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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?
Asked by Narayanan Last Modified
Answer 
Harshit Bhardwaj
(i)
There will be as many ways as there are ways of filling 3 vacant places
in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125
(ii)
In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5.
Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.
Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60
read less(i) when repetition of digits are allowed .
given,
1, 2 , 3 , 4, 5 . since there are five digits
Therefore , number of ways to fill each place of 3 digit number are 5
Hence,
by Fundamental principle of counting,
total number of ways = 5 × 5 × 5 = 125
(ii) when repetition isn't allowed.
here, five digits are available .
therefore,
number of ways to fill the unit place = 5
number of ways to fill tenth place = 4
number of ways to fill hundred place = 3
hence,
by Fundamental principle of counting,
total number of ways = 5 × 4 × 3 = 60
read less
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