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Wow, that's right!
2)The two digits are different. We can choose one in 7 ways and another one in 6 ways (because we said they were different), but the order in which we choose them doesn't matter (eg. choosing a 4 and a 7 is the same as choosing a 7 and a 5) so we will have to divide the number by 2. The number of ways in which we choose the two digits is . Now we have 6 digits (2 ones, 2 twos and two different digits). Again, we can permute the number in 6! ways, but, again, changing the 2 ones' places with each other doesn't change anything so we need to divide by 2. Same goes for the 2 twos, so we end up dividing 6! by 4. Notice that this time we are not dividing by 2 for the two remaining digits because they aren't the same. The number of permutations for this case is
Now that we know the results for the two cases and are sure we didn't miss any number in these cases, we get the final answer by adding the two results, which gives us:
I am using the powerful Combinations and Permutations Calculator, but