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Can't figure this out

  • 11-11-2018 10:04am
    #1
    wilser
    Registered Users, Registered Users 2 Posts: 1,112 ✭✭✭


    Hi
    Hoping someone can help with this.
    An running a kids wheelchair basketball team and am trying to give each child a near enough equal time on court.

    3 wheelchair players on court at a time,
    7 wheelchair players to pick from,
    2 halves of play 7 minutes long.

    Wheelchair players are allowed to bring siblings or friends with them to play.

    2 able bodied players can play in a wheelchair at any time (5 players on court per team)
    5 able bodied players to pick from

    Was just going to split it between able bodied and wheelchair as I think it gets too complicated otherwise. Any help with this would be great as I keep getting lost along the way with the maths.
    Thanks


Comments

  • #2
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    Hi,
    If you have 7 kids to choose 3 at a time from, there are 7*6*5 / (3* 2 * 1) or 35 ways to pick them.
    Each kid will be in 15 of these above subsets. (Basically, if you put one kid aside, then you can pick 2 from 6 to be his team mates, or in this case 6*5 / (2*1)).

    So, given that there are 14 minutes of play, then each kid in a wheelchair can get 14 / 35 * 15 or 6 minutes of play.

    Here's a sample set of times that would work (assuming that changing a player over will require the game to be paused)
    Players 1-7 are down the table, game minutes are across. 
    	1	2	3	4	5	6	7	8	9
1	x	x			x	x			x
2		x	x			x	x		
3			x	x			x	x	
4	x	x			x	x			x
5			x				x		
6				x				x	x
7	x			x	x			x	

    	10	11	12	13	14
1	x				
2	x	x			
3		x	x		
4	x				
5		x	x	x	x
6			x	x	x
7				x	x

    Edit : I had to split the table into two blocks, otherwise the data wrapped around and made things confusing.

    For the 5 able-bodied kids, there are 5*4/(2*1) or 10 ways to pick them. Each kid will be in 4 of those subsets, so picking two at a time using the same logic leads to 14 / 10 * 4 or 5.6 minutes of game time.

    I'm a bit tight for time at the moment, but if I get a chance after lunch I'll try to do something similar. I'll proabably round the able-bodied kids' time to 5 minutes each to make it a bit easier.


  • #3
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    For the able-bodied kids (three of them get 6 mins and the other two get 5, it's still reasonably fair!).
    1	2	3	4	5	6	7	8	9
1	x	x				x	x		
2		x	x				x	x	
3			x	x				x	x
4				x	x				x
5	x				x	x			

    10	11	12	13	14
1		x	x		
2			x	x	
3				x	x
4	x				x
5	x	x


  • #4
    wilser
    Registered Users, Registered Users 2 Posts: 1,112 ✭✭✭wilser


    Well fair play, the way I was doing it was nothing like that.
    The kids travel a fair bit to get to the games, so I was just trying to give them all a fair amount of time each on court.
    Thanks for taking the time to put this together, greatly appreciated, even if I don't fully understand it 😀


  • #5
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    No bother, it's only a suggested list but it fits the criteria of giving each child an equal amount of time (while mixing up who they're playing with etc).

    The maths I used is called combinatorics.
    Basically, in this case, we need to choose 3 kids from a group of 7. For the first kid, we have 7 choices, the second kid we have 6 and for the 3rd kid we have 5 choices so we have 7*6*5 or 210 combinations of choosing 3 from 7. But we're not quite done yet - when we pick the 3 it doesn't matter if we pick Alfred, Brendan or Charlie first, or Charlie then Alfred then Brendan. There are 6 ways to order the three kids, so we divide the 210 by 6 to come up with 35 as the number of different ways we can pick 3 kids from 7.


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