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Re: How many five-digit numbers can be formed using the digits 5
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07 Aug 2018, 06:13
Explanation
This problem relies on the fundamental counting principle, which says that the total number of ways for something to happen is the product of the number of options for each individual choice.
The problem asks how many five-digit numbers can be created from the digits 5, 6, 7, 8, 9, and 0. For the first digit, there are only five options (5, 6, 7, 8, and 9) because a five-digit number must start with a non-zero integer.
For the second digit, there are 5 choices again, because now zero can be used but one of the other numbers has already been used, and numbers cannot be repeated. For the third
number, there are 4 choices, for the fourth number there are 3 choices, and for the fifth number there are 2 choices. Thus, the total number of choices is (5)(5)(4)(3)(2) = 600.
Alternatively, use the same logic and realize there are 5 choices for the first digit. (Separate out the first step because you have to remove the zero from consideration.) The remaining five digits all have an equal chance of being chosen, so choose four out of the remaining five digits to complete the number. The number of ways in which this second step can be accomplished is \(\frac{5!}{1!}= (5)(4)(3)(2)\).
Thus, the total number of choices is again equal to (5)(5)(4)(3)(2) = 600.