Since the repetition of digits is not allowed then the number of 7 digit numbers formed = 9 x 9 x 8 x 7 x 6 x 5 x 4 = 544320
Explanation:
The first digit of the number can't be 0 so we have 9 digits to choose from.
And the remaining last places(here 6) can be filled by using the remaining 9 digits i.e. 9P6 = 9!/(9-6)! = 9 x 8 x 7 x 6 x 5 x 4
So, the result is 9 x 9P6 .
Six.
You can select 12 numbers for the first digit, 11 numbers for the second digit, and 10 numbers for the third digit; so 12*11*10 = 1320 sets of 3 numbers can be made out of 12 different numbers.
210
Through the magic of perms and coms the answer is 729
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
66
10,000
15
a lot
Six.
You can select 12 numbers for the first digit, 11 numbers for the second digit, and 10 numbers for the third digit; so 12*11*10 = 1320 sets of 3 numbers can be made out of 12 different numbers.
2
210
Through the magic of perms and coms the answer is 729
9
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.