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How many 3 digit numbers can be formed from the digits 1 2 3 4 5 6 7 8 9 if each digit is used only once?
Wiki User
∙ 9y ago
Multiply the number of possible starting numbers by the number of possible middle numbers by the number of possible end numbers to get your result.
In example, The possible starting numbers are 1, 2, 3, 4, 5, 6, 7, 8, and 9. Let's say I picked nine as the starting number. Since your question states that each number can only be used once, we eliminate nine from the selection of middle and end numbers. Now, the choices for the possible middle and end numbers are 1, 2, 3, 4, 5, 6, 7, and 8.
Possible starting numbers= 9
Possible middle numbers= 8
Multiply 9 by 8. You get 72 different choices for a two digit number.
Let's say that the middle number I picked was two. We then remove the number two from the possible choices for the final numbeselections: 1, 3, 4, 5, 6, 7, and 8.
Possible outcomes for two digit number= 72 (which is 9 times 8)
Possible end numbers = 7
Multiply 72 by 7 to get the possible outcomes for a three digit number with each digit used only once.
72 times 7 = 504. You have 504 possible outcomes.
Wiki User
∙ 9y ago
Wiki User
∙ 16y agoA33 =3x2x1=6 234 243 324 342 432 423
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