Anonymous

# How many ways can 1 2 3 5 5 8 be arranged and how many of these numbers could be divided by 15?

just as the title asks how many ways can 1 2 3 5 5 8 be arranged and how many numbers can be divided by 15?

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Hi,

In order to be divisible by 15, a number must end in either 5 or 0, and the sum of all its digits must be divisible by 3. Since 0 was not included in the numbers to be arranged, then 5 must be in the last position. That leaves the other 5 digits to be arranged in 5! ways. This makes 120 distinguishable arrangements of those 6 digits.

I hope that helps!! :-)

To figure out how many ways it can be arranged, it's just 6!. However, since 5 appears twice, it is 6!/2! or 360.

In order for it to be divisible by 15, the last digit has to equal 5 (since 1+2+3+5+5+8=24, and that's already divisible by 3)

From the rest of the digits, it's just 5! or 120.

• M3
Lv 7

a) the number can be arranged in 6!/2! = 360 ways

b) all numbers ending with 5 will also be divisible by 15

so they will number (2/6)*360 = 120

I think they can be arranged 6! different ways.

well the answer to the first question is.....first you must use a permutation or a combinattion(sorry i forgot which ones which)....which is where you do 1*2*3*4*5*6 because there are six numbers... the second one....ya gotta figure it out!

Source(s): my brain!!!