Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the value of 22C19?

A. 770
B. 1540
C. 3080
D. 4620
E. 6160

When calculating combinations, it's sometime useful to apply the following rule: nCr = nC(n-r). In other words, "n choose r" is equal to "n choose n - r"
So, for example: 10C7 = 10C3
12C9 = 12C3
8C7 = 8C1
etc.

Here's why this property works:
Let's say you have 22 friends, and you must invite 19 of them to attend your party.
This can be accomplished in 22C19 ways.
However, another way to select 19 party attendees, is to select 3 of the 22 people to NOT attend the party. Then, by default, the remaining 19 people get to attend the party.
So, selecting 3 of the 22 people to NOT attend the party is the same as selecting 19 of the 22 people TO ATTEND the party.
In other words, 22C19 = 22C3

Best of all, it's much easier to calculate 22C3 than it is to calculate 22C19

22C3 = (22)(21)(20)/(3)(2)(1)
= 1540

Answer: B

Cheers,
Brent

regor60 wrote: It is ? Aren't they the same thing ?

22!/19!(22-19)! vs 22!/3!(22-3)!

Aren't we going to factor out the 19! either way?

Good point!
I should have mentioned that I use a shortcut for calculating combinations that allows me to perform most calculations in my head.

For example, for 22C3, I write the first 3 values of 22! in the numerator, and I write the 3! in the denominator
So, 22C3 = (22)(21)(20)/(3)(2)(1)

Likewise, for 11C2, I write the first 2 values of 11! in the numerator, and I write the 2! in the denominator
So, 11C2 = (11)(10)/(2)(1) = 55

Here's a video explaining the whole technique: https://www.gmatprepnow.com/module/gmat ... /video/789

Cheers,
Brent

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

What is the value of 22C19?

A. 770
B. 1540
C. 3080
D. 4620
E. 6160

22C19 is:

22!/[(22 - 19)! x 19!] = 22!/(3! x 19!) = (22 x 21 x 20)/3! = 11 x 7 x 20 = 77 x 20 = 1540

Answer: B