Question

Help with two questions

Answer 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?

Answer 2

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 \(\dbinom{14}{10}*11!\)

Answer 3

Why is it combination, and not permutation @ganeshie8?

Answer 4

NVM, I got it @ganeshie8. Thanks :)

Answer 5

I hope you can do the second problem on your own

Answer 6

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

Answer 7

I mean 8, not 9 btw

Answer 8

\(\dbinom{8}{3}*5!\) looks good!


as you can see we're doing it in two steps :
1) choose the 5 players first
2) arrange them in order

Answer 9

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?

Answer 10

sure..

Answer 11

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

Answer 12

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

Answer 13

Correct.

You tie MAT in a bag. Then you will see just 3 objects

Answer 14

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 \(3!*3!\)

Answer 15

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

Answer 16

do you have an example ?

Answer 17

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

Answer 18

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.

Answer 19

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 \(\dbinom{6}{2}*3!*2!\)

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

Yes

Answer 24

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:
\(4C4 \times 5! \times 2!\)

Answer 25

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

Answer 26

@ganeshie8?