How many 3-digit numbers can be formed from the digits 5, 6, 7, 8 and 9 assuming that
(i) Repetition of the digits is allowed?
(ii) Repetition of the digits is not allowed?
How many 3-digit numbers can be formed from the digits 5, 6, 7, 8 and 9 assuming that
(i) Repetition of the digits is allowed?
(ii) Repetition of the digits is not allowed?
The correct option is C 125 and 60
(i) 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.
Now 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.
Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125
(ii) 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.
Now 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 are 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
(i) 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.
Now 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.
Hence, the total number possible 3-digit numbers =5 × 5 × 5 =125
(ii) 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.
Now 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 are 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
