Question

in how many ways can the digit 2,3,5,7and 9 be placed to form a 3digit number so that the higher order digit is always greater than lower order digits?(all are different digit)

Answer 1

here we have 9,7,5,3,2
since higher order digit is greater than lower order digit , so
at the hundrens place we have only 3 option (9,7, and ,5)

hence the possible no. that can be formed is :
975. 753. 532
973. 752
972. 732
953
952
932
total ways = 10