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27 Apr 2020, 14:35
Kudos
Bookmarks
Hi everyone,
The traditional format for the: nCr combinations is
n!
(r!*(n-r)!)
In studying the Magoosh GMAT Flash cards, I came across a problem that states:
nC2 = ?
The answer is
n(n-1)
2
Obviously, this appears like it does not line up with the initial formula. Question, how did we get from the answer format of formula 1, to the Flashcard answer?
Many thanks to you all,
PersonGuy
The traditional format for the: nCr combinations is
n!
(r!*(n-r)!)
In studying the Magoosh GMAT Flash cards, I came across a problem that states:
nC2 = ?
The answer is
n(n-1)
2
Obviously, this appears like it does not line up with the initial formula. Question, how did we get from the answer format of formula 1, to the Flashcard answer?
Many thanks to you all,
PersonGuy
Archived Topic
Hi there,
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Thank you for understanding, and happy exploring!
27 Apr 2020, 16:47
Kudos
Bookmarks
PersonGuy
Hello, PersonGuy. The flashcard is fine. Think about it. Plug in a value, say, 5, for n and work through the combination both ways.
Method 1:
\(\frac{5!}{2!(5-2)!}\)
\(\frac{(5*4*3*2*1)}{(2*1)(3*2*1)}\)
\(\frac{(5*4)}{(2*1)}\)
\(\frac{(5*4)}{2}\)
I will stop there, because it is easier to compare to the formula given on the flashcard.
Method 2:
\(\frac{(5)(5-1)}{2}\)
\(\frac{(5*4)}{2}\)
Notice that this last step matches exactly what we had derived before. It is just a shorter way of arriving at the same value. You could test any integer you wanted for n, and you would notice a similar pattern develop.
I hope that explanation suffices. If you have further questions, feel free to ask.
- Andrew
03 May 2020, 06:29
Kudos
Bookmarks
PersonGuy
I'd suggest understanding how to solve these types of problems without a formula, since the GMAT isn't really testing if you can memorize formulas and plug things into them, but in the case you mention, r = 2. So plugging r = 2 into the formula, we get:
n! / [ (2!) (n-2)! ]
Now, n! is the product of the integers from n down to 1, inclusive, so
n! = (n)(n-1)(n-2)(n-3) ... (2)(1)
Notice that n! 'contains' (n-2)! (which is the product of the integers from n-2 down to 1, inclusive), so in our original fraction, we can cancel (n-2)! from n!, leaving us with (n)(n-1). So
n! / [ (2!) (n-2)! ] = (n)(n-1) / 2! = (n)(n-1)/2
