permutation, combination

Thanks

Hi ,

for a no. to be divisible by 3 the nos should add up to a multiple of 3.
Given the digits (0,1,2,3,4,5)
Only the combinations given below will add up to a multiple of 3 :
1. (1,2,3,4,5)
2. (0,1,2,4,5)

Case 1:
these nos can be arranged in 5! ways i.e 120

Case 2:
Since if 0 is the 1st digit it will not form a 5 digit no. they can be arranged in :4*4*3*2*1

hence the total permutations is :120 +96 =216

Hi ,

for a no. to be divisible by 3 the nos should add up to a multiple of 3.
Given the digits (0,1,2,3,4,5)
Only the combinations given below will add up to a multiple of 3 :
1. (1,2,3,4,5)
2. (0,1,2,4,5)

Case 1:
these nos can be arranged in 5! ways i.e 120

Case 2:
Since if 0 is the 1st digit it will not form a 5 digit no. they can be arranged in :4*4*3*2*1

hence the total permutations is :120 +96 =216

I noticed most people solved this problem exactly the same way.

I did the last part a little different.

Case 1: 5! = 120 - this is the easy part

Case 2:

I kind of understand why people will do 4*4!. However if I were on the exam I probably would not have realized I needed to multiply by four.

Personally, I saw that 0 couldn't be the first number for a 0 1 2 4 5 combination. I could see that I needed to find the number of combinations then subtract the ones that didnt fit the criteria of being divisible by 3. So I said there are 5! combinations of 0 1 2 4 5. However we must subtract the number of combinations where 0 is the first number.

total combinations - 0 first combinations = 5! - 4! = 120 - 24 = 96

Then I added 120 + 96 = 216

Ian Stewart wrote:

umaa wrote:How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)

If a number is divisible by 3, the sum of its digits is divisible by 3. We need to choose five digits from {0,1,2,3,4,5} which add to a multiple of 3. We could choose:

{1, 2, 3, 4, 5}

or

{0, 1, 2, 4, 5}

If we use the digits {1, 2, 3, 4, 5}, we have 5 choices for the first digit, 4 for the second, etc- 5! = 120 numbers we can make in total.

If we use the digits {0, 1, 2, 4, 5}, we only have 4 choices for the first digit (because it cannot be zero), 4 choices for the second digit, 3 for the third, etc- 4*4! = 96 numbers we can make in total.

120+96 = 216.

but a number like 10234 is not divisible by 3, the omission of any of the digits 1,2,4,5 while including 0 will be a number thats not divisible by 3.