SOLUTION: 1.) A dance instructor asks each student to do 4 out of 10 dance routines. Of the 10 dance routines, 2 are easy, 5 are moderately difficult and 3 are difficult. In how many ways ca

Algebra ->  Permutations -> SOLUTION: 1.) A dance instructor asks each student to do 4 out of 10 dance routines. Of the 10 dance routines, 2 are easy, 5 are moderately difficult and 3 are difficult. In how many ways ca      Log On


   

Question 1177694: 1.) A dance instructor asks each student to do 4 out of 10 dance routines. Of the 10 dance routines, 2 are easy, 5 are moderately difficult and 3 are difficult. In how many ways can a student select each of the following for the 4 dance routines?
A. 4 moderately difficult routines
B. 4 easy or moderately difficult routines
C.2 moderately difficult and 2 difficult routines
D. 1 easy and 3 difficult routines
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Let's solve each part of the problem step-by-step.
**Understanding Combinations**
We'll use combinations (nCr) to calculate the number of ways to choose routines. The formula for combinations is:
* nCr = n! / (r! * (n-r)!)
Where:
* n = total number of items
* r = number of items to choose
**A. 4 Moderately Difficult Routines**
* Total moderately difficult routines: 5
* Routines to choose: 4
* Number of ways = 5C4 = 5! / (4! * (5-4)!) = 5! / (4! * 1!) = 5
**B. 4 Easy or Moderately Difficult Routines**
* Total easy routines: 2
* Total moderately difficult routines: 5
* Total easy or moderately difficult routines: 2 + 5 = 7
* Routines to choose: 4
* Number of ways = 7C4 = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
**C. 2 Moderately Difficult and 2 Difficult Routines**
* Moderately difficult routines: 5
* Difficult routines: 3
1. Choose 2 moderately difficult routines: 5C2 = 5! / (2! * 3!) = (5 * 4) / (2 * 1) = 10
2. Choose 2 difficult routines: 3C2 = 3! / (2! * 1!) = 3
* Total number of ways = 10 * 3 = 30
**D. 1 Easy and 3 Difficult Routines**
* Easy routines: 2
* Difficult routines: 3
1. Choose 1 easy routine: 2C1 = 2! / (1! * 1!) = 2
2. Choose 3 difficult routines: 3C3 = 3! / (3! * 0!) = 1
* Total number of ways = 2 * 1 = 2
**Answers**
* **A. 4 moderately difficult routines:** 5 ways
* **B. 4 easy or moderately difficult routines:** 35 ways
* **C. 2 moderately difficult and 2 difficult routines:** 30 ways
* **D. 1 easy and 3 difficult routines:** 2 ways