A five-digit number divisible by 3 is to be formed using

Can anyone please explain to me how do you calculate the total number of ways if 0 is in the first place? I was able to get to 120 for the first step but I am confused with the 0 part.

{0} - {1, 2, 4, 5}

0 is fixed at the first place. Four remaining numbers, {1, 2, 4, 5} can be arranged in 4! ways.

This makes a lot of sense! 4!= 24, I was wondering why did you multiply 24 by 4 to get 96? Could you please explain this part as well? Would greatly appreciate it!

Thanks a lot!

We don't multiply this by 4.

We have:
{the total number of combinations} - {the number of combinations we don't want} = {the number of combinations we want}

5! - 4! = 4!(5 - 1) = 4!*4. So, we just factor out 4! from 5! - 4! to get 4!(5 - 1).

Hope it's clear.

Understood! Thanks a lot Bunuel! Kudos given!

A number is divisible by 3 if the sum of the digits of the number is divisible by 3 .

Case 1) 0,1,2,4,5
The sum of the 5 digits is 0+1+2+4+5 = 12 (divisible by 3).

For the first place we have 4 numbers (0 excluded) ,for the second place we have 4 (now 0 included) ,for third 3 ,for fourth 2 and for fifth 1.
Hence the number of choices we have is 4×4×3×2×1=96.

Case 2) 1,2,3,4,5
The sum of the 5 digits is 1+2+3+4+5 = 15 (divisible by 3).

Here the number of choices we have is 5×4×3×2×1=120

Hence in total we have 120 +96 = 216 choices

We are given 6 digits, in which we have to use 5 of them to form a five-digit number, without repetition, that is divisible by 3. In order for the number to be divisible by 3, we need the sum of its digits to be divisible by 3. Notice that the sum of the 6 digits is 15, which is divisible 3. So, in order for the sum of 5 of them to also be divisible by 3, we must remove one of them that is divisible by 3. That is, we can remove either the digit 3 or the digit 0. Thus, our options are {1, 2, 3, 4, 5} or {0, 1, 2, 4, 5}.

In the first option, there are 5! = 120 ways to form a five-digit number that is divisible by 3. In the second option, however, since 0 can’t be the first digit, there are only 4 x 4 x 3 x 2 x 1 = 96 ways to form a five-digit number that is divisible by 3. Therefore, we have a total of 120 + 96 = 216 ways to form a five-digit number that is divisible by 3.

Answer: C

Official Solution

Credit: Veritas Prep

Blindly applying the formula we just discussed would yield n=6 and k = 5, so 6!/(6-5)! which is just 6! and therefore 720. Answer choice D is waiting for us if we go down this route, and it seems like a safe guess if you have no idea how to proceed, so a fair number of students may be tempted by this option. However, the formula gives you all possible permutations of these numbers, including, for example, 01234, which does not meet the second criterion of the question: That the numbers be divisible by 3. This restriction will eliminate many of the possibilities, so 720 possibilities (and 3152 even more so) is overshooting the target.

So what do we do? Go through all 720 possibilities and whittle the list down for every possibility that isn’t divisible by 3? That’s a good way to spend an unproductive afternoon! Why don’t we try and limit the number of possibilities by seeing which combinations of digits give us something divisible by 3 and then going through the permutations for those digits only? The rule for divisibility by 3 is that the sum of the digits must be divisible by 3. The fact that each digit must be unique means we only have six possibilities of digits, each possibility ignoring one of the six digits provided.

Digits 0,1,2,3,4 only: sum 10. Does not work.
Digits 0,1,2,3,5 only: sum 11. Does not work.
Digits 0,1,2,4,5 only: sum 12. Works!
Digits 0,1,3,4,5 only: sum 13. Does not work.
Digits 0,2,3,4,5 only: sum 14. Does not work.
Digits 1,2,3,4,5 only: sum 15. Works!

So we only really need to calculate the permutations of {1,2,3,4,5} and {0,1,2,4,5}. The first one is easier, as we can just take 5!/0! (0! Is 1 by definition) which gives us 120 possibilities. The second one contains a famous GMAT trick, which is that it cannot begin by 0 and still be a 5-digit number. Just like 7 is a one-digit number and not a five-digit number by adding an arbitrary number of 0’s at the beginning of it (James Bond’s understudy 00007). This logic dictates that the second set has only four possible numbers in the first position, and then 4 again for the second (any of the four that wasn’t put in the first spot) and then three, two, one as normal. Instead of the standard 5! we get 4 * 4! which is 96. As such, the correct answer is 120 + 96 = 216.

Answer choice B.

The sum of the digits need to be divisible by 3 in order for the 5 digit number to be divisible by 3.

0,1,2,3,4,5. By adding 1,2,3,4,5, we get 15 which is divisible by 3. Total 5 digit numbers which can be formed using the digits 1,2,3,4,5 without repetition is 5! (5x4x3x2x1) = 120

Additionally, 0,1,2,4,5 which has a sum of 12 (divisible by 3) can also be used to form 5 digit numbers. Since 0 cannot be it the ten thousands place, total number of 5 digit numbers possible is 4x4x3x2x1. = 96.

Total 5 digit numbers = 120 + 96 = 216. Ans C.