Combinatorics Problem: Sending 15 Postcards to 15 Friends in Unique Ways

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Homework Help Overview

The problem involves combinatorics, specifically calculating the number of ways to distribute 15 postcards of 3 different types, with 5 postcards of each type, to 15 friends, ensuring each friend receives one postcard.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial approach of using the formula 15!/(3*5!), questioning the validity of the multiplication by 3 and the factorial of 5. There are considerations about the arrangement of identical postcards and how that affects the calculation.

Discussion Status

The discussion is ongoing with participants exploring different interpretations of the problem and the appropriate combinatorial formulas. Some guidance has been offered regarding the general rule for selecting items of different types, but no consensus has been reached on the correct approach yet.

Contextual Notes

Participants are considering variations of the problem, such as changing the number of friends and types of postcards, which may influence the application of combinatorial principles.

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Homework Statement



You have 3 types of postcards. There are 5 of each type. How many ways can you send the 15 postcards to 15 friends, if each friend receives 1.


The Attempt at a Solution



I thought it would merely be 15!/(3*5!)
 
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Are you sure about the 3 * 5! ?
 
CompuChip said:
Are you sure about the 3 * 5! ?
because there are three sets of five identicals
 
hi swtlilsoni! :smile:
swtlilsoni said:
because there are three sets of five identicals


if there were 10 friends, and 2 sets of five identicals, would you use 10!/2*5! ? :wink:
 
Ohh okay so it would be 5!3!
 
Can you explain that to us, or are you just guessing now? :)
 
it's because it has to be multiplied. For every rearrangement of five identicals, there are two more rearrangements of the others
 
hi swtlilsoni! :smile:

(just got up :zzz: …)
swtlilsoni said:
it's because it has to be multiplied. For every rearrangement of five identicals, there are two more rearrangements of the others

the general rule for selecting a of one type, b of another, … z of another, from n altogether (with a+b+… +z = n), is:

n!/a!b!…z!​

for only two types, that reduces to the familiar:

n!/a!b! = n!/a!(n-a)! = nCa :wink:
 

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