Four Practice Problems:

Permutations and

Combinations

(Click the right arrow to begin.)

Problem 1

QUESTION: In how many ways can 3 men each be assigned to a hotel room if there are 8 hotel rooms available? (One room is as good as any other.)

 

(Click the down arrow for the solution.)

Solution 1

  • We're drawing X = 3 rooms from a mother set of n = 8 rooms.
  • The order of the rooms is of no concern, so it's a combination.

 

 

 

(Click the right arrow for the next problem.)

_nC_X = \dfrac{n!}{X!(n-X)!} = \dfrac{8!}{3!5!} = \dfrac{40,320}{6 \times 120} = 56
nCX=n!X!(nX)!=8!3!5!=40,3206×120=56_nC_X = \dfrac{n!}{X!(n-X)!} = \dfrac{8!}{3!5!} = \dfrac{40,320}{6 \times 120} = 56

Problem 2

QUESTION: Bob has prepared 7 dishes for a special dinner for his new girlfriend. She tells him that she can only eat 4 of them. So now Bob has to decide what to serve her and in what order.

In how many different ways can Bob serve a different sequence of 4 dishes from a line-up of  7 different dishes?

 

(Click on the down arrow for the solution.)

Solution 2

  • The order in which Bob serves the dishes matters, so it's a (partial) permutation.
  • n = 7 and X = 4

 

 

 

(Click the right arrow for the next problem.)

_nP_X = \dfrac{n!}{(n-X)!} = \dfrac{7!}{(7-4)!} = \dfrac{5040}{6} = 840
nPX=n!(nX)!=7!(74)!=50406=840_nP_X = \dfrac{n!}{(n-X)!} = \dfrac{7!}{(7-4)!} = \dfrac{5040}{6} = 840

Problem 3

QUESTION: Bob has 4 kids, but only 2 pieces of candy. (The candy is the same.) So he decides to just give the candy to the first 2 kids he sees when he gets home.

In how many ways can Bob give 2 of his kids candy?

 

(Click the down arrow for the solution.)

Solution 3

 

  • Since the order that the two chosen kids gets the candy doesn't matter (e.g., Billy then Bobby = Bobby then Billy), it's a combination with n = 4 and X = 2.

 

 

 

(Click the right arrow for the last problem.)

_nC_X = \dfrac{n!}{X!(n-X)!} = \dfrac{4!}{2!2!} = \dfrac{24}{2 \times 2} = 6
nCX=n!X!(nX)!=4!2!2!=242×2=6_nC_X = \dfrac{n!}{X!(n-X)!} = \dfrac{4!}{2!2!} = \dfrac{24}{2 \times 2} = 6

Problem 4

QUESTION: Bob has 5 employees but only 3 parking spots for them. The one next to the door is the best spot; the worst is the one by the dumpster. These 5 employees race every day to get to work first to grab these parking spots.

How many ways can the 5 employees grab those 3 parking spots?

(Click the down arrow for the solution.)

 

 

 

Solution 4

  • Since it matters which parking spot you get, the order matters. This is a permutation.
  • n = 5, X = 3

 

 

 

(That's it. Close the window when you're done.)

_nP_X = \dfrac{n!}{(n-X)!} = \dfrac{5!}{(5-3)!} = \dfrac{120}{2} = 60
nPX=n!(nX)!=5!(53)!=1202=60_nP_X = \dfrac{n!}{(n-X)!} = \dfrac{5!}{(5-3)!} = \dfrac{120}{2} = 60

1. Probability - Permutation and Combination Problems

By smilinjoe

1. Probability - Permutation and Combination Problems

Multinomial arrangements of mother sets sorted into two groups (= binomial arrangements).

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