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

(i) repetition of the digits is allowed?


(ii) repetition of the digits is not allowed?

Let the 3-digit number be ABC, where C is at the units place, B at the tens place and A at the hundreds place.


(i) When repetition is allowed:


The number of digits possible at C is 5. As repetition is allowed, the number of digits possible at B and A is also 5 at each.


Therefore, The total number possible 3-digit numbers =5 × 5 × 5 =125


(ii) When repetition is not allowed:


The number of digits possible at C is 5. Let’s suppose one of 5 digits occupies place C, now as the repletion is not allowed, the possible digits for place B is 4 and similarly there are only 3 possible digits for place A.


Therefore, The total number of possible 3-digit numbers=5 × 4 × 3=60


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