Probability - Gravner - 16.a

Question

A group of $18$ Scandinavians consists of $5$ Norwegians, $6$ Swedes, and $7$ Finns. They are seated at random around a table. Compute the following probabilities:

(a) that all the Norwegians sit together,

Answer 1

Circular permutation = $\left (n-1 \right )!$

Total permutations = $17!$

(a) that all the Norwegians sit together 

Total permutations where all the Norwegians sit together $= 13! \times 5!$ 

but why  $= 13! \times 5!$? 

all Norwegians sitting on 5 consecutive seats so they can be arranged $5!$ times and everyone including Norwegians as a group can be arranged in $((6 + 7 + 1 )-1)! = 13!$

So Probability that all the Norwegians sit together = $\large \frac{13! \times 5!}{17!}$

(b) that all the Norwegians and all the Swedes sit together

Total permutations where all the Norwegians and all swedes sit together = $8! \times 6! \times 5! $ 

So Probability that all the Norwegians and all swedes sit together = $\large \frac{8! \times 6! \times 5!}{17!}$

(c) that all the Norwegians, all the Swedes, and all the Finns sit together.

Total permutations where all the Norwegians, all the Swedes, and all the Finns sit together = $(3-1)!$

So Probability that all the Norwegians, all the Swedes, and all the Finns sit together = $\large \frac{2! \times 6! \times 5! \times 7!}{17!}$