# How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that:

1) Repetition is not allowed.

2) Repetition is allowed.

Last updated date: 20th Mar 2023

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Answer

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Hint: Fundamental principle of counting: According to the fundamental principle of counting if a task can be done in “m” ways and another task can be done in “n” ways, then the number of ways in which both the tasks can be done in mn ways.

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.

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.

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