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I don't get why the combination is multiplied by two? i solved by writing out the different possibilities
Available: 2, 3, 4, 5
You need to make 3 digit numbers which are multiples of 5. ___ ___ ___
For the number to be a multiple of 5, it must end with either 0 or 5. We only have 5 available so the number must end with 5.
___ ___ 5
Now you select 2 numbers out of the leftover 3 numbers (2, 3 and 4) in 3C2 ways and arrange them in the 2 places in 2! ways which gives you 3C2*2! = 6
or you can say that you can select a number for the hundreds place in 3 ways and select a number for the tens place in 2 ways (since numbers cannot be repeated). So you can make the number in 3*2 = 6 ways
Re: How many different three-digit multiples of 5 can be [#permalink]
26 Jul 2015, 13:10
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