Hello - I had difficulty completing this problem.
There are 6 senior consultants and 4 junior consultants.
A team is to be comprised of 3 senior consultants and 1 junior consultant.
How many different possible teams are there.
Also, what is involved in solving a parabola problem as regards x and y intercepts. Is there a chapter in the MGMAT prep that deals with this issue?
Thanks
GMAT Forum
3 postsPage 1 of 1
- beaver82
- Guest
Senior consultant:
IIIOOO
(6!)/(3!*3!)=4*5=20
(this is actually in the MGMAT Guide, since you are either in or out, you get 3 INs and 3 Outs, and you divide any repeating Is and Os).
Junior Consultant:
IOOO
4!/3!=4, same concept as senior one.
Senior AND Junior:
20*4=80 (# of total possible teams).
Please correct me if I made a mistake.
Eric
IIIOOO
(6!)/(3!*3!)=4*5=20
(this is actually in the MGMAT Guide, since you are either in or out, you get 3 INs and 3 Outs, and you divide any repeating Is and Os).
Junior Consultant:
IOOO
4!/3!=4, same concept as senior one.
Senior AND Junior:
20*4=80 (# of total possible teams).
Please correct me if I made a mistake.
Eric
- JonathanSchneider
- ManhattanGMAT Staff
- Posts: 370
- Joined: Sun Oct 26, 2008 3:40 pm
Nice work, Eric!
Notice that Eric broke down the choices into two separate choices, as indicated by the problem. Tough combinatorics and probability problems often benefit from seeing what the "simplest" choices truly are.
As to parabolas, I don't think they're tested on the GMAT. Did you see one in a problem?
Notice that Eric broke down the choices into two separate choices, as indicated by the problem. Tough combinatorics and probability problems often benefit from seeing what the "simplest" choices truly are.
As to parabolas, I don't think they're tested on the GMAT. Did you see one in a problem?
3 posts Page 1 of 1