A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Ahsome

  • one year ago

Help with two questions

  • This Question is Closed
  1. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A cricket team of 11 players is to be selected, in batting order, from 15. How many different arrangements are possible if: Mark must be in the team but he can be anywhere from 1 to 11? AND The starting 5 in a basketball team is to be picked, in order, from the 10 players in the squad. In how many ways can this be done if: Jamahl and Anfernee must be in the starting 5?

  2. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For first problem, since Mark must be in the team, you just need to choose \(10\) other players from the remaining \(14\) players. You can do this in \(\dbinom{14}{10}\) ways. After the \(11\) players are selected, you can arrange them in batting order in \(11!\) ways. So the total possible different arrangements is \(\dbinom{14}{10}*11!\)

  3. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Why is it combination, and not permutation @ganeshie8?

  4. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    NVM, I got it @ganeshie8. Thanks :)

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I hope you can do the second problem on your own

  6. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Would it just be: \[ \dbinom{9}{3}\times5! \]

  7. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I mean 8, not 9 btw

  8. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\dbinom{8}{3}*5!\) looks good! as you can see we're doing it in two steps : 1) choose the 5 players first 2) arrange them in order

  9. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yup. I got confused, cause using permutation * 5! gave a completely different result :P I think I get it now :). Can I just ask another question?

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sure..

  11. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    How many permutations of the letters in the word MATHS are therem wgere M, A and T appear together?

  12. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I did \(3!\times 3!\)

  13. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Correct. You tie MAT in a bag. Then you will see just 3 objects |dw:1440822752665:dw|

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    3 objects can be permuted in \(3!\) ways and for each of that, you can permute the stuff inside bag : \(\{M,~A,~T\}\) in \(3!\) so the total permutations such that M,A,T are together is \(3!*3!\)

  15. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Thought so. Thank you! Although it doesn't seem to work when I do it on permutations with different N and R values...

  16. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    do you have an example ?

  17. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A rowing team of 4 rowers is to be selected in order from 8 rowers. In how many of these ways do 2 rowers, Jane and Lee, sit together in the boat? But this requires different logic

  18. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    same logic Look at all the selections in which Jane and Lee are there. Then there are \(\dbinom{6}{2}\) ways to choose remaining \(2\) other rowers.

  19. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Put Jane and Lee in a bag. You will see 3 objects which you can permute in \(3!\) ways Also the two people inside bag can be permuted in \(2!\) ways so the total arrangements in which Jane and Lee are together is \(\dbinom{6}{2}*3!*2!\)

  20. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ahh, gotcha. We didn't do the 3! for the second question we did, cause we didn't merge them right?

  21. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    we want four rowers two of them are tied in a bag

  22. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yup. That's why we considered the 2!, cause it could be in different positions, right?

  23. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes

  24. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    And would that apply to this question: The number of permutations of the letters in POCKET where P and O are together. Merge the P and O together, you get 5 places. I then did: \(4C4 \times 5! \times 2!\)

  25. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Since we have 5 places. Four of them must be occupied by the other 4 letters. 4C4 = 1. We can then arrange the 5 letters in 5! ways, and arrange the two merged in 2! ways

  26. ahsome
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @ganeshie8?

  27. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.