How many 4 digit no can be formed with two distinct digjit?
Answers
Answered by
0
case 1: if one of those three digits is 1, remaining two digits has the other number, then possibilities is 9*(3C1)=9*3 = 27. here 3C1 is the ways of choosing one digit for 1 among n2,n3,n4, which is multiplied with 9,which is the n.o of possible numbers {0,2,3,4,5,6,7,8,9} at the remaining 2 digits.
Similar questions