Help with two questions

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- Ahsome

Help with two questions

- schrodinger

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- 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

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

Why is it combination, and not permutation @ganeshie8?

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- Ahsome

NVM, I got it @ganeshie8. Thanks :)

- ganeshie8

I hope you can do the second problem on your own

- Ahsome

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

- Ahsome

I mean 8, not 9 btw

- 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

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

sure..

- Ahsome

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

- Ahsome

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

- ganeshie8

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

- 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

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

- ganeshie8

do you have an example ?

- 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

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

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

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

- ganeshie8

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

- Ahsome

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

- ganeshie8

Yes

- 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

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

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