In a group of 6 people, each person sends an email and receives an email amongst themselves. The different number of ways of doing this is N then the last digit of N is (if it is given that no one sends email back to himself)

Question

In a group of 6 people, each person sends an email and receives an email amongst themselves. The different number of ways of doing this is N then the last digit of N is (if it is given that no one sends email back to himself) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Number of ways =6!((1/2 !) - (1/3 !) + (1/4 !) - (1/5 !) + (1/6 !)) =265

Q. In a group of $6$ people, each person sends an email and receives an email amongst themselves. The different number of ways of doing this is $N$ then the last digit of $N$ is (if it is given that no one sends email back to himself)

Number of ways
$= 6! (\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \frac{1}{5 !} + \frac{1}{6 !})$
$= 265$

Q. In a group of 6 people, each person sends an email and receives an email amongst themselves. The different number of ways of doing this is N then the last digit of N is (if it is given that no one sends email back to himself)

Number of ways =6!(2!1​−3!1​+4!1​−5!1​+6!1​) =265

Q. In a group of $6$ people, each person sends an email and receives an email amongst themselves. The different number of ways of doing this is $N$ then the last digit of $N$ is (if it is given that no one sends email back to himself)

Number of ways
$= 6! (\frac{1}{2 !} - \frac{1}{3 !} + \frac{1}{4 !} - \frac{1}{5 !} + \frac{1}{6 !})$
$= 265$