UGC NET CS 2015 June Paper3
October 16, 2023UGC NET CS 2009 DecPaper2
October 16, 2023NICNIELIT ScientistB 2020
Question 118

The number of 4 digit numbers which contain not more than two different digits is:
576


567


513


504

Question 118 Explanation:
The first (nonzero) digit of the number thousands digit (A)can be any one of nine.
The second digit is hundred digit(B) which is used if there are two digits can be any one of the nine digits different from the first.
Now consider the Hundreds Tens and Units digits in the case that there are two digits used in the number.
We have two possibilities A or B to fill each place – but we exclude AAA as not involving two digits, so there are 23−1=7 possible patterns with exactly two different digits and 9×9 ways of choosing the pair of digits in the first place.
Then there are nine possibilities with just one digit.
Total possibilities are 9×9×7+9=576
The second digit is hundred digit(B) which is used if there are two digits can be any one of the nine digits different from the first.
Now consider the Hundreds Tens and Units digits in the case that there are two digits used in the number.
We have two possibilities A or B to fill each place – but we exclude AAA as not involving two digits, so there are 23−1=7 possible patterns with exactly two different digits and 9×9 ways of choosing the pair of digits in the first place.
Then there are nine possibilities with just one digit.
Total possibilities are 9×9×7+9=576
Correct Answer: A
Question 118 Explanation:
The first (nonzero) digit of the number thousands digit (A)can be any one of nine.
The second digit is hundred digit(B) which is used if there are two digits can be any one of the nine digits different from the first.
Now consider the Hundreds Tens and Units digits in the case that there are two digits used in the number.
We have two possibilities A or B to fill each place – but we exclude AAA as not involving two digits, so there are 23−1=7 possible patterns with exactly two different digits and 9×9 ways of choosing the pair of digits in the first place.
Then there are nine possibilities with just one digit.
Total possibilities are 9×9×7+9=576
The second digit is hundred digit(B) which is used if there are two digits can be any one of the nine digits different from the first.
Now consider the Hundreds Tens and Units digits in the case that there are two digits used in the number.
We have two possibilities A or B to fill each place – but we exclude AAA as not involving two digits, so there are 23−1=7 possible patterns with exactly two different digits and 9×9 ways of choosing the pair of digits in the first place.
Then there are nine possibilities with just one digit.
Total possibilities are 9×9×7+9=576
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