If the numbers that we are checking divisibility for are between 1-100, there are 3 numbers.
30, 60, and 90 are all divisible by both 3 and 10 because both 3 and 10 are factors of these 3 numbers.
Answer:
(2,4)
Step-by-step explanation:
When graphed the lines intersect at point (2,4) which is the solution.
If the letters are considered distinct, then the number of permutations is
.
If we count either C as the same character, then we would be double-counting - to correct this, we would simply divided by the number of ways we can choose C from the available characters, or
.