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# Word Question

on
Both are same question but why one use the permutation and other combination ? Thanks.

1. How many different strings can be formed by rearranging the
letters in the word ABABA ?
C(5, 3) = 10

AAABB
AABAB
AABBA
ABAAB
ABABA
ABBAA
BAAAB
BAABA
BABAA
BBAAA

2. How many different ways are there to rearrange the letters in the word GOURMAND?
P(8,8)

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Commented:
Because the ababa string has repeated letters so you don't need to differentiate between aa and aa (same but reversed).
gourmand has all individual letters so every permutation is different.
ababa would produce many repeats.
Consulting and Network/Security Specialist
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Commented:
First one has double/tripple letters, so there's a smaller number of different combinations. (n! / (a! * (n-a)!)

For the second, there's no double letters, so you have 8 different letters, so 8! is the solution there ... first place can be one of 8 letters, second one of seven, and so on ...
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Commented:
Say you have a certain number of letters. In how many different ways can these things be arranged in a row? A permutation of some number of objects means the collection of all possible arrangements of those objects.

Note: 8 items have a total of 40,320 different combinations.

http://www.webmath.com/k8perm.html

The box comes up with an error when I enter
G
O
U
R
M
A
N
D
but the answer appears to be 40,320.

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Commented:
I can see where this might be a confusing question.

Strings are always PERMUTATIONS because order always matters.  They are not combinations.

1. How many different strings can be formed by rearranging the
letters in the word ABABA ?

The answer here is     5!/(3! * 2!)

The 5! is because there are five letters.   The 3! is from the  three A's.  The 2! is from the two B's.
It is a coincidence that the answer 5!/(3! * 2!)  is the same as  C(5, 3)  or  C(5, 2).
It only works for words that are made up of two distinct letters.

The number of permutations for the word  ABABAT  would be   6!/(3! * 2!)
There is no way to put this in the P(n,m)  or  C(m, n)  form.