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Coach Miller is filling out the starting lineup for his indoor soccer

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Joined: 30 Mar 2013
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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New post 23 Sep 2014, 08:42
amkabdul wrote:
Guys... since we have 6 positions in the field which are to be filled with 10 players.

Look at this: Filling the goal keeper position in 2 ways.
Next: the rest of the 5 positions are to be filled with 8 players and each player at a different position would be 8p5 But since we have 2 positions each for the defence and mid. divide it with 2! twice.

So answer is : 2* (8P5/(2!*2!)) = 3360

A very simple and easy approach based on the basics of P 'n C

Kudo me if you like this.


Agreed about the permutations approach. I was under the impression that we use permutations when there are places specified in the group, as there are here. I dont understand the combinations approach at all. For eg, in how many ways can you select a president and VP from a group of 6. Should be 6*5=30, or 6P2= 30.

Can someone please point me in the direction where I can understand when to apply C and when to apply P. I understand that P is applied when order is imp, but the problem is, that when selecting president and VP from a group, order shouldn't be important. AB or BA is the same, and therefore, even those should be solved by using combinations. I hope I'm explaining my confusion well enough for someone to help me :oops:
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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New post 23 Sep 2014, 10:38
2
usre123 wrote:

the problem is, that when selecting president and VP from a group, order shouldn't be important. AB or BA is the same


If you're choosing a President and Vice-President, and you choose A for President and B for VP, that's very different from choosing B for President and A for VP -- you have a different President! So when you're choosing a President and VP, order is very important, and AB and BA are not the same selection.
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Re: Coach Miller is filling out the starting lineup for his indoor soccer  [#permalink]

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New post 25 Jan 2017, 11:22
rvthryet wrote:
A Coach is filling out the starting lineup for his indoor soccer team. There are 10 boys on the team, and he must assign 6 starters to the following positions: 1 goalkeeper, 2 on defence, 2 in midfield, and 1 forward. Only 2 of the boys can play goalkeeper, and they cannot play any other positions. The other boys can each play any of the other positions. How many different groupings are possible?

A. 60
B. 210
C. 2580
D. 3360
E. 151200


We are given that there are 10 boys available to fill the following positions:

1 goalkeeper, 2 defenders, 2 midfielders, and 1 forward.

We are also given that only 2 of the boys can play goalkeeper.

Thus, we can select the goalkeeper in 2C1 = 2 ways.

We now have 8 boys left and need to select 2 defenders from those 8 boys:

8C2 = (8 x 7)/2! = 28 ways

We now have 6 boys left and need to select 2 midfielders:

6C2 = (6 x 5)/2! = 15 ways

We now have 4 boys left and need to select 1 forward:

4C1 = 4

Thus, the number of ways to select 6 starters is:

2 x 28 x 15 x 4 = 3,360

Answer: D
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Re: Coach Miller is filling out the starting lineup for his indoor soccer   [#permalink] 25 Jan 2017, 11:22

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