SOLUTION: If repetition is allowed, how many three-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6? If repetition is not allowed, how many three-digits numbers can be formed

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Question 1134616: If repetition is allowed, how many three-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6?
If repetition is not allowed, how many three-digits numbers can be formed using the digitsn1, 2, 3, 4, 5, 6?
The notes aren’t making sense. Thank you.

Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
If repetition is allowed, how many three-digit numbers can be formed using the
digits 1, 2, 3, 4, 5, 6?
We can pick the first digit in any of 6 ways,

For each of the 6 ways we can pick the first digit,
we can pick the second digit any of the same 6 ways.
So there are 6×6 or 36 ways to pick the first two digits.
 
For each of the 6×6 or 36 ways we can pick the first two digits,
we can pick the third digit any of the same 6 ways.
So there are 6×6×6 or 216 ways to pick the three digits.

Answer: 216

If repetition is not allowed, how many three-digits numbers can be formed using
the digits 1, 2, 3, 4, 5, 6?
We can pick the first digit in any of 6 ways,

For each of the 6 ways we can pick the first digit,
we can only pick the second digit any of 5 ways, since we can't use the digit we
picked for the first digit.
So there are 6×5 or 30 ways to pick the first two digits.
 
For each of the 6×5 or 30 ways we can pick the first two digits,
we can only pick the third digit any of 4 ways, since we can't use either of the
two digits we picked for the first and second digits.
So there are 6×5×4 or 120 ways to pick the three digits.

Answer: 120

Edwin

Answer by ikleyn(52777)   (Show Source): You can put this solution on YOUR website!
.

            After having very detailed solution and explanation from Edwin,
            it will be,  probably,  useful for you to have the answer in the short and compact form.


1.  If repetition is allowed, the answer is  .


2.  If repetition is NOT allowed, the answer is  6*5*4.


These formulas are SELF-EXPLANATORY and EASY-TO-MEMORABLE.


---------------

To see many other similar solved problems, look into the lesson
    - Miscellaneous problems on permutations, combinations and other combinatoric entities
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.


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