Four people put their hats on the table as they arrive. When they leave, each person picks up one hat. It happens so no-one has picked up his own hat. In how many ways can this have happened?
Four people put their hats on the table as they arrive.
When they leave, each person picks up one hat.
It happens so no-one has picked up his own hat.
In how many ways can this have happened?
This have happened in 9 ways.
Formula:
\(\text{The number of permutations with all the elements $\\$ displaced from their original position for $n = 4$ is:}\)
\(\begin{array}{|rcll|} \hline && ^4C_2(4-2)! -\ ^4C_3(4-3)! +\ ^4C_4(4-4)! \\\\ &=& ^4C_2\cdot 2! -\ ^4C_3\cdot 1! +\ ^4C_4\cdot 0! \\\\ &=& 6\cdot 2 - 4\cdot 1 + 1\cdot 1 \\\\ &=& 12 - 4 + 1 \\\\ &\mathbf{=}&\mathbf{ 9 } \\ \hline \end{array}\)
Each of the 4 people has a choice of 3 other hats, not including his/her hat.
Therefore, there are: 4 x 3 = 12 ways that this can happen.
Four people put their hats on the table as they arrive.
When they leave, each person picks up one hat.
It happens so no-one has picked up his own hat.
In how many ways can this have happened?
This have happened in 9 ways.
Formula:
\(\text{The number of permutations with all the elements $\\$ displaced from their original position for $n = 4$ is:}\)
\(\begin{array}{|rcll|} \hline && ^4C_2(4-2)! -\ ^4C_3(4-3)! +\ ^4C_4(4-4)! \\\\ &=& ^4C_2\cdot 2! -\ ^4C_3\cdot 1! +\ ^4C_4\cdot 0! \\\\ &=& 6\cdot 2 - 4\cdot 1 + 1\cdot 1 \\\\ &=& 12 - 4 + 1 \\\\ &\mathbf{=}&\mathbf{ 9 } \\ \hline \end{array}\)