How many different four person committees can be formed from a total of 8 people?

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#

(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#((7),(3)) #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#

Answer link

EZ as pi

Apr 17, 2017

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

Explanation:

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#9xx8xx7xx6#different committees, however this will include the same combinations of people.
There are#4xx3xx2xx1#ways in which 4 people can be chosen.

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

#(9xx8xx7xx6xx5)/(4xx3xx2xx1) = 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:

#(7xx6)/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;

The term “combination” is thrown around loosely, and usually in the wrong way. Things like, “Hey, what’s the suitcase lock combination?” are said But what one really ought to be saying is “Hey, what’s the suitcase lock permutation?” So what’s the difference? And what exactly are a permutation and combination? They are two very different terms. Let’s learn about them in detail,

Permutation

A permutation is an act of arranging objects or given quantity maybe numbers from a group of objects or collection given in a particular order as per given conditions. Example – How many 2 letter words are there that can be formed by using the letters in the word LATE? Answer is 4P2 (pronounced as 4 p 2) = 4!/(4 – 2)! = 4!/(2)! = (4 × 3 × 2 × 1)/(2 × 1) = 24/2 = 12.

Combination

The combination is the way of selecting the objects or given quantity maybe numbers from a group of objects or collection from a group of objects or collection, in such a way that the order of the objects does not matter. For example – how many groups of 2 people can be selected from 4 people? Answer – 4 C 2(pronounced as 4 C 2) = 4!/(4 – 2)!(2)! =  4!/(2)!(2)! = (4 × 3 × 2 × 1)/(2 × 1)(2 × 1) = 24/4 = 6.

The formula for permutations and combinations

• The formula for permutations is: nPr = n!/(n – r)!
• The formula for combinations is: nCr = n!/[r! (n – r)!]

Find the number of committees of the size of 4 formed from 8 people. 

Answer:

Here, analyzing the problem given, the hint is to select 4 people from 8 people. The biggest confusion is whether to apply permutation or combination? So start thinking way that whether the order of people will make a difference? if yes then go for permutation otherwise it’s a combination that will solve the question.

In this question let’s first select p1, p2, p3, p4 in the written order and then select p2, p1, p4, p3. Does that make a different group here? The answer is no. Because irrespective of order the first, second, third, fourth person have been selected. So here, go with the combination as the order doesn’t matter.

Applying combination using formula, nCr = n!/[r! (n – r)!]

Here, n = 8 and r = 4

So 8C4 = 8!/(8 – 4)!(4)! = 8!/(4)!(4)! = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1)(4 × 3 × 2 × 1) = 70

Similar Problems

Question 1: Find the number of committees of the size of 3 formed from 5 people.

Answer: 

5C3 =  5!/(5 – 3)!(3)! = 5!/(2)!(3)! = (5 × 4 × 3 × 2 × 1) / (2 × 1)(3 × 2 × 1) = 10

Question 2: How many different committees of 3 members can be chosen out of 5 people in a group so that one particular person is always chosen?

Answer: 

Since one particular person is always to be taken from the available 5 people in the committee of the 3. So in fact, choose 2 persons from the remaining 4 and that can be done in C(4, 2) = 4C2 = 4!/2! 2! = 6 number of ways.

Question 3: How many committees of 5 consisting of 3 men and 2 women can be formed from 8 men and 6 women?

Answer: 

Well, one can form 8 choose 3 groups of men, and for each of those, one can choose any of the 6 choose 2 groups of women.

nCr = n!/[r! (n – r)!] 

= 6C2 = (6)!/((2!)(6 – 2)!) = 15 

= 8C3 = (8)!/((3!)(8 – 3)!) = 56

So, 56 × 15 = 840 possible combinations, assuming one doesn’t care about anything other than the number of men, the number of women.

If one of those people is to be the chair, then there are for each possible group, 5 possible chairs, so multiply that by 5, 840 × 5 = 4200

How many committees can be formed from 8 people?

n=8, r=3 n = 8 , r = 3 . Therefore, 56 committees of 3 people can be formed from 8 people.

How many committees of size 4 can be formed from a group of 9 people?

In 126 ways a committee of 4 be chosen from a group of 9 people.

How many ways can a committee consisting of 4 members be formed from 7 people?

1 Answer. A committee of 4 people be selected from a group of 7 people in 35 ways.

How many 4 Person committees can be selected from a group of 10 people?

We are to determine the number of 4 member committees that can be formed from 10 people. This is simple combination problem since order is not important here. So Number of ways = 10C4 = 210.