At a party there are four Germans, three Italians and two Frenchmen. Suppose there are chairs arranged in a row and determine the probability that the Germans: a) Sit together. b) They do not sit together.

### Part a) The Probability That the Germans Sit Together

To find the probability that the Germans sit together, we need to calculate the ratio of the number of arrangements where the Germans sit together to the total number of arrangements.

Given:
- Total number of people at the party, N = 9.
- Number of Germans, G = 4.
- Number of Italians, I = 3.
- Number of Frenchmen, F = 2.

The total number of ways to arrange all 9 people without any restriction is 9!.

The number of ways to arrange the 6 units (German unit, Italian unit, French unit) is 6!.

Within the German unit, the 4 Germans can be arranged among themselves in 4! ways.

Therefore, the probability that the Germans sit together is:

P(\text{Germans together}) = \frac{6! \times 4!}{9!}

Calculating this ratio, we get:

P(\text{Germans together}) = \frac{6! \times 4!}{9!} = \frac{720 \times 24}{362880} = \frac{17280}{362880} = \frac{1}{21}

So, the probability that the Germans sit together is \frac{1}{21} .

### Part b) The Probability That the Germans Do Not Sit Together

To find the probability that the Germans do not sit together, we subtract the probability that they do sit together from 1:

P(\text{Germans not together}) = 1 - P(\text{Germans together})

Substitute the calculated value for P(\text{Germans together}):

P(\text{Germans not together}) = 1 - \frac{1}{21}

Calculating this probability, we get:

P(\text{Germans not together}) = 1 - \frac{1}{21} = \frac{20}{21}

So, the probability that the Germans do not sit together is \frac{20}{21} .

### Answer:
a) The probability that the Germans sit together is \frac{1}{21} .
b) The probability that the Germans do not sit together is \frac{20}{21} .