Question

URGENT probability question!!! if x things are arranged linearly, with m of type A and n of type B, how many different arrangements are possible--guess this involves permutations, but otherwise, not sure HOW to start?

Answer 1

The problem does involve permutations. There are x! ways of arranging the objects but this has to be divided by m! and n! because the type A's can be arranged in m! ways and the B's can be arranged in n! ways without making a different arrangement.
Number of different arrangements = \[\frac{x!}{m!n!}\]

Answer 2

gee, thanks! would this act as a coefficient in this case where given type a of type A to choose from and b of type B to choose from? how and why does it work?

Answer 3

in that case it would be\[ (a+b)! \over a!b!\]

Answer 4

No. It would not be suitable in problems where selections are made. It relates only to permutations where the objects are of two different kinds.

Answer 5

Thanks for all replies!
For this equation, (a+b) / !a!b! what if we wanted to choose only x out of those a+b items total?

Answer 6

the question say there a a total 'x' items....
m of them are A,
n of them are B.

Answer 7

Yup, x in the sequence sorry for the confusion
meaning x total in the sequence of items we're arranging

Answer 8

We have a items from Type A, which within type A are distinct (e.g. one example, 9 colors to choose from a crayon box to line up), and b items from Type B. We want to arrange x of these total, take m of A and n of B.

Answer 9

let x mean just one thing in this post...
x is the total number of items .. and you are arranging all of them in the original question

Answer 10

Sure, x meaning total then, where x = a+b, let's say if we were to choose y of these x, where y = m+n?

Answer 11

if you have m Apples, and n Bananas

the number of ways you can arrange them in a line =
(m+n)!/ (m!n!)

let's say you have 3 A and 2 B
BBAAA
BABAA
BAABA
BAAAB

ABBAA
ABABA
ABAAB

AABBA
AABAB

AAABB

that's 10

Answer 12

Gee thanks!

Answer 13

But what happens when we have more choices? e.g. we're looking at 3 A, but we have 5 total distinct A: A1, A2, A3, A4, A5. supposing, we fix 2 B as if there B1 and B2 were all we have in an arrangement of 5:

we could have A1, A2, A3, B1, B2
A1, A2, A3, B2, B1
B1, B2, A1, A4, A5,
B2, A1, A5, B1, A3
etc.

then let's suppose we did not just suppose 2 B's--imagine there were 4 or so and were could choose a combination, 2 from 5 and then balance a permutation at the same time! a bit of a doozy for me!

Answer 14

* sorry in last para. 2 from 4

Answer 15

There is a Theorem relating to permutations that states:
If n things can be divided into c classes such that things belonging to the same class are alike while things belonging to different classes are different, then the number of permutations of these things taken all at a time is\[\frac{n!}{n _{1}!n _{2}!....n _{c}!}\]
\[n _{1}+n _{2}+.... +n _{c}=n\]

Answer 16

\[n _{j}\]is the number of things in the jth class

Answer 17

if there are X things ... and all of them are unique .... then the number of ways they can be arranged is just \(X!\)

...it doesn't mattter if some belong to A and others to B

Answer 18

@PaxPolaris Quite right.