Find the number of positive integers, which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 but using each digit not more than once in each number. How many of these integers are greater than 3000? What happened when repetition is allowed?

Answer

1440

Explanation

(i) When repetition is not allowed.
There are 6 digits 0,1,2,3,4,5 and we can use any number of digits.

For numbers greater than 3000, number of digits is greater than or equal to 4.
All the 6- and 5-digit numbers are greater than 3000. Now for 4-digit numbers `gt 3000`.

Hence, number greater than 3000 having 4 digits is `3xx .^(5)P_(3)=180`
So, total numbers `gt 3000` is 180+600+600=1380.
(ii) When repetition is allowed.
It is equivalent to permutation of 6 objects (digits) when any object is repeated any number of time, which is equal to `6^(6)`. But this includes one number 0. Hence, total numbers are `6^(6)-1=46655`.
Now, number of 5-digit numbers is `5xx6xx6xx6xx6=6480`.
And, number of 6-digit numbers is `5xx6xx6xx6xx6xx6=38880`.
Again, 4-digit numbers are to be greater than 3000.

Hence, total number of 4-digit numbers is `3xx6xx6xx6=648`
But these 648 numbers have 3000 as a number which should be excluded. So, the number of 4-digit numbers `gt 3000` is 648-1=647. Therefore, total numbers `gt 3000` are 647+6480+38880=46007.