Question

If $4$Americans, $3$French people, and $3$British people are to be seated in a row, how many seating arrangements are possible when people of the same nationality must sit next to each other?

Short answer

The required no. of seating arrangements are$5184$.

Step-by-step solution

Step 1. Given information.

It is given that,

No. of Americans =$4$

No. of French =$3$

No. of British =$3$

All these people are to be seated in a row.

Step 2. Find the required no. of seating arrangements.

Three groups can be arranged in$=3!\text{ways}=3\times 2\times 1=6\text{ways}$

Four Americans can be arranged in = $4!\text{ways}=4\times 3\times 2\times 1=24\text{ways}$

Three French can be arranged in $=3!\text{ways}=3\times 2\times 1=6\text{ways}$

Three British can be arranged in $=3!\text{ways}=3\times 2\times 1=6\text{ways}$

Total no. of waysrole="math" localid="1648406231161" $=6\times 24\times 6\times 6=5184$

Therefore, the required no. of seating arrangements are$5184$.