Provide an appropriate response. From 10 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible? O 151,200 O 2520 O 5040 O 210
Provide an appropriate response. From 10 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible? O 151,200 O 2520 O 5040 O 210
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![**Combinatorics Problem**
*Provide an appropriate response.*
From 10 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible?
- ○ 151,200
- ○ 2520
- ○ 5040
- ○ 210
*Explanation:*
This problem involves combinations, where the order of selection does not matter. The number of ways to choose a committee of 4 from 10 candidates can be calculated using the combination formula \(\binom{n}{r}\), which is:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
In this case, \(n = 10\) and \(r = 4\):
\[
\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
Thus, the correct answer is 210 different committees.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fac9c6941-fcbd-4d0f-a0fa-57d2e8e41463%2Fabbe5b3b-2fd8-4a3f-9dd9-4e9c16b8148e%2F7rlr7xx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Combinatorics Problem**
*Provide an appropriate response.*
From 10 names on a ballot, a committee of 4 will be elected to attend a political national convention. How many different committees are possible?
- ○ 151,200
- ○ 2520
- ○ 5040
- ○ 210
*Explanation:*
This problem involves combinations, where the order of selection does not matter. The number of ways to choose a committee of 4 from 10 candidates can be calculated using the combination formula \(\binom{n}{r}\), which is:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
In this case, \(n = 10\) and \(r = 4\):
\[
\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210
\]
Thus, the correct answer is 210 different committees.
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