million number

[k.pankaj.r]

How many numbers greater than a million can be formed with the digits 2, 3, 0, 3, 4, 2, 3?

OA given is 360

The method given is since million is a seven digit number and we 3 three times and 2 two times
so 7!/2!3! minus the no. with 0 at first place i.e 6!/2!3!
which is [7!/2!3!] - [6!/2!3!]
i tried to solve it by this method
since we have 7 places o fill
at first place we have 6 options barring 0
at second place we have 6 options again
at third place we have 5 , at fourth have 4, at 5 place we have 3 options, at sixth we have 2 options
at seventh we have finally 1 option.
but we have 3 repeating three times and 2 repeating two times.
so (6X6X5X4X3X2X1)/(3X2)
but it coes out to be 720
what am i doing wrong.
please help....

[rijul007]

How many numbers greater than a million can be formed with the digits 2, 3, 0, 3, 4, 2, 3?

Case 1:
2XXXXXX

No of arrangements = 6!/3! = 4*5*6 = 120

Case 2:
3XXXXXX

No of arrangements = 6!/2!2! = 3*2*5*6 = 180

Case 3:
4XXXXXX

No of arrangements = 6!/2!3! = 2*5*6 = 60

Total = 120 + 180 + 60 = 360

[rijul007]

How many numbers greater than a million can be formed with the digits 2, 3, 0, 3, 4, 2, 3?

OA given is 360

The method given is since million is a seven digit number and we 3 three times and 2 two times
so 7!/2!3! minus the no. with 0 at first place i.e 6!/2!3!
which is [7!/2!3!] - [6!/2!3!]
i tried to solve it by this method
since we have 7 places o fill
at first place we have 6 options barring 0
at second place we have 6 options again
at third place we have 5 , at fourth have 4, at 5 place we have 3 options, at sixth we have 2 options
at seventh we have finally 1 option.
but we have 3 repeating three times and 2 repeating two times.
so (6X6X5X4X3X2X1)/(3X2)
but it coes out to be 720
what am i doing wrong.
please help....

The red part should have been
(6X6X5X4X3X2X1)/(3!X2!)

[k.pankaj.r]

thanks rijul...helped a lot...