Waste less time on Facebook — follow Brilliant.
×

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!!

Note by Siddharth Iyer
4 years, 7 months ago

No vote yet
0 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 4 years, 7 months ago

Log in to reply

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

Siddharth Iyer - 4 years, 7 months ago

Log in to reply

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 - 4 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...