i think that answer must be 1/60

S Q U A R E
Requirements:
(1) R is third letter
(2) QES are next to each other

Since R is the third letter (_ _ R _ _ _), the only way QES can be next to each other are in the final three spots. Otherwise, one is separate from the others (which fails a requirement).

The # of permutations of QES in the final three spots is
3P3 = 3!/0! = 6/1 = 6 permutations.

Since R is constant in the third spot, the final four spots have 6 total permutations.

The first two spots have to contain the unused letters U, A.
2P2 = 2!/0! = 2/1 = 2 (this is just as easily written out)

2 * 6 = 12 ways total.

Hi BuckeyeT,

QES are next to each other but i dont think the order is important.So we will get a seperate 3! words from that combination.Please let me know if i am wrong.



S Q U A R E.

_ _ R (Q E S)

i grouped QES together so QES = x

total no of permutations = 3! * 3! = 36 ways

tkarthi4u wrote:Hi BuckeyeT,

QES are next to each other but i dont think the order is important.So we will get a seperate 3! words from that combination.Please let me know if i am wrong.



S Q U A R E.

_ _ R (Q E S)

i grouped QES together so QES = x

total no of permutations = 3! * 3! = 36 ways

nix