For every integer k such that 1 LE k LE 6 determine how many

For every integer k such that 1 LE k LE 6, determine how many natural numbers less that 10^6 have exactly k distinct digits in their expression in base 10 system.

Solution

for k = 1, only 1 distict digit numbers are

1 to 9 one digit - 9 numbers

2 digit numbers - also 9 numbers, 11,22....99

similarly for 3,4,5,6 digits

total numbers 9*6 = 54

for k = 2

we want exactly two distict numbers

for two digits

we have 9* 8 + 9*1(when second digit is zero) = 72 numbers

for three digits

when first two digits same

9*8 = 72

when last two digits same

9*8 + 9*1( when last two zero) = 81

when first third same

9*8 + 9*1(when second is zero) = 81

total = 234

Or what we can do is

calculate for 100-199 and multiply by 9 (due to symmetry)

for 4 digits calculate for 1000-1999 and multiply by 9

and so on