How many 3 digits can be formed using 2 3 5 6 7 with repetitions?
Answers
Answer:
Step-by-step explanation:
The first digit has 9 options, i.e. 1–9.
The second digit has 9 options as well, i.e. 0–9 without the one you have chosen in the first digit.
The third digit has 8 options, i.e. 0–9 without the two digits already appeared.
The number of 3-digit numbers without repetition of digits is then 9×9×8=648.
Another way to look at it:
900 is the number of 3-digit numbers formed using the digits with repetition. There are 9 numbers, i.e. 111, 222, …, 999 that has 3 numbers being the same.
Calculating the numbers where there are only 2 numbers being the same is a little tricky, but not impossible.
We have the options aab, aba, and baa, where a and b are different integers. For each type there is 9×9=81 of them. (a or b has 9 options, and there are 9 remaining options for non-first digits)
So altogether, there should be 900−9−9×8×3=648 integers.