number lock of suitcase$ = 4$.As per given statement there is no repetition of digit between From $0$ to $9$ i.e. first digit can be between $0$ to $9$.Thus no of digit $ = 10$.

Complete step-by-step answer:

Since, we have given that repetition is not allowed.

Now, the first wheel can have any one of the tens digits from $0$ to $9$. So, a wheel can have any of the $10$ digits.

Since, repetition is not allowed, so the second wheel can have any of the remaining $9$ digits.

Similarly, the third wheel can have any of the remaining 8 digits.

And the fourth wheel can have any of the remaining $7$ digits.

So, number of four digit lock code that can be formed without repetition of digit$ = 10 \times 9 \times 8 \times 7 = 5040$

So, a total four digit number formed$ = 5040$ .

But since, the lock can open with only one of the all four-digit numbers.

Hence, required probability$ = \dfrac{1}{{5040}}$.

Note: Whenever we face such a type of question the key concept is that. Since there is no repetition one digit is used cannot be repeated again and as per there are a number of wheels in the number lock of suitcase$ = 4$ there are just 4 digits and cannot be exceeded.

Complete step-by-step answer:

Since, we have given that repetition is not allowed.

Now, the first wheel can have any one of the tens digits from $0$ to $9$. So, a wheel can have any of the $10$ digits.

Since, repetition is not allowed, so the second wheel can have any of the remaining $9$ digits.

Similarly, the third wheel can have any of the remaining 8 digits.

And the fourth wheel can have any of the remaining $7$ digits.

So, number of four digit lock code that can be formed without repetition of digit$ = 10 \times 9 \times 8 \times 7 = 5040$

So, a total four digit number formed$ = 5040$ .

But since, the lock can open with only one of the all four-digit numbers.

Hence, required probability$ = \dfrac{1}{{5040}}$.

Note: Whenever we face such a type of question the key concept is that. Since there is no repetition one digit is used cannot be repeated again and as per there are a number of wheels in the number lock of suitcase$ = 4$ there are just 4 digits and cannot be exceeded.

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