De-Arrangements

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!!

#Combinatorics #MathProblem

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Comments

There 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 - 8 years, 2 months ago

In case we wanted exactly 4 persons, then would the answer be 30?

Siddharth Iyer - 8 years, 2 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.

Francisco Rivera - 8 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$