Yahoo Answers: Answers and Comments for How many ways can a 5person committee be selected from a group of 8 people? [Mathematics]
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From Anonymous
enGB
Tue, 24 May 2011 20:59:51 +0000
3
Yahoo Answers: Answers and Comments for How many ways can a 5person committee be selected from a group of 8 people? [Mathematics]
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From Bob: It depends on how you define a committee. If t...
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Tue, 24 May 2011 21:06:37 +0000
It depends on how you define a committee. If there are different positions then the order DOES matter. However, if each spot is equal, then things cannot be repeated. Meaning:
A B C D E
Would be the same as
B A C D E
And as such would not be counted again. When order is not important, it's called a combination, whereas you used permutations, which is when order IS important. Your teacher may have been thinking about combinations.
If nPk is the number of permutations when there are k spots and n things and nCk is the number of combinations when there are k spots and n things, then:
nPk = n!/(nk)!
nCk = n!/((nk)!k!)
Imagine you have 3 chairs. Chairs 1, 2, and 3. You have 7 unique letters, a  g. How can we arrange them?
abc
 d
 e
 f
 g
cb
 d
 e
 f
 g
db
 c
 e
 f
 g
eb
 c
 d
 f
 g
fb
 c
 d
 e
 g
gb
 c
 d
 e
 f
bac
 d
 e
 f
 g
Etc, etc...
Do you see any patterns? There are 3 levels of letters, in a way. There are the letters farthest to the left, in the middle, and farthest to the right.
For EVERY second level letter, there were 5 letters to the far left.
For EVERY first level letter, were 6 letters in the middle.
There were 7 first level letters because we had 7 different letters.
7, 6, and 5. Why those numbers? Well, if you're a letter and you're in the second level, there're only 71 possibilities because there's already a letter in the first level!
If you're a letter in the third level, there're only 72 possibilities because there are letters in the first and second level already!
So, number of ways to arrange 7 letters into 3 spots is 7 * 6 * 5. I see a pattern: It's 7 multiplied by the itself and the two numbers below it. In math, a factorial, denoted by n!, is a positive integer multiplied by every whole number one less than it until 1. So, 5! would be 5 * 4 * 3 * 2 * 1, 3 factorial would be 3 * 2 * 1, etc...
This sort of looks like 7 factorial, except we're only taking the first 3 terms. How do we write that mathematically?
7 * 6 * 5 * 4 * 3 * 2 * 1
If we could somehow cancel out the 4 * 3 * 2 * 1, then we'd be good! Well, that's just 4!, and if you try this with other examples, you'll notice that it's:
(number of things to choose from  number of spots)!
Let's define the number of things to choose from as n, and the number of spots as k.
Then:
nPk = n!/(nk)!
Because We're taking the first k terms of n!.
How does this apply to combinations? Well, combinations are just permutations where things cannot be repeated. So how many ways can things be repeated?
Well, if there are 3 spots, then things can look like:
abc
acb
bac
bca
cab
cba
There are 6 ways to arrange it. Another way of thinking of it is there are a total of 3 things to choose from with 3 spots, so it's 3P3, and using the formula we discovered, that's just 3!/1 (0! is defined to be 1). 3! is 6.
The bottom term will always cancel out, so the number of ways things can be repeated will always be (number of spots)!.
If we divide the number of permutations by the number of times they were repeated, then we'll get the number of unique configurations!
So:
nCk = n!/((nk)!k!)
(^ copied and pasted from another of my answers)
So, plugging it into that formula, we get:
8C5 = 8!/((85)!5!)
8C5 = 8!/((3!5!)
8C5 = 8*7*6*5*4 / 5*4*3*2*1
8C5 = 8*7*6*5*4 / 5*4*3*2*1
8C5 = 6720/ 120
8C5 = 56
56 different ways.

From husoski: Your answer would have been right if there wer...
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Tue, 24 May 2011 21:06:46 +0000
Your answer would have been right if there were, say, 5 different named positions on the committee. To fix your answer, divide by the number of ways to order 5 items: 5! = 120.
6720/120 = 56 ways.
The number of ways to pick k members from a set of n distinct items, without regard to order, is:
C(n,k) = n C k = [n*(n1)*...*(nk+1)] / (1*2*...*k)
You got the numerator part of this. The denominator is the k! part. Another way to write it:
C(n,k) = n! / [k! * (nk)!]

From Anonymous: 8*7*6*5*4 assumes that the ordering is importa...
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Tue, 24 May 2011 21:04:52 +0000
8*7*6*5*4 assumes that the ordering is important. However, the order in the 5person committee is not important. Hence, you use 8 choose 5 = 8! / (5! (85)!).

From Anonymous: 8c5=56
8!/(8!5!)5!
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Tue, 24 May 2011 21:04:34 +0000
8c5=56
8!/(8!5!)5!

From David: it is a combination problem
8c5
8! / [5!...
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Tue, 24 May 2011 21:03:34 +0000
it is a combination problem
8c5
8! / [5!(8  5)!] = 56
56 different 5 person committees can be chosen from 8 people

From kosier: Committees could desire to be chosen via whoev...
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Mon, 05 Dec 2016 04:00:35 +0000
Committees could desire to be chosen via whoever has the utmost reward. it won't be random "drawn out of a hat" determination. it incredibly is stupid and could make for an unproductive committee. in case you go with to choose for the best and maximum useful committee, then get rid of any gender circumstances, and seem on the resume of all the conceivable applicants and decide the best 5.