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How many distinct three-digit positive multiples of 5 can be formed from the digits 2, 3, 4, and 5 without repeating any digit?
A. 3
B. 6
C. 10
D. 12
E. 18
A. 3
B. 6
C. 10
D. 12
E. 18
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16 Dec 2013, 21:46
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legitpro
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
General Discussion
28 Oct 2007, 13:31
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never mind i think i got it.
since we can only use 2,3,4, and 5. units digit must be 5
so for possible combinations of xy5
it would be 235, 325, 245, 425, 345, 435
answer 6
since we can only use 2,3,4, and 5. units digit must be 5
so for possible combinations of xy5
it would be 235, 325, 245, 425, 345, 435
answer 6
