The number of numbers that can be formed by the digits 1,2,3,4,3,2,1. The odd digits are at odd places is

My answer is 18

(4!/2!2!)*(3!/2!)

for all those, like me, that can never remember all the combinatorics formulas.... i suggest that instead of memorizing them, just count...
especially on gmat - there is never a difficult counting question. you just need to count very cautiously not to make mistakes... here is how i do it:

- the odd numbers are at odd places... so we have 4 places (a,b,c,d) to put the numbers 1,1,3,3. ask yourself where can i put the 1s?
the answer is you can do it at places: ab,ac,ad,bc,bd,cd... total of 6 places.

of course, once you put the 1s down, the rest are 3s so there is no freedom there... so the total number of possibilities of ordering the odd numbers is 6.

where can we put the even numbers? oh, there are 3 places left (e,f,g).
where can i put the 4? well, this is simple. there are 3 possibilities. and once you put the 4, no freedom left, the other places are 2.

so we have 6 combinations for odd. 3 combination of even. the two are independent of each other.....

6*3 = 18.
yaeh.... that's the answer. and i didn't use factorials at all. nor should i divide anything... just simple (and cautious) counting, using common sense reasoning (and one math principle - that if you have independent choices, the total number of combinations is the product of the two).

if you practice counting it won't take you so much time as i took here to explain. however it will make you more confident (you never ask yourself do i use the right formula). and you end up doing less mistakes.

on a philosophical note - i want to add, that GMAT math is not about knowing lots of math to solve questions, it is about knowing how to use as little math as possible needed to solve questions.

amit.