Kenneth Rosen Edition 6th Exercise 5.5 Question 19 (Page No. 380)

Question

Suppose that a large family has 14 children, including two
sets of identical triplets, three sets of identical twins, and
two individual children. How many ways are there to seat
these children in a row of chairs if the identical triplets or
twins cannot be distinguished from one another?

 

The answer is so # of ways = 14! /( 3!*3!*2!*2!*2!) =302 702 400

 

But isn't  the total permutation of all the 14 students where similar objects are divided which gives all the arrangement of all the similar objects when they are neighborhood or not.

 

But the question had asked for similar objects sit side by side.

 

Please clarify

Comments

How many ways are there to seat these children in a row of chairs if the identical triplets or twins cannot be distinguished from one another?

it means howmany ways they can sit, but they didn't mention any restriction.

 

the identical triplets or twins cannot be distinguished from one another

means you can't identify them clearly.

let the two triplets are (a,a,a) and (b,b,b)  ---- in this you can't exactly know who a is among (a,a,a)

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I apologize for the confusion.

 

What I wanted to know if the perambulation of all this 14 people and given query which has asked "How many ways are there to seatthese children in a row of chairs if the identical triplets or twins cannot be distinguished from one another?" are they same.

 

As both gives the same answer.

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if the perambulation of all this 14 people

14 !

 

generally, by default persons are non-identical, let if they are balls ( by default identical ) or some other objects instead of people, we generally use $\frac{14!}{3! * 3! * 2!*2!*2!}$