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Finding 'N' when total permutations or combinations given

This topic has 3 member replies
msd_2008 Senior | Next Rank: 100 Posts
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Posted:
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Finding 'N' when total permutations or combinations given

Fri Dec 05, 2008 3:47 am
Guys,

Can anyone please explain to me how to find 'n' when total permutations or combinations have been given?
For eg:- 10 C 4 = 210. What if the problem said N C 4 = 210, find value of N?
Similarly, 10 P 4 = 5040. What if the question said N P 4 = 5040, find the value of N ?

Regards
MSD

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niraj_a Legendary Member
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Fri Dec 05, 2008 12:19 pm
true. if they need you to calc values then just work backwards from the answer choices, it will be much faster.

logitech Legendary Member
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Fri Dec 05, 2008 12:07 pm
parallel_chase wrote:
GMAT only tests logic , therefore you wont be required to calculate such values
Priceless intel!

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parallel_chase Legendary Member
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Fri Dec 05, 2008 11:54 am
Now those are pretty big values to calculate the combinations

Try with easy one, GMAT only tests logic , therefore you wont be required to calculate such values

Here is the basic concept:

5C2 = 10

NC2 = 10

N! / (N-2)! 2! = 10

N! / (N-2)! = 10* 2!

N! = N*(N-1)*(N-2)!

N*(N-1)*(N-2)! / (N-2)! = 20

N*(N-1) = 20

N^2 -N -20 = 0

N^2 -5N + 4N -20 = 0

(N-5) (N+4) = 0

either N can be 5 or -4, since we are calculating for combinations there can be no negative values, hence the answer is 5.

You can apply the same method in the original question.

NC4 = 210

N!/(N-4)!*4! = 210

N!/(N-4)! = 210 * 2!

N! = N*(N-1)*(N-2)*(N-3)*(N-4)!

N*(N-1)*(N-2)*(N-3)*(N-4)! / (N-4)! = 5040

N*(N-1)*(N-2)*(N-3) = 5040

You can solve for N to find its value.

If GMAT gives you such big numbers, which is very unlikely but nevertheless you can insert the answer options to find the solution.

Hope this helps.

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