# 2nd combinatorial problem

1. Jul 18, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
Decide in how many ways the letters of the word ABRACADABRA can be arranged in a row if C, R and D are not to be together.

2. Relevant equations
The number of ways of arranging n objects which include 'a' identical objects of one type, 'b' identical objects of another type,.... is
n!/(a!b!....)

n objects divided into m groups with each group having G1, G2, ..., Gm objects respectively has m! * G1! * G2! * ... *Gm!

3. The attempt at a solution
A: 5
B: 2
R: 2
C: 1
D: 1

11 letters in total.

There are a few identical letters so the total number of ways of permuting the objects accounting for the identical letters is 11!/(5!2!2!)

The total number of ways of arranging the letters such that C, R and D are together is:
8!4!/(5!2!2!) as there are 8 groups, with one group containing 4 letters. However given the identical letters, we divide by the same number as above.

So (11!-8!4!)/(5!2!2!)=81144

The answers suggested 78624

I can't see what is wrong with my reasoning.

2. Jul 18, 2007

### f(x)

total ways of arranging = 11! / (5!.2!.2!)

exceptions=> ways in which CRD are together = (9!.3!) / (5!.2!.2!)

why (9!.3!) / (5!.2!.2!) ? Consider CRD to be one grp and the rest 9 to be another. Then 9! ways or arranging those letters, 3! ways of arranging CRD, 5 A's are common, 2 B's. Another 2! in Dr. because there are two R's, and it doesnt matter which R is in the word CRD

3. Jul 18, 2007

### pivoxa15

Nice one f(x). I see now.