IIT JEE , AIEEE Forum - Mathematics , Algebra - Maths Olympiad ---> PnC

krishna.gopal

This thin is possible in two ways
1) There are 3 digits repeated twice
2) There are 1 digit which comes 4 times and another which comes twice.
For 1)
Number of ways of choosing 3 digits 5C3
Number of ways of arranging them = 6!/(2!)^3
For 2)
Number of ways of choosing 2 digits 5C2
Two ways to decide which one will come 4 times and which one will come twice
Number of ways of arranging them = 6!/(2!*4!)

so answer is 5C3*6!/(2!)^3+5C2*2*6!/(4!*2!)

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