Question 87547
At unit place only one of these are possible 3,5,or 7.


Case A (when 3 at unit place i.e fifth position) : 

for first position i.e  _xxxx only 2 or 4 is possible otherwise it will greater 

than 50,000


so, no. of ways to fill first position  = 2  (two digits have used )


now, no. of ways to fill second position = 5    ( only 5 digits are remaining)


no. of ways to fill third position i.e xx_xx  = 4 (only 4 digits are remaining)



no. of ways to fill fourth position = 3     (only 3 digits are remaining)

fifth position is already occupied by 3.

so, total no. of different numbers that can be formed = 2 * 5 * 4 * 3 =120
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Case B (when 5 at unit place i.e fifth position) : 

for first position i.e  _xxxx only 2, 3 or 4 is possible otherwise it will 

greater than 50,000

so, no. of ways to fill first position  = 3      (two digits have used )

now, no. of ways to fill second position = 5    ( only 5 digits are remaining)

no. of ways to fill third position i.e xx_xx  = 4 (only 4 digits are remaining)

no. of ways to fill fourth position = 3     (only 3 digits are remaining)

fifth position is already occupied by 5.

so, total no. of different numbers that can be formed = 3 * 5 * 4 * 3 =180



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Case C (when 7 at unit place i.e fifth position) : 

similarly,

so, no. of ways to fill first position  = 3      (two digits have used )

now, no. of ways to fill second position = 5    ( only 5 digits are remaining)

no. of ways to fill third position i.e xx_xx  = 4 (only 4 digits are remaining)

no. of ways to fill fourth position = 3     (only 3 digits are remaining)

fifth position is already occupied by 7.


so, total no. of different numbers that can be formed = 3 * 5 * 4 * 3 =180





now we add the value of these three  cases.

no. of different 5-digit numbers that are odd and less than 50,000 can be 

formed    =  120 + 180 + 180  = 480





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