Solve the given permutation problem

  • #1
chwala
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Homework Statement
How many ##5##-digit numbers are there that have ##5## different digits and are divisible by ##5##?
Relevant Equations
Permutation
My First approach on this;
The last digit should be a ##0## or a ##5##. Therefore the first ##4## slots can be filled in ##_9P_4×1## ways = ##3024##.
Secondly, we also note that ##0## cannot be on the first slot as this would imply ##4## digits instead of ##5##...further the last slot has to either be a ##0## or a ##5##. Therefore we are going to have; ##8×_8P_3×1##=##2688##. Thus we shall have ##3024+2688=5712## ways.

In my second approach;

I am thinking along the lines of the ##5## slots being filled initially by digits that are not divisible by ##5## i.e
The ##5## slots can be filled in ##9×_9P_4## ways= ##27, 216## ways.

The ##5## slots can be filled by numbers that are not divisible by ##5## in ;
##8×_8P_3×8=21,504## ways.

Therefore, numbers that are divisible by ##5## can be filled in ##27,216 - 21,504=5,712## ways. Bingo! :cool:
Any feedback is highly welcome or more insight that is.
 
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  • #2
Alternatively:

If the number ends in ##0##, three are ##9\times 8 \times 7 \times 6## possibilities.

If the number ends in ##5## there are ##8\times 8 \times 7 \times 6## possibilities.
 
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