Maths Quiz

[I_hate_portal]

really Off Topic but I have a quiz question that was sent out in work.

"Imagine a football team with 17 players. 9 are English, 8 are Foreign.

How many different 11 man squads are possible if you implement the rule that you have to have 5 Englishmen and 6 Foreigners in the squad"

My collegeues and i can't even agree on how the problem should be solved never mind the actually answer. I have one in mind but would like to see if anyone else gets the same.

 

[zedzadic]

Hmm, now, I won't bother checking my answer out with real numbers, but I would assume you want the most simple, albiet not easy, or quick method. You're looking for something on the line of Combinatorics [http://en.wikipedia.org/wiki/Combinatorics#Analytic_combinatorics]. I would suggest using Newton's Binomial, where the main argument(?)[May be incorrect.I've never done math in english before.) is 17 and the integrer is 11 [I'm guessing the names, again, may be false]. Now, this won't sort out the combinations you need, seeing as you only get all possible variants, therefore, the most simple solution is to actually get a pencil and look for the ones you need. Alternatively, you could incorporate the mathematical formula in a computer algorith and set the two groups as being different types (char, numerical) then use it to sort out the solutions by putting a few conditions on the return function [note: I mean only in C++, don't know any other language well enough].

 

[Kailat777]

That's actually pretty simple, you just choose 5 of the 9 Englishmen and 6 of the 8 Foreigners and put them together. Thus, the solution is (9C5)*(8C6)=9!*8!/(5!*4!*6!*2!), or 3,528.

 

[zedzadic]

Lol, you're right, I never enven considered that!

 

[Datalord]

since you need to pick 5 out of 9 englishmen, and 6 out of 8 foreigners, its:
for the englishmen: for the foriegners
((9*8*7*6*5)/(5*4*3*2*1))*((8*7*6*5*4*3)/(6*5*4*3*2*1))
=
(9*8*8*7*7*6*6*5*5*4*3)/(6*5*5*4*4*3*3*2*2)=9*8*8*7*7*6/4*3*2*2
=3528 ways

 

[Datalord]

While this thread is still here, anyone else on the escapist qualify for the AIME this year, or an equivalent test outside of the USA

 

[I_hate_portal]

ha, cheers. my collugues had me convinced that it was a trick question. u worked it out a lot easier than i did.

 

[Kailat777]

AIME? I'm much too old for that, my friend. I'll be writing and publishing my thesis next year, if that counts as an equivalent test.

Also, no prob on the problem. The trick to these things usually isn't finding a way to solve, but finding an easy way to solve. It would get more difficult if you had to make groups of 11 with, say, at least 3 Englishmen and and least 3 Foreigners. Feel free to play with that if you want.

 

[KendayTheAbsolute]

6

 

[Nargleblarg]

Only 8 I think

 

[SenseOfTumour]

That's actually pretty simple, you just choose 5 of the 9 Englishmen and 6 of the 8 Foreigners and put them together. Thus, the solution is (9C5)*(8C6)=9!*8!/(5!*4!*6!*2!), or 3,528.

If that's pretty simple, dont bother calling me when you've got some complex stuff to work out.

I'm actually fairly confident in my mental math abilities, but I certainly never dealt with anything like that in school.

 

[Gooble]

I'd have been able to work that out 2 years ago, but all my A-level maths knowledge has drained away from my brain :'(

 

[Datalord]

I am taking most of these answers to mean a no to my question, cept for Kailat777

 

[I_hate_portal]

I'd have been able to work that out 2 years ago, but all my A-level maths knowledge has drained away from my brain :'(

Your not kidding, I was (or always thought i was) really good at problem solving, but it took me ages to work this out, pretty much because I hadn't used formula like Kailat and datalord.

6

My mate said 6 and I understand how you get it but you have to take into account all the variations that are possible.

 

[Datalord]

6

Hth did you get 6?

Only 8 I think

Hth did you get 8?