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Permutation and Combination

Can any one please explain this question ? Thanks.

A committee of three is chosen from a group of 20 people. How many different committees are possible, if

a. the committee consists of a president, vice president, and treasurer ?
b. there is no distinction among the three members of the committee ?

Answer: P(20,3)=6,840 (b) 2.6.5=60
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mustish1
Asked:
mustish1
3 Solutions
 
mustish1Author Commented:
I think the formula on part a is P(n,r)=n!/(n-r)!
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ozoCommented:
You are correct about the first part.
The second part should be C(20,3)=n!/(r!(n-r)!)
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SuperdaveCommented:
The first one is 60*59*58, i.e. pick one of 60 people for president, pick one of 59 remaining people for vice-president, pick one of the remaining 58 for treasurer.

The second part is choosing all combinations of three people from 60.  People generally use the formula ozo gave to compute that.  You could also get it by dividing your first answer by 6 to account for the 6 (3!) ways you could rearrange the three positions on a committee that you are now going to count as one committee.
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mustish1Author Commented:
According to my formula the answer is still wrong
P(n,r)=n!/(n-r)!
P(20,3) = 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*4*3*2*1/(20-3)!

so its a 20*19*18=6,840

Can you please show it the part 2 I still dont get it.
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mustish1Author Commented:
Sorry i made a mistake in my text

According to the first formula
P(n,r)=n!/(n-r)!
P(20,3) = 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*4*3*2*1/(20-3)!

so its a 20*19*18=6,840

Can you please show it the part 2 I still dont get it.
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TommySzalapskiCommented:
From Ozo's post
The second part should be C(20,3)=n!/(r!(n-r)!)

So that's 20!/(3!*17!)

As Superdave said, you'll notice that this is the same as P(20,3)/3!

If I have a president, VP and treasurer I can rearrange those three guys P(3,1) = 6 different ways. Since in part 2 the people don't have titles, all 6 of those permutations are the same three guys.
So you take the 6840 number and divide it by 6 to get the total number of possible unordered groups of 3.

But Ozo and Superdave already said all this. Perhaps it is clearer now?
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