since each digit should appear atleast twice

so 4 cases are possible

 

so no. of ways of selcting those three digits = 5 C 3 = 10

no . of of arranging = 6!/2!*2!*2! = 90

so total nos = 10 * 90 = 900

 

so no. of ways of selecting = 5 C 2 = 10

no. of ways of arranging = 6!/3!*3! = 20

so total nos. = 200

 

no of ways in which the no. which  appears four times can be selected is 5C1 = 5 ways

 

no . of ways in which the other no. can be selected from the remaining set of four nos. is 4C1 = 4 ways

 

arrangement can be done in 6!/4!*2!

so total nos =  5 * 4 * 6!/4!*2! = 300

 

case 4: all digits are same

total nos 5

 

so adding all cases total nos = 900 + 200 + 300 + 5 = 1405