First constraint=2 leftmost digits even. Therefore 4*5 ways of forming the 2 leftmost digits as the number cannot start with 0.

Constraint within this constraint: 4 should not occur twice. We see that 4 can occur twice only once . Therefore tre are 4*5 - 1 ways of forming the 2 leftmost digits=19 ways

Third constraint: The other 3 digits are odd. Therefore 5*5*5 ways of forming the 3 rightmost digits=125 ways

Total number of possible combinations given the constraints is 19*125= 2375

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Srinivasan Vaidyaraman

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