permutation and combination

wintersolstice

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Nov 5, 2011
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Joshua tries to recall the 6-digit pin number for his ATM card.
(Assuming that the number 000000 is even and valid.)
How many possible numbers can there be if he remembers that the digits do not repeat and there are exactly 3 odd digits?

what I tried to is:

odd digit: 1,3,5,7,9
even digit: 0,2,4,6,8

_ _ _ _ _ _

6 positions, exactly 3 non-repetitive odd digits so will have 3 non-repetitive even digits also, so 5C1x4C1x3C1x5C1x4C1x3C1=3600
but the answer given is 72000.
i tried to multiply by 6!, thinking that there's arrangement, but answer will then exceed.
what's wrong with my approach?
any help is much appreciated.
thanks. =)
 
Joshua tries to recall the 6-digit pin number for his ATM card.
(Assuming that the number 000000 is even and valid.)
How many possible numbers can there be if he remembers that the digits do not repeat and there are exactly 3 odd digits?

what I tried to is:

odd digit: 1,3,5,7,9
even digit: 0,2,4,6,8

Choosing the odd numbers: 5C3 = 10

Choosing the even numbers: 5C3 = 10

Ways to arrange 6 numbers: 6! = 720

(10)(10)(720) = 72000
 
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