Question 1002206
nCr formula is n! / (r! * (n-r)!


4C2 formula becomes 4! / (2! * (4-2)! which becomes 4! / (2! * 2!) which becomes (4*3*2!) / (2!*2!) which becomes (4*3)/2! which becomes 6.


your answer should have been 6.


let the 4 distinct objects be a, b, c, d


the possible combinations, taken 2 at a time, are:


ab
ac
ad
bc
bd
cd


that's 6 possible combinations.


you used the correct formula, but you must have applied it wrong, or you explained it wrong.


4C2 = 4! / (2! * 2!) = (4*3*2*1) / (2*1*2*1)


get rid of the 1's because they don't add anything to it, and you get:


4C2 = (4*3*2) / (2*2) which becomes (4*3*2) / 4 which becomes 3*2 = 6


false is the correct answer because 6 is not equal to 2.


your last statement, however, doesn't make sense because you say:


*note: I used the combination formula nCr and I got 2. So "False." Just wanted to be sure if it was the right application I made. Thank you 


if you got 2, then you should have said true.


perhaps you meant you didn't get 2?


you got the answer right but your explanation was confusing.