First slide

Permutations

Question

The number of ways in which we can arrange the digits 1, 2, 3, …, 9 such that the product of five digits at any of the five consecutive positions is divisible by 7 is

Moderate

A

$7!$
B

$^{9} P_{7}$
C

$8!$
D

$5 (7!)$

Solution

Let an arrangement of 9 digit number be $x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} x_{8} x_{9}$ .
Note that we require product of each of $(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}); (x_{2}, x_{3}, x_{4}, x_{5}, x_{6}); \dots; (x_{5}, x_{6}, x_{7}, x_{8}, x_{9})$ is divisible by 7.
This is possible if the 5th digit is 7.
Therefore, we can arrange the 9 digits in desired number of ways in $8!$ ways.

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