How many three digit numbers may be formed from 1, 2, 3, 4, 5 and 6 if repetition is allowed

A 3-digit number has three places which from left to right are hundred's place, ten's place and unit's place.

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Here we have to form 3-digit numbers using the digits 2, 3, 4, 5, 6.

The hundred's place can be filled in by using any one of the given 5 digits in 5 ways.

Since, repetition of digits is allowed each ten's place and unit's place can be filled in by any one of the given 5 digits in 5 ways.

Marlene S. answered • 01/24/16

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Hi Thea

With 7 numbers, we can create three digit numbers like this.

a) With repetitions would suggest with replacement.

We'd have 7 options for the first digit, 7 options for the 2nd digit and 7 choices for the 3rd digit

73 = 343 combinations

b) Without repeated numbers, would suggest without replacement

We'd have 7 options for the first digit, 6 options for the 2nd digit and 5 choices for the 3rd digit

7•6•5 = 210 combinations

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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`.