daviesj wrote:

Each digit 1 through 5 is used exactly once to create a 5-digit integer. If the 3 and the 4 cannot be adjacent digits in the integer, how many 5-digit integers are possible?

(A) 48

(B) 66

(C) 72

(D) 78

(E) 90

Such questions are best solved using the complement approach. You need to find arrangements in which 3 and 4 are not adjacent. Instead, work on finding the arrangements in which they are together since it is much easier to do.

Number of arrangements using 5 distinct digits = 5!

Number of arrangements in which 3 and 4 are adjacent - consider 3 and 4 together as one group. Now you have 4 numbers/groups to arrange which can be done in 4! ways. In each of these arrangements, 3 and 4 can be arranged as 34 or 43. So we need to multiply 4! by 2 to give all arrangements where 3 and 4 are adjacent to each other e.g. 34251, 43251, 23415, 32415 ...

Number of arrangements in which 3 and 4 are not adjacent = 5! - 2*4! = 72

Check out my detailed post on this concept:

http://www.veritasprep.com/blog/2011/10 ... ts-part-i/
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