how many 7-digit numbers are there containing two 3s and three 8s if neither of the other two digits are 3 or?

I am assuming that other digits, besides 3's and 8's, could also be repeated.

The main issue is whether or not the first digit is 3 or 8.

The number of ways of choosing which two of the seven digits are 3's is (7 choose 2) = 21, then the number of ways of choosing which three of the remaining five digits are 8's is (5 choose 3) = 10. So the number of ways of placing the 3's and 8's is 21*10 = 210.

However, if the first digit is not 3 or 8, then the number of ways of choosing which two of the other six digits are 3's is (6 choose 2) = 15, then the number of ways of choosing which three of the remaining four digits are 8's is (4 choose 3) = 4. So the number of ways of placing the 3's and 8's is 15*4 = 60, if the first digit is not 3 or 8.

So if the first digit is 3 or 8, the number of ways of placing the 3's and 8's is 210 - 60 = 150.

If the first digit is not 3 or 8, then after the 3's and 8's are placed, there are 7 possibilities for the first digit (since it can't be zero) and 8 possibilities for the other remaining digit. So, in this scenario, the number of 7-digit numbers is 60*7*8 = 3360.

If the first digit is 3 or 8, then after the 3's and 8's are placed, there are 8 possibilities for each of the two remaining digits. So, in this scenario, the number of 7-digit numbers is 150*8*8 = 9600.

The total number of possible 7-digit numbers is 3360 + 9600 = 12960.

Lord bless you today!