A community for students.
Here's the question you clicked on:
 0 viewing
xEnOnn
 3 years ago
Suppose I have the following sequence in an event:
BABCCABAA
There are 9 elements in the sequence and I want to find the number of arrangements I get can out of this 9 elements. The order does matter but because there are repeated elements such as 4A's, 3B's, etc, it becomes not as easy as just 9 factorial. For example, the following 2 are the considered only one arrangement:
\[BA_3BCCA_4BA_1A_2\]
and
\[BA_1BCCA_4BA_2A_3\]
So how can I find the number of arrangements when there are such repeated elements in it?
xEnOnn
 3 years ago
Suppose I have the following sequence in an event: BABCCABAA There are 9 elements in the sequence and I want to find the number of arrangements I get can out of this 9 elements. The order does matter but because there are repeated elements such as 4A's, 3B's, etc, it becomes not as easy as just 9 factorial. For example, the following 2 are the considered only one arrangement: \[BA_3BCCA_4BA_1A_2\] and \[BA_1BCCA_4BA_2A_3\] So how can I find the number of arrangements when there are such repeated elements in it?

This Question is Closed

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1It's simple, \( \large \frac{9!}{4! \times 3! \times 2!} = 1260 \)

jerwyn_gayo
 3 years ago
Best ResponseYou've already chosen the best response.0permutation. . i'll go with FFM

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.2Does this come from a formula? What's the rationale behind this equation?

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.2Also, what if order does not matter? ie, combinations?

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Put separate stickers on each of the four 'A', three 'B', and one of the two C's, to distinguish them. Now there are 9 distinct letters can be arranged in 9! ways. Remove the stickers, of all A's, then all B's and then all C's. Each time we remove the stickers, 4! arrangements collapse into 1 for A's, 3! arrangement collapses to 1 for B's and 2! to 1 for C. So the number of arrangements the given 9 letters is \[ \large \frac{9!}{4! \times 3! \times 2!} \]

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.2Based on your explanation, if the order does not matter, can I say it would then be this: \[\large \frac{9!}{(4! \times 3! \times 2!) \times 3!}\]

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1No that is incorrect interpretation for my explanation.

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.2But if the order does not matter, then I need to divide away the number of stickers I remove too, right? Although I remove 9 stickers in total, I can't divide by 9! again. Otherwise the equation will become a fraction.

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1What you mean by order does not matter, my answer assumes it means that A's and B's and C's are indistinguishable.

jerwyn_gayo
 3 years ago
Best ResponseYou've already chosen the best response.0at xenon, i think you can only do that to combination problems, this is permutation. .

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.2@jerwyn gayo oh yea...I think you are right. Thanks FFM for the help!! :)

xEnOnn
 3 years ago
Best ResponseYou've already chosen the best response.2Because I think if I want "order does not matter", it will just be 1.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.