John wants to form 5 digit numbers from the set of integers from 0 to 9 inclusive. Assuming he can reuse numbers, and the 5 digit number must start with and 8 and end with a 0, how many different numbers may he form?

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John wants to form 5 digit numbers from the set of integers from 0 to 9 inclusive. Assuming he can reuse numbers, and the 5 digit number must start with and 8 and end with a 0, how many different numbers may he form?

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Answer 1 · 2017-04-01T01:28:26

$1000$ $1000$ possibilities

Explanation:

The first number must be an $8$ $8$ , so there is only $1$ $1$ possibility for the first digit.
The second number can be anywhere between $0$ $0$ to $9$ $9$ , so there are $10$ $10$ possibilities for the second digit.
The third number can be anywhere between $0$ $0$ to $9$ $9$ , so there are $10$ $10$ possibilities for the second digit.
The fourth number can be anywhere between $0$ $0$ to $9$ $9$ , so there are $10$ $10$ possibilities for the second digit.
The fifth number must be a $0$ $0$ , so there is only $1$ $1$ possibility for the last digit.

We multiply these possibilities to get $1 \cdot 10 \cdot 10 \cdot 10 \cdot 1 = 1000$ $1*10*10*10*1=1000$ possibilities.

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John wants to form 5 digit numbers from the set of integers from 0 to 9 inclusive. Assuming he can reuse numbers, and the 5 digit number must start with and 8 and end with a 0, how many different numbers may he form?

1 Answer

Explanation:

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