Please help.
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.
B)
6!
a) To determine the number of ways a person can select three appetizers and two soups, you need to use the concept of combinations.
The formula for combinations is given by:
nCr = n! / (r!(n-r)!)
Where n represents the total number of items and r represents the number of items to be selected.
In this case, there are 6 appetizers and 5 soups. To choose 3 appetizers from 6, use the combination formula:
6C3 = 6! / (3!(6-3)!)
= 6! / (3!3!)
Simplifying further:
6! = 6 x 5 x 4 x 3 x 2 x 1
3! = 3 x 2 x 1
Plugging these values into the formula:
6C3 = (6 x 5 x 4) / (3 x 2 x 1)
= 20
Therefore, there are 20 ways to select three appetizers from a menu of six.
To choose 2 soups from 5, apply the combination formula using the same method:
5C2 = 5! / (2!(5-2)!)
= 5! / (2!3!)
= (5 x 4) / (2 x 1)
= 10
Hence, there are 10 ways to select two soups from a menu of five.
b) Let's address each scenario one by one:
1. The number of ways the three married couples can be seated in six seats in a row:
To calculate this, we will use the permutation formula as the ordering matters.
The formula for permutations is given by:
nP r = n! / (n - r)!
For this scenario, there are six seats available, and we need to seat six people. Therefore:
6P6 = 6! / (6 - 6)!
= 6! / 0!
= 6!/1
= 6! (since 0! = 1)
Simplifying further:
6! = 6 x 5 x 4 x 3 x 2 x 1
6! = 720
Therefore, there are 720 ways the three married couples can be seated in six seats in a row.
2. The number of ways the couples can be seated if each couple sits together with the husband to the left of the wife:
In this case, each couple can be treated as a single entity. Hence, there are three entities to be seated.
Now, consider these three entities as separate seats. The number of ways to arrange these three entities is given by:
3P3 = 3! / (3 - 3)!
= 3! / 0!
= 3!
Simplifying,
3! = 3 x 2 x 1
3! = 6
Therefore, there are 6 ways the couples can be seated if each couple sits together with the husband to the left of the wife.
3. The number of ways the couples can be seated if each couple sits together:
In this case, each couple is treated as a single entity, and there are three entities to be seated.
The number of ways to arrange these three entities is:
3P3 = 3! / (3 - 3)!
= 3! / 0!
= 3!
Therefore, there are 3 ways the couples can be seated if each couple sits together.
4. The number of ways the couples can be seated if all the men sit together and all the women sit together:
To solve this scenario, treat all the men as a single entity and all the women as another single entity. Now, we have two entities to be seated.
The number of ways to arrange these two entities is:
2P2 = 2! / (2 - 2)!
= 2! / 0!
= 2!
Hence, there are 2 ways the couples can be seated if all the men sit together and all the women sit together.