You are welcome.

]]>Many thanks.

D.P.]]>

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First, we see that 2 out of 6 digits are a 1, and that 2 out of 6 digits are a 2. That means that we are left with 2 places to choose a digit for. We have two cases for that:

1)The two digits we pick are the same. We can choose a digit in 7 ways, because we can only choose from digits 3,4,5,6,7,8 and 9. Then we permutate the digits we got to make a number from 2 ones, 2 twos and 2 of some other digit. We can do that in 6! ways, but we don't want to count the permutations that were counted as the result of the 2 ones changing places, so we divide by 2. The same goes for 2 and the other digit, so we end up dividing 6! by 8. The answer for this case is

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 isNow 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 cannot perform this problem:

How many 6 digit number could I write that:

1) no zeroes

2) exactly two times the digit "1"

3) exactly two times the digit "2"

Anyone could help me?

Which formula could sort this out?

Many thanks.

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