Question

A person writes letters to six friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in wrong envelopes?

Answer 1

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Answer 2

Answer for the letters can be placed in the envelopes so that at least two of them are in wrong envelopes is 719.

Given      

To find how many ways can the letters be placed in the envelopes so that at least two of them are in wrong envelopes.

Formula :                  

 n

∑            =  $n_{C_{n - r} } D_{r}$  

(r - 2)

             =  $n_{C_{n - 2} }D_{2} + n_{C_{n - 3} }D_{3} + n_{C_{n - 4} }D_{4} + n_{C_{n - 5} }D_{5} + n_{C_{n - 6} }D_{6}$

              $6_{C_{4} } . 2! ( 1 - \frac{1}{1!} + \frac{1}{2!} ) + 6_{C_{3} } . 3! ( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} ) + 6_{C_{2} } . 4! ( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} ) + 6_{C_{1} } . 5! ( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -\frac{1}{5!} ) + 6_{C_{0} } . 6! ( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} )$                                                                                                                          

              = 15 + 40 + 135 + 264 +265

              = 719.

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A person writes letters to six friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in wrong envelopes?  

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