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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?

Guest May 29, 2018

Best Answer 

 #2
avatar+20681 
+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:

see link: https://dd233cdb-a-62cb3a1a-s-sites.googlegroups.com/site/vravi68/2003JIAMNSVR.pdf?attachauth=ANoY7crVxJ9vuMj0rFmoEN_zsclHPQHKIpvsEFaZoYCp0RA8or4qPKuffYEwM1lsziada9eoD7zmTsB44rENxa_nvA3t_A_-rcUnBvaclgqWMMH7ohYxfRnND2BVnpVeNXOM5d7kYV4W_EYJ1-FUBBaKLWWq3-plEZg2-A56zBzCDjZYmg_PZzW3hwkte4ipwpRfxSCtM_ybHJv4sBAsQfF_GEID5SlzDQ%3D%3D&attredirects=0

 

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

 

laugh

heureka  May 30, 2018
 #1
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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.

Guest May 29, 2018
 #2
avatar+20681 
+2
Best Answer

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:

see link: https://dd233cdb-a-62cb3a1a-s-sites.googlegroups.com/site/vravi68/2003JIAMNSVR.pdf?attachauth=ANoY7crVxJ9vuMj0rFmoEN_zsclHPQHKIpvsEFaZoYCp0RA8or4qPKuffYEwM1lsziada9eoD7zmTsB44rENxa_nvA3t_A_-rcUnBvaclgqWMMH7ohYxfRnND2BVnpVeNXOM5d7kYV4W_EYJ1-FUBBaKLWWq3-plEZg2-A56zBzCDjZYmg_PZzW3hwkte4ipwpRfxSCtM_ybHJv4sBAsQfF_GEID5SlzDQ%3D%3D&attredirects=0

 

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

 

laugh

heureka  May 30, 2018

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