Tough combinatrics problem

Stockmoose16 wrote:

pseudononymous wrote:

Stockmoose16 wrote:
If you only subtract the places where the 3 Us are together, you're failing to count the places where 2 Us are together. Also, 7!/2! is derived from:

UUXXNXNU = 7!/2! ... you ADD this to the number of 3 U combinations to get the answer.

Can a GMAT INSTRUCTOR confirm this?

No, I think it's 8!/(3!*2!) - 7!/2! + 6!/2! = 1200

Why would you add back the instances with 3 consecutive U's?

because 7!/2! double counts UUU arrangements. For example:

(UU)UNNXXX
U(UU)NNXXX

By adding back 6!2!, you get rid of the double counting.

Please go through this webpage. I found it really helpful.

https://www.mansw.nsw.edu.au/members/ref ... no4yen.htm

PS: I have not read all the responses. Incase part c has already been explained pl ignore my reply.


Part a: Explained 8!/(2! * 3!)
Part b: Explained 6!/(2!)

Answer for part C:
total Us are 3
and other letters are 5

between 5 letters we have 6 places
so if all Us will come at these 6 places they will always be apart from each other

so it will be 6C3 = 20
so in 20 cases no 2 Us will be together.

Hope this helps