2IIM - CAT Discussion Forum

The following question appeared in CAT 2008. It is a simple question in permutation combination.

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?
(1) 499
(2) 500
(3) 375
(4) 376
(5) 501

Answer (2). 500.

4 options (1, 2, 3 or 4) for the first place and then 5 for the other 3.

The smallest number in the series is 1000 and the largest number is 4000.

The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.

The next 3 digits (hundreds, tens and units place) can take any of the 5 values - 0 or 1 or 2 or 3 or 4.

Hence, there are 3 * 5 * 5 * 5 or 375 numbers from 1000 to 3999.

Including 4000, there will be 365 + 1 = 376 numbers.

Good one. Actually made the same mistake made by crackcat.

Interesting question. I really like questions that have one easy "wrong" answer. On the same genre, here goes one more.

There is an old clock that sounds the gong on every hour. The number of times the gong sounds is equal to the time at that hour. As in, at 3 O'clock, the gong sounds thrice. If the clock takes 6 seconds between the first and last sounds when it rings at 4 O'clock, how long does it take to complete the 12 sounds when the time is 12 midnight?

Cheers all,

The clock takes 2 seconds between two successive sounds. So, for striking 12 times, it will take 2*11 seconds, or the right answer is 22 seconds. This one is not a great/difficult quesion. Just that it is tempting to mark the answer choice as 18 seconds.

Here is the question for the day from Permutations and combinations

The number of divisors of the type 4n+2 of the number 2520 is

a. 12
b. 13
c. 24
d. 16