Practice Final A2
rahulchandrupatla
The method professor showed us at the review session makes this question a lot easier, and I wanted to know if this is what others got!. So since 20 people, but 3 cant stand next to each other, using the fingers and spaces, there are 18 spaces where Alice, Bob, and Carl can be in which is (18c3) and then u can order those 3 people by 3!. Then we can order the other 17 people with 17!, all over #(omega) which is the way we can order 20 people, so 20!. So putting it all together, I got $\frac{(18 choose 3) 3!17!}{20!}$. Anybody got something similar/different?
betht123: March 19, 2015, 2:51 p.m.
I got the same!
tdoelker: March 19, 2015, 4:33 p.m.
I got the same too!
Lilyhui: March 19, 2015, 5:22 p.m.
Same!
arpanshah: March 19, 2015, 6:29 p.m.
I also got the same thing for this problem!