# 10 in how many ways can a committee of 4 be chosen from a group of 9 people? Ideas

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4 person committees are possible from a group of 9 people if ? | Socratic

## A) there are no restrictions? B) Both John and Barbara must be on the committee? C) Either John or Barbara (but not both) must be on the committee?

Hopefully you know combinatoric idea that $\left(\begin{matrix}n \\ k\end{matrix}\right)$ means “n-choose-k”.

(a)

The number of different ways of choosing 4 items from a group of 9 items is simply: #((9),(4)) = (9!)/(5!4!) = 126#

(b)

You have already chosen 2 of the committee so you are now looking at choosing the remaining 2 from a group of 7:

#((7),(2)) = (7!)/(5! 2!) =21#

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(c)

For all committees that has John onboard, you need to choose 3 more, but you are choosing from a group of 7 as Barbara cannot be on the same committee. So in that scenario you have $\left(\begin{matrix}7 \\ 3\end{matrix}\right)$ ways of doing it.

The same applies if Barbara is to be on the committee and John excluded.

So, overall, the total number of ways to do this is:

#2 xx ((7),(3)) = (2 xx 7!)/(4!3!) = 70#

#A) 126# different committees
#B) 21# different committees
#C) 70# different committees

Recall: The number of different ways of arranging $n$ numbers is #n!#

If 4 people are to be selected from 9 people:
There are 9 different choices for the first person.
There are 8 different choices for the second person.
There are 7 different choices for the third person.
There are 6 different choices for the fourth person.

This gives $9 \times 8 \times 7 \times 6$ different committees, however this will include the same combinations of people.
There are $4 \times 3 \times 2 \times 1$ ways in which 4 people can be chosen.

A) Therefore, for 4 people chosen from 9 there are

$\frac{9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 126$ different committees.

B) If two people must stand, there are 2 people to be chosen from the remaining 7. Applying the same thinking as above this gives:

$\frac{7 \times 6}{2} = 21$ different committees.

C) if either one or the other must stand there are then 3 people who must be chosen from the remaining$7$ people;

$\frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

But this can happen in 2 different ways, depending on whether John or Barbara are on the committee.

So, $2 \times 35 = 70$ different committees can be formed.

## Extra Information About in how many ways can a committee of 4 be chosen from a group of 9 people? That You May Find Interested

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### How many 4 person committees are possible from a group of …

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• Sumary: Hopefully you know combinatoric idea that ((n),(k)) means “n-choose-k”. (a) The number of different ways of choosing 4 items from a group of 9 items is simply: ((9),(4)) = (9!)/(5!4!) = 126 (b) You have already chosen 2 of the committee so you are now…

• Matching Result: 9×8×7×6×54×3×2×1=126 different committees. B) If two people must stand, there are 2 people to be chosen from the remaining 7. Applying the same …

• Intro: How many 4 person committees are possible from a group of 9 people if ? | Socratic A) there are no restrictions? B) Both John and Barbara must be on the committee? C) Either John or Barbara (but not both) must be on the committee? Hopefully you know combinatoric idea…
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### How many ways can 4 people be chosen from a group of 9 …

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• Sumary: You can put this solution on YOUR website! It depends. Does the order of the group of 4 people matter? Let’s say your group of 9 consists of Alice, Bob, Carly, David, Ernest, Fran,…

• Matching Result: The question is this: Is Alice, Bob, Carly, and David different from David, Carly, Bob, and Alice? If you were just selecting 4 committee members, then the …

• Intro: SOLUTION: How many ways can 4 people be chosen from a group of 9? Tell wheater the situation is permutation or combination Question 144794: How many ways can 4 people be chosen from a group of 9? Tell wheater the situation is permutation or combination Answer by solver91311(24713)   (Show Source):…

### in how many ways can a committee of four be … – SOLUTION

• Author: algebra.com

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• Sumary: there are 10 people total and you want to choose a committee of 4 out of the 10.

• Matching Result: if the committee is to have only harry on it, then you get to choose 4 out of 9 again. some of those committees will have the same members on them, so you will …

• Intro: SOLUTION: in how many ways can a committee of four be chosen from a group of ten if barry and harry refuse to serve together on the same committee? i believe the formula will be as follows: there are 10 people total and you want to choose a committee of…

### What is the number of different committees of 4 people that …

• Author: gmatclub.com

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• Sumary: What is the number of different committees of 4 people that can be selected from a group of 10 people? A. 20 B. 40 C. …

• Matching Result: Hello, Out of 10 people, we need to select 4 people and this can be done in 10C4 ways. 10C4 can be written as 10 …

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## Frequently Asked Questions About in how many ways can a committee of 4 be chosen from a group of 9 people?

If you have questions that need to be answered about the topic in how many ways can a committee of 4 be chosen from a group of 9 people?, then this section may help you solve it.

70 ways

### What percentage of a group of 10 people can be divided into committees of four?

Since order is not relevant in this situation, the number of ways to form a committee of four members from a group of ten people is to be determined. Therefore, Number of ways = 10C4 = 210.

495 ways

### How many ways are there to choose a committee of five out of nine people?

Consequently, there are a total of 41 options.

### How many different ways are there to choose a committee of four from a group of seven?

Therefore, using the formulas (A) and (B), we have established that there must be 35 different ways for a committee of four people to be chosen from a group of seven.

### Out of a group of 7, how many teams of four people can be selected?

There are 35 different ways to choose a committee of four people from a group of seven.

### If order is not crucial, how many ways are there to select a committee of four people from a group of six?

Therefore, they can be selected in 15 different ways.

### How many ways are there to choose 4 students from a group of 10?

The number of ways to select four items from a set of ten items is = 10? 9? 8? 7 4? 3? 2? 1 = 5040 24 = 210.

### How many different ways are there to choose a three-person committee out of nine?

9C3=9! 6! ×3! =9×8×76=84.

### Four students can fit into a row of seven chairs how many different ways?

Option D is correct because there are 96 different ways that 4 students can arrange themselves on 7 chairs so that no seat is left vacant between the two students.

### How many different ways are there for a selection of 6 people to fill 4 seats in a row?

As a result, there are 360 possible arrangements for 4 people to sit in a row of 6 seats. Note: The formula we used to solve this problem is nPr=n! (n?r)!