How many 3 digit numbers can be formed using 1, 2, 3, 4,5 without repetition

Complete step by step solution:
Case [1]: Repetition not allowed:
The number of ways in which the ones place can be filled = 5 ways.
The number of ways in which the tens place can be filled = 4 ways (because repetition is not allowed. So, the choice of one place cannot be used).
 The number of ways in which hundreds place can be filled = 3 ways.
Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 4\times 3=60$
Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 60 i.e 5! ways
Case [2]: Repetition is allowed:
The number of ways in which the ones place can be filled = 5 ways.
The number of ways in which the tens place can be filled = 5 ways (because repetition is allowed. So, the choice of one's place can be used).
 The number of ways in which hundreds place can be filled = 5 ways.
Hence according to the fundamental principle of counting the number of ways in which the three places can be filled to form a three-digit number = $5\times 5\times 5=125$
Hence the number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 = 125

Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = ${}^{5}{{P}_{3}}=\dfrac{5!}{\left( 5-3 \right)!}=\dfrac{5!}{2!}=\dfrac{120}{2}=60$ which is the same result as above.

I solved it this way because if a number needs to be formed with a certain number of digits (with no restrictions - which I think 'as often as desired' means), you assign a scale to the given constituent digits and then pick the same number of digits as the new number needs, going from highest to lowest. Then multiply these digits by each other to give the amount of different numbers that could be created.

However my answer of 60 possible numbers conflicts with the textbook's answer of 64.

The next part of the question was finding how many 3-digit numbers can be formed using 2, 3, 4 and 5 using at most one each. I was able to get this question, by changing 2, 3, 4 and 5 to 1, 2, 3 and 4; then multiplying 4 by 3 by 2 to give 24 possibilities.

In mathematics, permutation is known as the process of arranging a set in which all the members of a set are arranged into some series or order. The process of permuting is known as the rearranging of its components if the set is already arranged. Permutations take place, in more or less important ways, in almost every area of mathematics. They frequently appear when different commands on certain finite sets are considered.

What is a Combination?

A combination is an act of choosing items from a group, such that (not like permutation) the order of choice does not matter. In smaller cases, it is possible to count the number of combinations. Combination refers to the union of n things taken k at a time without repetition. In combination, you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Permutation Formula

In permutation r things are selected from a set of n things without any replacement. In this order of selection matter.

nPr = (n!) / (n-r)!

Here,

n = set size, the total number of items in the set

r = subset size , the number of items to be selected from the set

Combination Formula

In combination r things are selected from a set of n things and where the order of selection does not matter.

nCr = n!/(n−r)!r!

Here, 

n = Number of items in set

r = Number of items selected from the set

How many 3-digit even numbers can be formed by using the digits 1,2,3,4, and 5?

Solution:

If repetition is allowed  

A three digit even number is to be formed from given 5 digits 1,2,3,4,5.

Ones place can be filled by 2 or 4 since the number is to be even. So, there are 2 ways to fill ones place.

Solution : (i) When repetition of digits is allowed:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
(ii) When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of choosing thrid digit = 3
Total possible numbers `= 5 xx 4 xx 3 = 60`.

How many 3

(ii) repetition of the digits is not allowed? Solution: Answer: 60.

How many 3

But there are 4 different numbers. So the number of 3-number combinations are- (1,2,3),(1,2,4),(1,3,4),(2,3,4). Each can be arranged in 6 ways, so we get 24 ways totally.

How many 3

There are 504 different 3-digit numbers which can be formed from numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 if no repetition is allowed.

How many 3

There are 9 digits available to form 3-digits numbers without repetition. There are 9 choices for the hundreds digit, 8 choices for the tens digit, and 7 choices for the units digit. Therefore, 9 * 8 * 7 = 504 such 3-digits are possible.