how many 7-digit number can be formed using 1.2.3.4.5.6.7.8.9?

Question

how many 7-digit can be formed using 1,2.3,4,5,6,7,8,9.where digit 1 and 2 do not appear consecutively, in either order,and the digits are all distinct?

Answer 1

First, remove 1 and 2 from the set. There are 7!=5040 ways of arranging the remaining 7 digits.

Next, we combine 1 with any 6 out of 7 of the remaining digits. There are 7 positions for digit 1 and there are 5040 ways of arranging 6 out of 7 digits so that gives us 7*5040=35280. Similarly for combining 2 with 6 out of 7 of the digits 3-9. Another 35280.

Now we have to use 1 and 2 combined with any 5 out of 7 of digits 3-9. There are 2520 arrangements 5 out of 7 distinct digits, and there are 15 ways to place 1 and 2 in order within these arrangements so that 1 and 2 are not adjacent. There are 15 ways to place 2 and 1 in order within the arrangements, so we need to add in 30*2520=75600. The total number comes to 5040 (not using 1 and 2) + 70560 (using 1 or 2 with 6 digits from 3-9) + 75600 (using 1 and 2 non-adjacently with 5 digits from 3-9) = 151200.

Answer 2

7-digit can be formed using 1,2,3,4,5,6,7,8,9.where digit 1 and 2 do not appear consecutively

Removing 1 and 2 = 7! = 5040

Hence,
5040 (not using 1 and 2) + 70560 (using 1 or 2 with 6 digits from 3-9) + 75600 (using 1 and 2 non-adjacently with 5 digits from 3-9) = 151200