How would one solve this question:

There are 6 friends with 6 different letters E1, E2, E3, E4, E5, E6 to be posted to their respective friends. In how many ways can exactly 3 of the friends receive the right letter? I know that we have to do this problem using de-arrangements but I don't know how to. Please help!!

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TopNewestThere are \( \displaystyle \binom{6}{3} \) ways to choose the three friends that will get the correct letter. As for the other friends there are two permutations that do not permute any friend to his/her letter (these are permutations with all cycles size greater than 1, which means in this case there is only one cycle size three). Therefore the answer is \[ \displaystyle \binom{6}{3} \cdot 2 = 40 \] – Francisco Rivera · 4 years, 4 months ago

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– Siddharth Iyer · 4 years, 4 months ago

In case we wanted exactly 4 persons, then would the answer be 30?Log in to reply

– Francisco Rivera · 4 years, 4 months ago

Not quite. If you wanted 4 people then there would be \( \displaystyle \binom{6}{2} = 15 \) ways to choose the two people that get the wrong letter. However, there is only one permutation of length 2 that has no cycle of 1, therefore the answer would just be 15.Log in to reply