anonymous
  • anonymous
Just want someone to check over my answer! How many different ways can the letters of the word RIVAL be scrambled so that the 2 vowels can be together?
Mathematics
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anonymous
  • anonymous
Just want someone to check over my answer! How many different ways can the letters of the word RIVAL be scrambled so that the 2 vowels can be together?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
nvm it's \[6! / 2!3!\]
anonymous
  • anonymous
= 60?
mathstudent55
  • mathstudent55
Think of the letters AI being 1 letter. How many ways can you arrange the letters R, V, L, AI? Then think of the letters IA being 1 letter. How many ways can you arrange R, V, L, IA? Then add the two results. |dw:1438646050051:dw|

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anonymous
  • anonymous
im so confused.. so would i just do 4!?
mathstudent55
  • mathstudent55
Think of the simple problem: How many ways can you arrange the letters of the "word" RLVX
mathstudent55
  • mathstudent55
In this case, the answer would be simply the counting principle, 4!, right? Your case is similar to this case. Since the two letters A and I must be together, they count as a unit. Call the letters A and I together just X. How many ways can you arrange the letters of the "word" RLVX? Answer: 4! Since AI can be together as AI or as IA, you need to do the counting principle for each way of grouping A and I, so you get 2 * 4!.
anonymous
  • anonymous
ahah aight thank youuu
mathstudent55
  • mathstudent55
You're welcome.

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