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# How many 6 digit numbers can be formed from the digits 0,1,3,5,7 and 9 which are divisible by 10 and no digit is repeated?

$\begin{array}{1 1}(A)\;125\\(B)\;135\\(C)\;120\\(D)\;140\end{array}$

A number is divisible by 10 if zero occurs at the unit's place.
Now we have to fill up 5 place with the digits 1,3,5,7 and 9
This can be done in 5!
$5!=5\times 4\times 3\times 2\times 1$
$\;\;\;\;=120$ ways
$\therefore$ Required number of numbers=120
Hence (C) is the correct answer.