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Q. 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.

Permutations and Combinations

Solution:

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) $
$=36 \cdot 7 !-2 \cdot 6 ! \cdot 2 !=6 \cdot 6 ![42-7]=6 \cdot 35 \cdot 6 !=151200$