Solving a Practice Exam Question: Arranging Letters in ROCKET

  • Context: MHB 
  • Thread starter Thread starter nari
  • Start date Start date
  • Tags Tags
    Exam Rocket
Click For Summary
SUMMARY

The discussion focuses on calculating the number of arrangements of the letters in the word "ROCKET" with the condition that the vowels (E and O) must remain together. The correct formula for this scenario is 5! x 2, where 5! accounts for the arrangement of the five "letters" formed by treating the vowels as a single unit (R, OE, C, K, T). The factor of 2 represents the two possible arrangements of the vowels themselves (EO or OE).

PREREQUISITES
  • Understanding of factorial notation and its application in permutations
  • Basic knowledge of combinatorial principles
  • Familiarity with grouping elements in arrangements
  • Concept of treating multiple items as a single unit in permutations
NEXT STEPS
  • Study permutations with restrictions in combinatorics
  • Learn about the principles of grouping in arrangements
  • Explore advanced factorial applications in probability
  • Practice similar problems involving arrangements of letters and numbers
USEFUL FOR

Students preparing for exams in combinatorics, educators teaching permutation concepts, and anyone interested in solving arrangement problems in mathematics.

nari
Messages
4
Reaction score
0
Hi! There is this question in my practice exam:

How many ways can all the letters in the word ROCKET be arranged so that the vowels are always together?

The answer is 5! x 2.

I understand where the 2 is coming from. What I don't understand is the 5. Shouldn't it be 4!, since we already selected two out of six letters?
 
Physics news on Phys.org
nari said:
Hi! There is this question in my practice exam:

How many ways can all the letters in the word ROCKET be arranged so that the vowels are always together?

The answer is 5! x 2.

I understand where the 2 is coming from. What I don't understand is the 5. Shouldn't it be 4!, since we already selected two out of six letters?
Think of the E and the O as being glued together. They then count as a single "letter", and you have 5! ways of arranging the five letters R OE C K and T. The 2 just tells you whether the E comes before or after the O.
 
Opalg said:
Think of the E and the O as being glued together. They then count as a single "letter", and you have 5! ways of arranging the five letters R OE C K and T. The 2 just tells you whether the E comes before or after the O.

So then the options would be: eo/oe + r o/e c k t?
 
nari said:
So then the options would be: eo/oe + r o/e c k t?

I don't quite follow the whole logic here but the beginning part is true, the letter pairs are "eo" or "oe". For simplicity let's just called these two letters P (for pair). Opalg suggested the same thing but maybe seeing it like this will help.

How many ways can you arrange RPCKT?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K