How many 3 digit numbers can be formed if repetition is not allowed

Total three digits numers are 900

We have total 10 digits which are 0,1,2,3,4,5,6,7,8,9

But a leading zero does not have any value so we cant have zero at 100th place

So for 100th place we have 9 digits, for tenth place we have 10 digits and for units place we have 10 digits

So total 3 digits number are 9*10*10 =900

If we talk about even numbers then we have only 5 digits fot units place 0,2,4,6,8 So total 3 digits even numbers would be 9*10*5 =450

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A. Repetition is not allowed and such that they are all:
1. Greater than 200
The hundreds unit must be greater than 1, so there are 4 choices, 4 remaining choices for tens, and 3 choices for units:
4 x 4 x 3 = 48 possible numbers
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2. Ending in 5
One choice for units, 4 remaining for tens, 3 for hundreds:
1 x 4 x 3 = 12 possible numbers
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3. odd numbers
Three choices for units, 4 remaining for tens, 3 for hundreds:
3 x 4 x 3 = 36 possible numbers
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B. Repetition is allowed such that they are all:
1. even numbers
Two choices for units, 5 for tens and 5 for hundreds:
2 x 5 x 5 = 50 possible numbers
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2. Less than 500
Three choices for hundreds, 5 for tens, 5 for units:
3 x 5 x 5 =75 possible numbers
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3. having 1 or 5 in their hundreds place
2 choices for hundreds, 5 for tens, five for units:
2 x 5 x 5 = 50 possible numbers

Explanation:

The first digit of the 3-digits can take 7 distinct values: 1, 2, 3, 4, 5, 7, 9. As repetition is allowed, the second digit can also take 7 distinct values, and the third can take 7 distinct values aswell, giving a total of #7 * 7 * 7 = 343# distinct combinations of numbers.

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-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.

Answer

Verified

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.

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