Question

All the five digit numbers in which each successive digit exceeds its predecessor

Answer 1

# Answer = 126

# Explanation-

For a number to follow given condition, it should start from either 1,2,3,4,5.

1) Starting from 1-

8C4 = 8! / 4!×4!

8C4 = 8×7×6×5×4!/4!×4×3×2×1

8C4 = 70

2) Starting from 2-

7C4 = 7! / 4!×3!

7C4 = 7×6×5×4!/4!×3×2×1

7C4 = 35

3) Starting from 3-

6C4 = 6! / 4!×2!

6C4 = 6×5×4!/4!×2×1

6C4 = 15

4) Starting from 4-

5C4 = 5! / 4!×1!

5C4 = 5×4!/4!×1

5C4 = 5

5) Starting from 5-

4C4 = 1

Total such numbers = 70+35+15+5+1 = 126 numbers.

Hope this is useful...