CAT 2008 | Question: 01

Question

How many integers, greater than $999$ but not greater than $4000,$ can be formed with the digits $0, 1, 2, 3$ and $4,$ if repetition of digits is allowed?

• A. $499$

• B. $500$

• C. $375$

• D. $376$

• E. $501$

Answer 1

376 will be the right answer.

The Most Significant Digit(or leftmost digit) can not be 0, other wise resultant number will become less than 999.

For example: 0432, 0111 etc.

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Also if the Most Significant Digit is 4, then rest all the digits at other position must be 0, otherwise the resultant number will become greater than 4000.

For example : 4111, 4001 etc

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If the Most Significant Digit is any one of the 1, 2 & 3, then there is no restrictions  for the digits in other places.

So 

Suppose the 4 digit number looks like @&&& where @ is the Most Significant Bit then,

1) When @ = 0 , we will have no choices for &'s.

2) When @ = 1 or 2 or 3, we will have all the five choices for each "&" in &&&, so for the numbers of type "1&&&" we will have 125 (= 5x5x5) choices, eg 1123, 1420 etc.Similarly for the numbers of type "2&&&" and "3&&&" we will have 125 and 125 choices respectively, so we have total 375(=125 + 125 + 125) choices if leftmost digit is a 1, or a 2 or a 3.

3) Finally when @  = 4, all the &'s must be 0 in order to make 4000, so here we have only 1 choice.

So we have total 376(= 0 + (3 x 125) + 1) choices / numbers.