Combinatorics Questions:

SDPY15

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Alrighty, so I'm working through some pre-test questions and I ran across two which have really stumped me.


  • If you have a team that enters a best out of 7 series, how many different ways can you win in exactly 6 games.
    • I originally though this question could be done by taking 7!/(6!)(1!) where 7=n, 6=wins, and 1=losses, to give an answer of 7, but I was wrong. Could I just use nPr to get 5040, or would that be wrong too?



  • If a baseball team is made up of 7 girls and 8 boys and there are 9 people on the field at one time (at least 4 of which must be girls) then how many different ways could you arrange a team?
    • In this question I used the nCr formula then went girlsCr times boysCr to get 1960, but this was wrong. I asked my teacher and he said that I missed out on the fact that there could be more then 4 girls on the field. This lead me to more confusion but I believe that by adding on a regressive scale (as done above) I could go (7C4 x 8C5) + (7C5 x 8C4) + (7C6 x 8C3) + (7C7 x 8C2)=3850 combinations. This seems to make sense because the more girls added to the field the fewer options for arrangement. Nevertheless I'm still unsure and would be grateful for any help.
Thanks Yall!
 
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Alrighty, so I'm working through some pre-test questions and I ran across two which have really stumped me.


  • If you have a team that enters a best out of 7 series, how many different ways can you win in exactly 6 games.
    • I originally though this question could be done by taking 7!/(6!)(1!) where 7=n, 6=wins, and 1=losses, to give an answer of 7, but I was wrong. Could I just use nPr to get 5040, or would that be wrong too?
    • "Best-of 7" series, takes 4 games to win. "Exactly 6 games" means the losers won 2 games, and also that the winners won game #6. So you have to consider ordering 5 games, where one team won 3 and the other team won 2.


  • If a baseball team is made up of 7 girls and 8 boys and there are 9 people on the field at one time (at least 4 of which must be girls) then how many different ways could you arrange a team?
    • In this question I used the nCr formula then went girlsCr times boysCr to get 1960, but this was wrong. I asked my teacher and he said that I missed out on the fact that there could be more then 4 girls on the field. This lead me to more confusion but I believe that by adding on a regressive scale (as done above) I could go (7C4 x 8C5) + (7C5 x 8C4) + (7C6 x 8C3) + (7C7 + 8C2)=3850 combinations. This seems to make sense because the more girls added to the field the fewer options for arrangement. Nevertheless I'm still unsure and would be grateful for any help.
    • 4 of the 9 are girls, and the other 5 players are drawn from 3 girls and 8 boys. Just consider the 5.
    • Once you have determined the number of girls, all the rest are boys - so you should NOT multiply combinations of girls times combinations of boys. We don't care which girl or which boy.
    • In the 5, what is probability of 0 girls? 1 girl, 2 ...
Thanks Yall!
.
 
If you have a team that enters a best out of 7 series, how many different ways can you win in exactly 6 games.
You lose two among the first five and win the sixth game.

If a baseball team is made up of 7 girls and 8 boys and there are 9 people on the field at one time (at least 4 of which must be girls) then how many different ways could you arrange a team?

You can have at least four girls on the team and at most seven.

\(\displaystyle \displaystyle\sum\limits_{k = 4}^7 {\dbinom{7}{k}\dbinom{8}{9-k}}\)
 
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Maybe it would be better if I worded the first question differently:

You have a team that enters a best out of 7 series. How many ways can you win if you play 6 games?
 
Maybe it would be better if I worded the first question differently:
You have a team that enters a best out of 7 series. How many ways can you win if you play 6 games?
That is not wording the question differently.
It is completely changing the meaning.


Which do you intend to ask?
 
Preferably the second. :) Sorry for the misunderstanding.

That means the you play six games regardless of when you win the four games.
You could win on the fourth game and play two more.
You could win on the fifth game and play one more.
Or you could win on the sixth game.

I don't think that is really what it means. Is it?
 
Ok, but for the first question I still don't understand how to set it up! Would I just answer it by using 5C3?
Yes.

You have to win exactly 3 of the first 5 games, and you must win game 6.
 
That means the you play six games regardless of when you win the four games.
You could win on the fourth game and play two more.
You could win on the fifth game and play one more.
Or you could win on the sixth game.

I don't think that is really what it means. Is it?

Hmm, I don't know. As far as I'm concerned we have a best out of 7 for 6 games and you need to get at least four scores to have 1 win. In other words, how many ways can you arrange those scores in six games?

Now that I ramble on about it I think i may just have to go 6C4 to get 15 but I don't know how accurate this is.
 
Hmm, I don't know. As far as I'm concerned we have a best out of 7 for 6 games and you need to get at least four scores to have 1 win. In other words, how many ways can you arrange those scores in six games?
Now that I ramble on about it I think i may just have to go 6C4 to get 15 but I don't know how accurate this is.


Look, I have written or edited hundreds of questions like this one.
Your OP states: If you have a team that enters a best out of 7 series, how many different ways can you win in exactly 6 games.

I think that it means exactly that.


You may find this page, Problem of points, useful.
Contrary to what it says, the history goes back to the 11th century CE.
Although, it is correct that a complete solution was not found until the 17th century CE.
 
Hmm, I don't know. As far as I'm concerned we have a best out of 7 for 6 games and you need to get at least four scores to have 1 win. In other words, how many ways can you arrange those scores in six games?

Now that I ramble on about it I think i may just have to go 6C4 to get 15 but I don't know how accurate this is.
Whenever someone wins a 4th game, the series is OVER. Thus the winner of the series has to win the last game played - no choice. Only the first 5 games are subject to re-ordering.
 
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