lilai3
  • lilai3
Can someeeeoneeee pleasepleaseplease help meeee???????????? Schedule: How many different schedules can Sheila create if she has to take English, math, science, social studies, and art next semester? Assume that there is only one lunch period available. Band Concerts: The band is having a holiday concert. In the first row, the first trumpet is always furthest to the left. How many ways are there to arrange the other 4 people who need to sit in the front?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Schedule is 6 different catagories including lunch. Since lunch can only be at one time. Take that away. That leaves 5 catagories to be arranged 5 different ways. 5 x 5 = 25. I think this is right anyways. I could be totally wrong :P The band concert I assume is done the same way. The furthest left chair is constant and will not change. So that leaves 4 people to be arranged 4 different way. 4 x 4 = 16. Someone else make sure I didn't goof this. Thanks
ash2326
  • ash2326
In the first part leaving lunch there are 5 categories or 5 subjects. Generally lunch is at a fixed time. So 5 subjects can be arranged in 5! ways=5*4*3*2*1=120 ways So she can have 120 different schedules Second pat 4 people can be arranged in 4! ways= 4*3*2*1=24 ways
lilai3
  • lilai3
whaaaa all different answers??!?!?!?!??

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anonymous
  • anonymous
Five classes to go into five spots. Consider the first hour class first. She has five choices for this class. After picking this class, there are four classes that can go into second hour. Then three, then two. Last hour class gets what's left. Multiply the possible choices to get the 5!=120 answer above.
anonymous
  • anonymous
MMMMM yes. These are correct and I was mistaken.
lilai3
  • lilai3
*shrugs* come one... everyone makes MISTAKES..... lord 4gives u.. teehee :D
lilai3
  • lilai3
anyway, is the band concert one 16?
anonymous
  • anonymous
No. The positioning is just like the previous problem. We have four musicians and four chairs. We have four choices for who gets the first chair. There are three for the second, and two for the third. The fourth chair goes to whoever is left. Multiply to get all the combinations. 4*3*2 = 4! = 24 possible arrangements.
lilai3
  • lilai3
thank you, @AnimalAin! <333 :DDD

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