(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Homework Help: 2nd combinatorial problem

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