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?

Short answer

302,702,400 ways to seat 14 children when there are 2 identical triplets and 3 pair of identical twins.

Step-by-step solution

Step 1: Use the formula for string factorial

Total number of arrangements of n objects if al are different = n!

If r! of them are same number of ways are given by

\(\begin{array}{l}\frac{{n!}}{{{n_1}!{n_2}!..{n_k}!}} = \frac{{n!}}{{r!}}\\r! = {n_1}!{n_2}!..{n_k}!\end{array}\)

The n! Is string length

r! are outcome of strings

Step 2: Solution of strings of 20-decimal digits

Let’s, applied r! and n! values,

There are 2 set identical triplets and 3 sets of identical twins

Here,

\(\begin{array}{l}n! = 14!\\r! = 3!,3!,2!,2!,2!\\\frac{{n!}}{{r!}} = \frac{{14!}}{{3!.3!.2!.2!.2!}}\\\frac{{n!}}{{r!}} = 302,702,400\end{array}\)