How Many 3 Digit Numbers Can Be Formed With Given Digits :
Here we are going to see some practice questions based on the concept fundamental principle of counting.
Question 1 :
Count the numbers between 999 and 10000 subject to the condition that there are (i) no restriction. (ii) no digit is repeated. (iii) at least one of the digits is repeated
Solution :
The numbers lie between 999 to 10000 must be a 4 digit number.
__ __ __ __
Thousands place :
Other than 0, we have 9 options to fill up the thousands place.
Hundreds place :
Including 0, we have 10 options.
Tens place :
Including 0, we have 10 options.
Unit place :
Including 0, we have 10 options.
Hence total numbers lie between 999 and 10000 is
9 ⋅ 10 ⋅ 10 ⋅ 10 = 9000
Thousands place :
Other than 0, we have 9 options to fill up the thousands place.
Hundreds place :
Here we may accept 0, but we should not use the number that we have already used in the thousands place. So, we have 9 options.
Tens place :
We have 8 options.
Unit place :
We have 7 options.
So, the total numbers are
9 ⋅ 9 ⋅ 8 ⋅ 7 = 4536
At least one of digits is repeated means, we may have one repeated digit, two repeated digits or more.
In order to find the answer for this question, we have to subtract no digits repeated from total numbers.
Hence the answer is 9000 - 4536 = 4464.
Question 2 :
How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if (i) repetition of digits are not allowed? (ii) repetition of digits are allowed?
Solution :
___ ___ ___
Case 1 :
If we use 0 in the unit place, we will have only one option.
In tens place, we will have 5 options.
In hundreds place, we will have 4 options.
Case 2 :
If we use 5 in the unit place, we will have only one option.
In hundreds place other than 0 and 5, we will have 4 options.
In tens place, we will have 4 options including 0.
Hence total numbers formed = 20 + 16 = 36 numbers
Unit place :
Since the required numbers are divisible by 5, it ends with 0 or 5. So, we have 2 options.
Hundreds place :
Other than 0, we will have 5 options.
Tens place :
Sine the repetition is allowed , we have 6 options.
Hence total numbers to be formed = 5 ⋅ 6 ⋅ 2
= 60 numbers
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