6 seats are available around the table but only 4 people are having dinner.
In how many different ways can these people sit ?
The number of groups of 4 seats that can be formed with 6 seats is:
\dbinom{6}{4}=\dfrac{6!}{4!\cdot2!}=\dfrac{4!\cdot5\cdot6}{4!\cdot 1\cdot2}=15
In each group of 4 seats, 4 people can sit in "permutation of 4" ways:
4!=1\cdot2\cdot3\cdot4=24
4 people can sit around a table with 6 seats in the following number of ways:
24\cdot15=360
There are 360 different ways for these people to sit around the table.
There are 50 people at a meeting. Every person shakes hands with everyone else.
How many handshakes were there?
The number of groups of 2 people that can be formed with 50 people when order doesn't matter is:
\dbinom{50}{2}=\dfrac{50!}{2!\cdot48!}=\dfrac{48!\cdot49\cdot50}{48!\cdot 1\cdot2}=1\ 225
There were 1225 handshakes.
Paul wants to set a passcode for his new safe coded box. The passcode has 4 digits and he can only use the numbers 3, 6, 9 and the letter c.
In how many ways can he set the passcode?
Since the order of the numbers and letter matters, we must use permutations of 4 elements taken as 4:
\dfrac{4!}{\left(4-4\right)!}=1\cdot2\cdot3\cdot4=24
There are 24 different ways for Paul to set the passcode.
From a group of 6 men and 8 women, how many teams of 2 men and 3 women can be formed?
The number of groups of 2 men that can be formed with 6 men is:
\dbinom{6}{2}=\dfrac{6!}{2!\cdot4!}=\dfrac{4!\cdot5\cdot6}{4!\cdot1\cdot2}=15
The number of groups of 3 women that can be formed with 8 women is:
\dbinom{8}{3}=\dfrac{8!}{3!\cdot5!}=\dfrac{5!\cdot6\cdot7\cdot8}{5!\cdot1\cdot2\cdot3}=56
Therefore, the number of teams of 2 men and 3 women that can be formed is:
15\cdot56=840
The number of teams of 2 men and 3 women that can be formed is 840.
How many basketball teams of 5 players can be formed with 10 players without regard to position played?
Since the position played does not matter, use combinations:
\dbinom{10}{5}=\dfrac{10!}{5!\cdot5!}=\dfrac{5!\cdot6\cdot7\cdot8\cdot9\cdot10}{5!\cdot1\cdot2\cdot3\cdot4\cdot5}=252
There are 252 teams that can be made.
A police station is choosing a captain and an assistant. There are 10 police officers at this station.
In how many ways can police officers be chosen for these two positions?
Since it matters what position each officer could take, use permutations of 10 elements taken as 2:
\dfrac{10!}{\left(10-2\right)!}=\dfrac{10!}{8!}=\dfrac{8!\cdot9\cdot10}{8!}=90
There are 90 different ways the police officers can be chosen.