In the given problem, apply your knowledge in algebraic expression and basic mathematics especially in solving word problem about combinatorics.
In how many ways can a committee of 9 students be chosen from 8 juniors and 10 seniors if there must be 5 seniors in the committee?
a committee compose of 9 students
8 juniors
10 seniors
How many ways can a committee compose of 9 students and must be 5 seniors?
The formula for combination is
combination (nCr) = n!/r!(n-r)!
9 students → 5 seniors + 4 students (either junior or senior)
nCr (seniors) = 10C5 = 10!/5!(10-5)!
nCr = 10C5 = 252
nCr (4 students) = 18C4 = 18!/4!(18-4)!
nCr = 18C4 = 3060
nCr (committee) = 10C5 × 18C4
nCr = 252 × 3060
nCr = 771 120
771 120 ways
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Verified answer
In the given problem, apply your knowledge in algebraic expression and basic mathematics especially in solving word problem about combinatorics.
Problem
In how many ways can a committee of 9 students be chosen from 8 juniors and 10 seniors if there must be 5 seniors in the committee?
Given
a committee compose of 9 students
8 juniors
10 seniors
Asked
How many ways can a committee compose of 9 students and must be 5 seniors?
Solution
The formula for combination is
combination (nCr) = n!/r!(n-r)!
9 students → 5 seniors + 4 students (either junior or senior)
nCr (seniors) = 10C5 = 10!/5!(10-5)!
nCr = 10C5 = 252
nCr (4 students) = 18C4 = 18!/4!(18-4)!
nCr = 18C4 = 3060
nCr (committee) = 10C5 × 18C4
nCr = 252 × 3060
nCr = 771 120
Answer
771 120 ways
Learn more about combination here at brainly.ph/question/13262774
#CarryOnLearning
#BRAINLYFAST