a). How many ways can a person select three appetizers and two soups if there are six appetizers and five soups on the dinner menu?
b). Three married couples have bought tickets for six seats in a row for a movie.
- HOw many ways can they be seated?
- How many ways can they be seated if each couple is to sit together with the husband to the left of the wife.
- HOw many ways can they be seated in each couple is to sit together.
- How many ways can they be seated if all the men are to sit together and all the women are to sit together.
a) To find the number of ways a person can select three appetizers and two soups, we can use the concept of combinations.
The number of ways to select three appetizers from a menu of six can be found using the formula for combinations: nCr = n! / (r! * (n-r)!), where n is the total number of items and r is the number of items to be chosen.
So, for selecting three appetizers from six, the calculation would be: 6C3 = 6! / (3! * (6-3)!), which simplifies to 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20.
Similarly, for selecting two soups from five, the calculation would be: 5C2 = 5! / (2! * (5-2)!), which simplifies to 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10.
To find the total number of ways for both selections to occur together, we multiply the number of ways for each selection: 20 * 10 = 200.
Therefore, there are 200 ways for a person to select three appetizers and two soups from a menu of six appetizers and five soups.
b) Let's consider each scenario separately:
1. Number of ways they can be seated without any restrictions:
To find the number of ways they can be seated without any restrictions, we can use the concept of permutations. The formula for permutations is nPr = n! / (n-r)!, where n is the total number of items and r is the number of items to be arranged.
So, for six seats, the calculation would be: 6P6 = 6! / (6-6)! = 6! / 0! = 6! / 1 = 6!.
Therefore, there are 6! = 720 ways the six people can be seated without any restrictions.
2. Number of ways they can be seated if each couple sits together with the husband to the left of the wife:
In this scenario, a couple is considered as a single entity. So, we have three entities: three couples.
To arrange three entities, the calculation would be: 3! = 3 factorial = 3 * 2 * 1 = 6.
Additionally, within each couple, the husband must be seated to the left of the wife. So, the number of ways each couple can be seated within themselves is 2! = 2 factorial = 2 * 1 = 2.
Therefore, the total number of ways they can be seated if each couple sits together with the husband to the left of the wife is: 6 * 2 * 2 * 2 = 48.
3. Number of ways they can be seated if each couple sits together:
In this scenario, again, a couple is considered as a single entity. So, we have three entities: three couples.
To arrange three entities, the calculation would be: 3! = 3 factorial = 3 * 2 * 1 = 6.
Within each couple, they can be seated in two ways: husband-wife or wife-husband.
Therefore, the total number of ways they can be seated if each couple sits together is: 6 * 2 * 2 * 2 = 48.
4. Number of ways they can be seated if all the men sit together and all the women sit together:
In this scenario, we treat all the men as a single entity and all the women as another single entity. So, we have two entities: men and women.
The entities themselves can be arranged in 2! = 2 factorial = 2 * 1 = 2 ways.
Within each entity, the members can be seated in factorial ways.
Therefore, the total number of ways they can be seated if all the men sit together and all the women sit together is: 2 * 3! * 3! = 2 * 6 * 6 = 72.
Thus, the number of ways they can be seated in different scenarios is as follows:
- Without any restrictions: 720 ways
- Each couple sits together with the husband to the left of the wife: 48 ways
- Each couple sits together: 48 ways
- All the men sit together and all the women sit together: 72 ways.