There are 7 letters in GUMTREE, and the 2 E's are indistinguishable, so there are 7!/2! = 2520 distinguishable arrangements total. Exactly half of them, or 1260, have G somewhere left of U, (and the other half have G right of U). In turn, by the same token, exactly half of those 1260, or 630, have M right of U. Answer: 630 Edwin