Question

How many odd numbers, greater than $600000$ can be formed from the digits $5,6,7,8,9,0$ if
(a) Repetitions are allowed,
(b) Repetitions are not allowed.

Answer 1

Hint: In order to calculate the number of the numbers greater than \[600000\] formed from the digits \[5,6,7,8,9,0\], we must notice that we are dealing with 6-digit numbers. Also, there are two cases given in the question, which we will solve on by one.

Complete step by step answer:
(a) Repetitions are allowed.
 \[5,6,7,8,9,0\]: Six digits
The first place can be filled by \[6,7,8,9\] i.e. in \[4\] ways as the number is to be greater than $600000$. The last place can be filled in by $5,7,9$ i.e. $3$ ways as the number is to be odd and because of repetition. Hence, the number of ways of filling the first place and the last place is \[4\times 3=12~\] ways
We have to fill in the remaining \[4\] places of the six-digit number i.e. second, third, fourth, and fifth place. Since repetition is allowed each place can be filled in \[6\] ways. Hence the \[4\] places can be filled in \[6\times 6\times 6\times 6=1296~\]ways.
Hence, by fundamental theorem, the total numbers will be \[1296\times 12=15552\].

(b) Repetition not allowed.

$1^{st}$ Place	Last (Number to be odd)
$6$	$5,7,9=3$ ways
$8$	$5,7,9=3$ ways
$7$	$5,9=2$ ways (not 7)
$9$	$5,7=2$ ways (not 9)

The total number of ways of filling the first and the last place under the given condition is \[3+3+2+2=10~\]ways.
(It was \[12\] when repetition was allowed).
Having filled in the first and the last places in \[10\] ways, we can fill in the remaining \[4\] places out of \[4\] digits (repetition not allowed) in \[4!=24\] ways.

Thererefore, the total number of ways is \[10\times 24=240\].

Additional Information:
A permutation is an arrangement of objects in a definite order. The members or elements of sets are arranged here in a sequence or linear order. Permutation can be classified in three different categories: Permutation of n different objects (when repetition is not allowed), Repetition, where repetition is allowed, Permutation when the objects are not distinct (Permutation of multisets)

Note:
It should be taken into consideration that we need to solve it first to find all the possible numbers. Thereafter for when repetition is not allowed. This question can be also solved with different methods using the permutation formula or as per the requirement.