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[#15] 2 Min. Challenge : Counting Methods - Students

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[#15] 2 Min. Challenge : Counting Methods - Students [#permalink] New post 09 Mar 2004, 15:48
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Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Rules

1. Time yourself
2. Solve this problem on a seperate sheet.
3. Write down the solution and your time.


In how many ways can 4 groups of 2 each be selected from a group of 8 students?

1. 1980
2. 2520
3. 3050
4. 3670
5. 3980
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 [#permalink] New post 09 Mar 2004, 16:20
Less than 10 seconds

8!/16
or 8C2 * 6C2 * 4C2 * 2C2 = 2520
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 [#permalink] New post 09 Mar 2004, 19:01
2520. It took more than two mins.

anand, how did you arrive at 8!/16 ??
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 [#permalink] New post 09 Mar 2004, 19:09
8C2 * 6C2 * 4C2 * 2C2
= (8*7)/2 * (6*5)/2 * (4*3)/2 * 2/2
= (8*7*6*5*4*3*2*1)/(2*2*2*2) = 8!/16
Since 7! = 5040 we have 8! = 8 * 5040

8!/16 = ( 8 * 5040 ) / ( 8 * 2) = 5040/2 = 2520
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 [#permalink] New post 09 Mar 2004, 19:27
Agree with Anandnk. Formula took me less than 10 seconds but with calculation, maybe 30 seconds. I suspect anandnk to be cooking the books with "less than 10 seconds" :)
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 [#permalink] New post 09 Mar 2004, 19:29
10 seconds because I know all factorials upto 7
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 [#permalink] New post 09 Mar 2004, 19:57
Quote:
In how many ways can 4 groups of 2 each be selected from a group of 8 students?

1. 1980
2. 2520
3. 3050
4. 3670
5. 3980


Do u know pals,
Grouping can be done this way also, formula is
8! / (2! * 2! * 2! * 2!) = 2520

And i am sure it wont take more than 2 minutes, correct !

A general group formula goes like this: (p+q+r)! / (p! * q! *r!), for 3 groups

Hope this will help,
Dharmin :-D
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 [#permalink] New post 09 Mar 2004, 20:06
Thanks anand.
Dharmin, your method is the easiest.
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Re: [#15] 2 Min. Challenge : Counting Methods - Students [#permalink] New post 10 Mar 2004, 00:15
praetorian123 wrote:
Rules

1. Time yourself
2. Solve this problem on a seperate sheet.
3. Write down the solution and your time.

In how many ways can 4 groups of 2 each be selected from a group of 8 students?
1. 1980
2. 2520
3. 3050
4. 3670
5. 3980


[2] 2520 is the correct answer. good work.

I quote the legend akamai here:

Selecting multiple groups is EXACTLY the same as selecting one group. Let's say you are selecting 2 people from a group of 8. You are actually selected one group of 2 and one group of 6 (Which is an intuitive way of proving why 8C2 = 8C6). In the denominator of combinations with multiple groups, simply put the factorials of the number of people in each group.

2C8 = choose 2 to be IN and choosing 6 to be OUT or 8!/(2!6!)

To choose 4 groups of 2 from 8, the formula is simply 8!/(2!2!2!2!)

If you think of combinations as "adjusted" permutations, we have 8! ways to arrange 8 people, but each of the 4 pairs is an equivalent combination so we need to divide by 2^4.

More to Come...keep solving
Praetorian
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 [#permalink] New post 10 Mar 2004, 09:12
Great postings!

Yes, I was also able to solve this in < 2mins. 8!/2!*2!*2!*2!

I actually studied this concept of groups yesterday.

I recommend "Probability Without Tears" by Derek Rowntree. Great book. Completely explains probability/combinations/permuations in a very clear and straightforward manner. You can borrow it from your local library if you don't want to buy it.
  [#permalink] 10 Mar 2004, 09:12
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[#15] 2 Min. Challenge : Counting Methods - Students

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