Question

How many 7 digit number are there such that the digits are distinct integer taken from the set $S=\{1,2,3,4,5,6,7,8,9\}$ and such that the integer 5 and 6 do not appear consecutively in either order.

Answer 1

Answer: 151200

Total number of ways $={}^9{C}_7\cdot 7!$
now there are 6 pairs of consecutive places e.g. ab, bc,...... when we can place 56 or 65 .
This can be done in $6\cdot 2=12$ ways and remaining places can be filled in ${}^7{C}_5\cdot 5$ !.
Here number of ways in which all the seven places can be filled with consecutive 65 or 56
$=12\cdot {}^7{C}_5\cdot 5\text{ ! }$
Hence the required number of ways
$={}^9{C}_7\cdot 7!-\left({}^7{C}_5\cdot 5!\cdot 12\right)$
$=36\cdot 7!-2\cdot 6!\cdot 2!=6\cdot 6![42-7]=6\cdot 35\cdot 6!=151200$