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# Math Question

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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?

May 29, 2018

#2
+21978
+2

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:

$$\text{The number of permutations with all the elements \\ displaced from their original position for n = 4 is:}$$
$$\begin{array}{|rcll|} \hline && ^4C_2(4-2)! -\ ^4C_3(4-3)! +\ ^4C_4(4-4)! \\\\ &=& ^4C_2\cdot 2! -\ ^4C_3\cdot 1! +\ ^4C_4\cdot 0! \\\\ &=& 6\cdot 2 - 4\cdot 1 + 1\cdot 1 \\\\ &=& 12 - 4 + 1 \\\\ &\mathbf{=}&\mathbf{ 9 } \\ \hline \end{array}$$

May 30, 2018

#1
0

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.

May 29, 2018
#2
+21978
+2

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:

$$\text{The number of permutations with all the elements \\ displaced from their original position for n = 4 is:}$$
$$\begin{array}{|rcll|} \hline && ^4C_2(4-2)! -\ ^4C_3(4-3)! +\ ^4C_4(4-4)! \\\\ &=& ^4C_2\cdot 2! -\ ^4C_3\cdot 1! +\ ^4C_4\cdot 0! \\\\ &=& 6\cdot 2 - 4\cdot 1 + 1\cdot 1 \\\\ &=& 12 - 4 + 1 \\\\ &\mathbf{=}&\mathbf{ 9 } \\ \hline \end{array}$$

heureka May 30, 2018