# 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
$_nC_X = \dfrac{n!}{X!(n-X)!} = \dfrac{8!}{3!5!} = \dfrac{40,320}{6 \times 120} = 56$

## 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
$_nP_X = \dfrac{n!}{(n-X)!} = \dfrac{7!}{(7-4)!} = \dfrac{5040}{6} = 840$

## 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
$_nC_X = \dfrac{n!}{X!(n-X)!} = \dfrac{4!}{2!2!} = \dfrac{24}{2 \times 2} = 6$

## 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
$_nP_X = \dfrac{n!}{(n-X)!} = \dfrac{5!}{(5-3)!} = \dfrac{120}{2} = 60$

By smilinjoe

# 1. Probability - Permutation and Combination Problems

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

• 2,763