Finding 'N' when total permutations or combinations given

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.

parallel_chase wrote:GMAT only tests logic , therefore you wont be required to calculate such values

Priceless intel!

true. if they need you to calc values then just work backwards from the answer choices, it will be much faster.