Perms/Combs (1 Viewer)

[rand_althor]

If you had a group of 20 unique objects, and were asked to choose 5 of those objects (where order does not matter), would the number of ways of doing this be ²⁰C₅=15504 or 20x19x18x17x16=1860480? I've been told it is the first option, but I don't see why the second option is wrong. Choosing 1 object from 20 (can be done in 20 ways), then 1 from 19 (can be done in 19 ways) and so on til you choose 1 out of 16 makes sense to me. Can someone please clarify why the second option is incorrect.

 

[InteGrand]

If you had a group of 20 unique objects, and were asked to choose 5 of those objects (where order does not matter), would the number of ways of doing this be ²⁰C₅=15504 or 20x19x18x17x16=1860480? I've been told it is the first option, but I don't see why the second option is wrong. Choosing 1 object from 20 (can be done in 20 ways), then 1 from 19 (can be done in 19 ways) and so on til you choose 1 out of 16 makes sense to me. Can someone please clarify why the second option is incorrect.

Your second option takes the order into account. It says that {A,B,C,D,E} is different to {E,A,C,D,B}, which is not the case for a combination (your second one is a permutation, which is when the order DOES count).

 

[dan964]

sorry for no latex
nPr = n!/(n-r)!
20P5 = 20!/(20-5)!
= 20!/15!
= 20*19*18*17*16*15!/15!
= 20*19*18*17*16 (which is your second option)

nCr=nPr/r!
= (20*19*18*17*16)/5!

You need to divide by the 5! in order to remove duplicate combinations (order is irrelevant)