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P and C problem

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P and C problem

by vikrambhatia86 » Wed May 26, 2010 3:06 am
guys, have a look at this problem:

How many different 5 person teams can be formed from a group of x individuals.?

a) if there had been x+2 individuals in the group,exactly 126 - 5 different person teams could have been formed.
b) If there had been x+1 individuals in the group, exactly 56 different 3 -person teams could have been formed.

The answer to this is D as we can calculate x by either of the 2 options ,but in the explanation available to me ,combination has been used to calculate x and not permutation.
I feel this is a permutation problem and we have to find unique teams and we will use x+2! P 5 =126 and x+1 ! P 3 =56 to find x.

Please comment .
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by thephoenix » Wed May 26, 2010 3:13 am
selection is always combination
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by gmatmachoman » Wed May 26, 2010 3:14 am
thephoenix wrote:selection is always combination
Adding on to that "arrangements in a pattern" is always permutations!
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by thephoenix » Wed May 26, 2010 3:18 am
gmatmachoman wrote:
thephoenix wrote:selection is always combination
Adding on to that "arrangements in a pattern" is always permutations!
any thoughts to solve it if it wud have been a PS problem
how to find x with statement 1
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by vikrambhatia86 » Wed May 26, 2010 3:22 am
thephoenix wrote:
gmatmachoman wrote:
thephoenix wrote:selection is always combination
Adding on to that "arrangements in a pattern" is always permutations!
any thoughts to solve it if it wud have been a PS problem
how to find x with statement 1
It turns out that if there
are 9 available individuals, then we could create exactly 126 different 5-person
teams (since 9! ÷ [(5!)(4!)] = 126). This value (9) represents x + 2. Thus x would
equal 7.
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by liferocks » Wed May 26, 2010 3:48 am
Just wanted to solve it.
From 1
product of 5 consecutive integers=126*5*4*3*2=2*7*9*5*4*3=9*8*7*6*5..so x=7

From 2
product of 3 consecutive integers=56*3*2=7*8*6..x=7

Any better method anyone?
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by selango » Wed May 26, 2010 4:44 am
IMO this approach can be used,

Since 5 to be selected from x+2 peoples,Its x+2C5 ways

-->(X+2)!/(X-3)!.5!=126


-->(x+2)*(x+1)*x/(x-3)!=126.5!



From here I cant proceed..any suggestions
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by liferocks » Wed May 26, 2010 5:00 am
selango wrote:IMO this approach can be used,

Since 5 to be selected from x+2 peoples,Its x+2C5 ways

-->(X+2)!/(X-3)!.5!=126


-->(x+2)*(x+1)*x/(x-3)!=126.5! =126*5*4*3*2=2*7*9*5*4*3=9*8*7*6*5..so x=7


From here I cant proceed..any suggestions
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by selango » Wed May 26, 2010 5:14 am
I understand this...But if v take x=7,then

1/(x-3)!=30

even if substitute 7 here,the equation cannot be solved..
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by kevincanspain » Wed May 26, 2010 5:16 am
vikrambhatia86 wrote:guys, have a look at this problem:

How many different 5 person teams can be formed from a group of x individuals.?

a) if there had been x+2 individuals in the group,exactly 126 - 5 different person teams could have been formed.
b) If there had been x+1 individuals in the group, exactly 56 different 3 -person teams could have been formed.

The answer to this is D as we can calculate x by either of the 2 options ,but in the explanation available to me ,combination has been used to calculate x and not permutation.
I feel this is a permutation problem and we have to find unique teams and we will use x+2! P 5 =126 and x+1 ! P 3 =56 to find x.

Please comment .
If I had to calculate, I would use trial and error

C(x+2,5)=126

(x+2)(x+1)x(x-1)(x-2)/5! = 126

(x+2)(x+1)x(x-1)(x-2) = 120(126)


Note that since 120(126) < 10^5, x must be less than 10

9(8)(7)(6)(5) = 9(7)(2)(3)(5)(8)=126(120)

Thus x= 7
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