Firm A has two mutual fund managers and Firm B has three managers. How many unique performane rankings for the two firms by the manager are possible if only the firm name, rather than manager name, is listed? A) 2 B) 3 C) 6 D) 10
I don’t understand the question. A) 2
second choice would be 6.
6 combinations
D
Great Dreary…D is correct…
any explanation? I am confused
can someone rephrase the question.
It’s a coloring problem… or a multinomial permutation problem. N!/c1! * c2! where N = c1+c2
I got it wrong as well. Here it is the answer: The number of ranking combinations if the fund manager names are used is: n!=5!=120 When only two company names are used for the ranking, the number of unique combinations decreases: Na+b! /(na!)(nb!)= 5!/(2!)(3!)= 120/12=10 Enjoy!
OK, it’s like this: AABBB or ABABB or ABBBA etc there are 5 managers, but we give teh spot to teh firm not the poor managers. You can rank as Firm A, Firm A, Firm B, etc…how many way can you do that?
Dreary Wrote: ------------------------------------------------------- > OK, it’s like this: > > AABBB or > ABABB or > ABBBA etc there are 5 managers, but we give teh > spot to teh firm not the poor managers. > > You can rank as Firm A, Firm A, Firm B, etc…how > many way can you do that? This is a lot more intuitive. Thanks Dreary!
Ugh, I was trying to use nCr, vs !. Thanks!
I think using combinaitons nCr is also okay for this question. As combination is a special form of multinomial formula as there are only 2 labels. So we have 5 combinations possible and we have 2 labels, so 5C2 = 5!/2!*(5-2)!= 5!/3!*2!= 10
AWESOME, thanks! I did 3nCr2 but now i see why it is 5nCr2.
5!/3!2! 10