Ahsome
  • Ahsome
Help with two questions
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Ahsome
  • Ahsome
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?
ganeshie8
  • ganeshie8
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 \(\dbinom{14}{10}*11!\)
Ahsome
  • Ahsome
Why is it combination, and not permutation @ganeshie8?

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Ahsome
  • Ahsome
NVM, I got it @ganeshie8. Thanks :)
ganeshie8
  • ganeshie8
I hope you can do the second problem on your own
Ahsome
  • Ahsome
Would it just be: \[ \dbinom{9}{3}\times5! \]
Ahsome
  • Ahsome
I mean 8, not 9 btw
ganeshie8
  • ganeshie8
\(\dbinom{8}{3}*5!\) looks good! as you can see we're doing it in two steps : 1) choose the 5 players first 2) arrange them in order
Ahsome
  • Ahsome
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?
ganeshie8
  • ganeshie8
sure..
Ahsome
  • Ahsome
How many permutations of the letters in the word MATHS are therem wgere M, A and T appear together?
Ahsome
  • Ahsome
I did \(3!\times 3!\)
ganeshie8
  • ganeshie8
Correct. You tie MAT in a bag. Then you will see just 3 objects |dw:1440822752665:dw|
ganeshie8
  • ganeshie8
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 \(3!*3!\)
Ahsome
  • Ahsome
Thought so. Thank you! Although it doesn't seem to work when I do it on permutations with different N and R values...
ganeshie8
  • ganeshie8
do you have an example ?
Ahsome
  • Ahsome
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
ganeshie8
  • ganeshie8
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.
ganeshie8
  • ganeshie8
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 \(\dbinom{6}{2}*3!*2!\)
Ahsome
  • Ahsome
Ahh, gotcha. We didn't do the 3! for the second question we did, cause we didn't merge them right?
ganeshie8
  • ganeshie8
we want four rowers two of them are tied in a bag
Ahsome
  • Ahsome
Yup. That's why we considered the 2!, cause it could be in different positions, right?
ganeshie8
  • ganeshie8
Yes
Ahsome
  • Ahsome
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: \(4C4 \times 5! \times 2!\)
Ahsome
  • Ahsome
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
Ahsome
  • Ahsome
@ganeshie8?

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