Advancement in the Hundreds

If we choose the digits of a three digit number with distinct digits, there's only 1 way to arrange them in increasing order --> all we need to do is count the ways to make such 3-digit numbers. Also, we can't choose 0 as one of our numbers, because it'll end up as the "first digit" and we'll actually have a two digit number. So we're actually choosing amongst the 9 digits 1,2,3,...9.

(9 choose 2) = 84.

This is my first ans ...... so for any mistake .... pardon me...

there are few points i want to mention about ques 1. we cannt use 0 as digit so only 9 digits left 2. when we cant use 1,9 as middle digit so middle digit will be form 2 to 8

when middle digit is 2 ==== there are 1 7 combinations 1{1} for left place and 7{3,4,5,6,7,8,9} for right place when middle digit is 3 ==== there are 2 6 combinations 2{1,2} for left place and 6{4,5,6,7,8,9} for right place when middle digit is 4 ==== there are 3 5 combinations 3{1,2,3} for left place and 5{5,6,7,8,9} for right place when middle digit is 5 ==== there are 4 4 combinations 4{1,2,3,4} for left place and 4{6,7,8,9} for right place when middle digit is 6 ==== there are 5*3 combinations 5{1,2,3,4,5} for left place and 3{7,8,9} for right place

>> notice now that combinations are repeating.... so (7+12+15)*2+16=84

Going down Column 1: starting with the lowest possible number of the 100s, 123. Add 1 to the 'ones' value.

Going down Column 2: Once you reach your maximum 'ones' value, add one to the 'tens' value and keep adding one.

Repeat until Column 7. When maximum 'tens' and 'ones' are reached, add 1 to the 'hundreds' value.

Next lowest possible value for 200s is 234. Start from Column 2.

Repeat steps until you reach largest possible number, 789.

No. of possible numbers = (7+6+5+4+3+2+1)+(6+5+4+3+2+1)+... = 1(7)+2(6)+3(5)+4(4)+5(3)+6(2)+7(1) = 84