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1. State if each scenario involves a permutation or a combination. Then find the number of possibilities:

(a) Castel and Joe are planning trips to three countries this year. There are 7 countries they would like to visit. One trip will be one week long, another two days, and the other two weeks.

(b) You are setting the combination on a three-digit lock. You want to use the numbers 123 but don't care what order they are in.

(c) A group of 25 people are going to run a race. The top 8 finishers advance to the finals.

Answer :

a. The number of possibilities that Castel and Joe are planning trips is [tex]35[/tex]

b. The number of possibilities of arranging three digit number is,6

c. The number of possibilities are [tex]1081575[/tex].

Permutation and combination :

a. We have to find the number of possibilities that Castel and Joe are planning trips to three countries this year out of 7 countries.

You need to pick 3 countries from a total of 7,

[tex]^{7} C_{3}=\frac{7!}{3!*4!} \\\\^{7} C_{3}=\frac{7*6*5*4!}{6*4!} =35[/tex]

b. The number of possibilities of arranging three digit number is,

[tex]3!=3*2=6[/tex]

c. A group of 25 people are going to run a race. The top 8 finishers advance to the finals.

you just want 8 people out of the 25 available.

the total amount of possible cases are,

[tex]^{25}C_{8}=\frac{25!}{8!*17!} =1081575[/tex]

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Rewritten by : Barada

Answer:

a) This is a permutation, there are 210 ways to do it

b) This is a permutation with 6 possibilities

c) This is a combination with a total of 1081575 possibilities

Step-by-step explanation:

a) Here the order does matter bacause it is not the same that they go to France in the weak trip than they go there for 2 days, as a result, this scenario is a permutation.

You need to pick 3 countries from a total of 7, but the order matters, so the total amount of ways to do this is

[tex] {7 \choose 3} * 3! = \frac{7!}{(7-3)!} = 210 [/tex]

b) Here each permutation of 123 will give you a possible (and a different one from the others) lock-combination, so this is again a permutation.

The total amount of ways we have to do the selectio of the lock-combination is the total amount of permutations of a set of 3 elements, in other words, it is 3! = 6.

c) If you only cares about who advances to the finals, then the order doesnt matter here, you just want 8 people out of the 25 available. This is therefore a combination, and the total amount of possible cases are

[tex] {25 \choos 8} = 1081575 [/tex] .

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