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How many ways can a committee of 3 be selected from 7 so that there is

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How many ways can a committee of 3 be selected from 7 so that there is  [#permalink]

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New post 20 Feb 2014, 13:06
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How many ways can a committee of 3 be selected from 7 so that there is a president, a vice president, and a secretary?

Answer is: 210
7x6x5=210
7P3 = 7X6X5 =7!/4!

Why is it 7!/4! ? I don't understand the 4! part.

Can someone explain that for me"?

Thanks!
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Re: How many ways can a committee of 3 be selected from 7 so that there is  [#permalink]

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New post 20 Feb 2014, 21:26
se7en14 wrote:
how many ways can a committee of 3 be selected from 7 so that there is a president, a vice president, and a secretary?

Answer is: 210
7x6x5=210
7P3 = 7X6X5 =7!/4!

Why is it 7!/4! ? I don't understand the 4! part.

Can someone explain that for me"?

Thanks!


Out of 7, 3 people need to be selected and arranged (into 3 different positions: President, VP, Sec)

So there are 2 ways to do it: 7 * 6* 5 (you say, select the president in 7 ways, now select VP in 6 ways and then select Sec in 5 ways) = 210

or you can use the permutation formula nPr such that nPr = n!/(n-r)!. nPr helps you select r people out of n people AND arrange those r people.
Above, we used 7P3 = 7!/(7 - 3)! = 7!/4! = 7*6*5 = 210

Now I am assuming that your question is why the formula is n!/(n-r)!

Say you have n people and you want to arrange them. You can do it in n! ways, right? Just our basic counting principle. Say there at 7 people and you want to arrange all 7 in 7 spots. You can do it in 7! ways ( = 7*6*5*4*3*2*1). 7 ways to fill the first spot, 6 ways to fill the second. 5 ways to fill the third, 4 ways to fill the fourth etc.

Now what if you have only 3 spots? You have to fill 3 only. You can do it in 7*6*5 ways. What about the rest of the 7-3 = 4 spots? (which is n - r) You have to ignore them. So if you do arrange people in 7 spots by using 7! in the numerator, you must divide out the extra n - r spots i.e. 4!. That is the reason you divide by 4! here.
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Re: How many ways can a committee of 3 be selected from 7 so that there is  [#permalink]

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New post 23 Sep 2014, 04:47
VeritasPrepKarishma wrote:
se7en14 wrote:
how many ways can a committee of 3 be selected from 7 so that there is a president, a vice president, and a secretary?

Answer is: 210
7x6x5=210
7P3 = 7X6X5 =7!/4!

Why is it 7!/4! ? I don't understand the 4! part.

Can someone explain that for me"?

Thanks!


Out of 7, 3 people need to be selected and arranged (into 3 different positions: President, VP, Sec)

So there are 2 ways to do it: 7 * 6* 5 (you say, select the president in 7 ways, now select VP in 6 ways and then select Sec in 5 ways) = 210

or you can use the permutation formula nPr such that nPr = n!/(n-r)!. nPr helps you select r people out of n people AND arrange those r people.
Above, we used 7P3 = 7!/(7 - 3)! = 7!/4! = 7*6*5 = 210

Now I am assuming that your question is why the formula is n!/(n-r)!

Say you have n people and you want to arrange them. You can do it in n! ways, right? Just our basic counting principle. Say there at 7 people and you want to arrange all 7 in 7 spots. You can do it in 7! ways ( = 7*6*5*4*3*2*1). 7 ways to fill the first spot, 6 ways to fill the second. 5 ways to fill the third, 4 ways to fill the fourth etc.

Now what if you have only 3 spots? You have to fill 3 only. You can do it in 7*6*5 ways. What about the rest of the 7-3 = 4 spots? (which is n - r) You have to ignore them. So if you do arrange people in 7 spots by using 7! in the numerator, you must divide out the extra n - r spots i.e. 4!. That is the reason you divide by 4! here.


Hello, why isn't it 7C3? because the committee will have people ABC. Doesnt matter how you arrange then, it's still people ABC on the committee. and that's why 7*5*6 is different from 7C3...could you explain why I must use 7P3, please?
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Re: How many ways can a committee of 3 be selected from 7 so that there is  [#permalink]

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New post 23 Sep 2014, 20:52
usre123 wrote:
VeritasPrepKarishma wrote:
se7en14 wrote:
how many ways can a committee of 3 be selected from 7 so that there is a president, a vice president, and a secretary?

Answer is: 210
7x6x5=210
7P3 = 7X6X5 =7!/4!

Why is it 7!/4! ? I don't understand the 4! part.

Can someone explain that for me"?

Thanks!


Out of 7, 3 people need to be selected and arranged (into 3 different positions: President, VP, Sec)

So there are 2 ways to do it: 7 * 6* 5 (you say, select the president in 7 ways, now select VP in 6 ways and then select Sec in 5 ways) = 210

or you can use the permutation formula nPr such that nPr = n!/(n-r)!. nPr helps you select r people out of n people AND arrange those r people.
Above, we used 7P3 = 7!/(7 - 3)! = 7!/4! = 7*6*5 = 210

Now I am assuming that your question is why the formula is n!/(n-r)!

Say you have n people and you want to arrange them. You can do it in n! ways, right? Just our basic counting principle. Say there at 7 people and you want to arrange all 7 in 7 spots. You can do it in 7! ways ( = 7*6*5*4*3*2*1). 7 ways to fill the first spot, 6 ways to fill the second. 5 ways to fill the third, 4 ways to fill the fourth etc.

Now what if you have only 3 spots? You have to fill 3 only. You can do it in 7*6*5 ways. What about the rest of the 7-3 = 4 spots? (which is n - r) You have to ignore them. So if you do arrange people in 7 spots by using 7! in the numerator, you must divide out the extra n - r spots i.e. 4!. That is the reason you divide by 4! here.


Hello, why isn't it 7C3? because the committee will have people ABC. Doesnt matter how you arrange then, it's still people ABC on the committee. and that's why 7*5*6 is different from 7C3...could you explain why I must use 7P3, please?


If you only want to select a group of 3 people out of 7, you use 7C3. But this situation is a little different. You want to choose 3 people for 3 different posts: President, VP, Secretary

So out of 7 people you need to choose 3 and arrange them in 3 positions.

Say we select A, C and D out of the 7 people.

Now different arrangements are possible:
President - A, VP - C, Secretary - D
President - C, VP - A, Secretary - D
etc

You need to account for these different arrangements. So you use 7P3.


Or another way to look at it is that you select 1 person for the President's post, another for VP's and yet another for Secretary's. You can do this in 7*6*5 ways.
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How many ways can a committee of 3 be selected from 7 so that there is  [#permalink]

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New post 24 Sep 2014, 12:11
Out of 7, 3 people need to be selected and arranged (into 3 different positions: President, VP, Sec)

So there are 2 ways to do it: 7 * 6* 5 (you say, select the president in 7 ways, now select VP in 6 ways and then select Sec in 5 ways) = 210

or you can use the permutation formula nPr such that nPr = n!/(n-r)!. nPr helps you select r people out of n people AND arrange those r people.
Above, we used 7P3 = 7!/(7 - 3)! = 7!/4! = 7*6*5 = 210

Now I am assuming that your question is why the formula is n!/(n-r)!

Say you have n people and you want to arrange them. You can do it in n! ways, right? Just our basic counting principle. Say there at 7 people and you want to arrange all 7 in 7 spots. You can do it in 7! ways ( = 7*6*5*4*3*2*1). 7 ways to fill the first spot, 6 ways to fill the second. 5 ways to fill the third, 4 ways to fill the fourth etc.

Now what if you have only 3 spots? You have to fill 3 only. You can do it in 7*6*5 ways. What about the rest of the 7-3 = 4 spots? (which is n - r) You have to ignore them. So if you do arrange people in 7 spots by using 7! in the numerator, you must divide out the extra n - r spots i.e. 4!. That is the reason you divide by 4! here.[/quote]

Hello, why isn't it 7C3? because the committee will have people ABC. Doesnt matter how you arrange then, it's still people ABC on the committee. and that's why 7*5*6 is different from 7C3...could you explain why I must use 7P3, please?[/quote]

If you only want to select a group of 3 people out of 7, you use 7C3. But this situation is a little different. You want to choose 3 people for 3 different posts: President, VP, Secretary

So out of 7 people you need to choose 3 and arrange them in 3 positions.

Say we select A, C and D out of the 7 people.

Now different arrangements are possible:
President - A, VP - C, Secretary - D
President - C, VP - A, Secretary - D
etc

You need to account for these different arrangements. So you use 7P3.


Or another way to look at it is that you select 1 person for the President's post, another for VP's and yet another for Secretary's. You can do this in 7*6*5 ways.[/quote]


Thank you. I can see now how silly my question was!
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Re: How many ways can a committee of 3 be selected from 7 so that there is  [#permalink]

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Re: How many ways can a committee of 3 be selected from 7 so that there is   [#permalink] 25 Apr 2019, 14:05

How many ways can a committee of 3 be selected from 7 so that there is

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