in how many ways can G,H,I,I, and J be arranged to compose a 5-letter code so that two "I" are not next to each other?
OA[spoiler]=60-24=36[/spoiler]
OA[spoiler]=60-24=36[/spoiler]
Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.
See classesLearn live with an expert or move at your own pace. Every option includes the complete TTP study system.
Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.
with Josh Braslow, 20+ years in test prep
Self-paced EA prep. Study on your schedule.
with
Complete access from day one. Study on your schedule.
Well total ways of arranging 5 letters with 2 common letters=5!/2!=60 and 4!=24DanaJ wrote:Total ways of arranging the 5 letters will be 5! = 60.
Then you need to subtract the cases where you have the two I's next to each other. In such cases, they'd create one single, indivisible group. So you can safely say that the number of cases in which you have II is the number of possible arrangements of (G), (H), (II) and (J) or possible arrangements of 4 objects, i.e. 4! = 24.
You're left with the answer you provided.